[PHP] important linuxtoday article: Will Open Source Lose the Battle for the web?
hmm just read the controversial linuxtoday article http://linuxtoday.com/news_story.php3?ltsn=2001-08-13-009-20-OP an excerpt: "but when will PHP grow to become something more than a web scripting language? Where is the PHP enterprise component architecture? What about clustering and failover? Where are the WSDL and UDDI implementations? Don't show me bits and pieces here and there. Show me a framework. Show me a reference implementation." Comments on the article? -- Vincent [EMAIL PROTECTED] Xaymaca Studios http://www.blendermania.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] Referring to Parent Folders
I don't know if this is a PHP config problem, or an Apache 1.3/PHP 4
problem, or what. When I include a script, say 2 folders up in the
structure, I'm having to move up to it instead of skipping to root and
moving down to it. Allow me to clarify, let's say my tree structure looks
like so [brackets indicate folder]:
[doc root]
-[php]
-- script.php
-- [some folder]
--- [another folder]
current.php
Let's then say that I want to include 'script.php' inside 'current.php'. As
it is now, I'm having to use include('../../script.php'). This is obviously
a nuisance, especially if I want to move things around. Now, I remember in
the past (different server config), using include('/php/script.php') would
work perfectly fine, but it does NOT work now, and I'd love for it to work.
Using /folder/file.ext works in regular HTML so I can only assume it is not
a problem with Apache. So my question is, is this simply a PHP config
problem?
PHP 4.3.4
Apache 1.3.something
Thanks!
Vinny
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[PHP] PHP querying mysql db for data limited to the last month
So I have this code I'm working with (pasted below) that queries a mysql db
table called timetracking. The goal of the page is to search the db for all
data based on a certain engineer, sorted by product and it takes pre-defined
values based on actions performed, sums them based on product and display's
the percentage of time an engineer has spent on each product. Everything
works great except I need to limit the results to the last months data only,
but everything I try seems to just break it. Can anyone push me in the right
direction a little? I have tried using BETWEEN in the SELECT statement, some
while statements and if statements, and all I do is keep breaking it. If
anyone has any ideas, it would be exceptionally helpful.
Thanks in advance,
Vinny
$monthago = date("Y-m-d h:i:s", mktime(date("h"), date("i"), date("s"),
date("m")-1, date("d"), date("Y")));
echo "Today = ", $today;
echo "One Month Ago = ", $monthago, "";
$query = "SELECT *, SUM(timespent) FROM timetracking WHERE engineer =
'$engineer' GROUP BY product";
$result = mysql_query($query) or die(mysql_error());
$result2 = mysql_query($query) or die(mysql_error());
echo "";
while($row = mysql_fetch_array($result)){
$total = $row['SUM(timespent)'] + $total;
}
while($row = mysql_fetch_array($result2)){
$perc = $row['SUM(timespent)'] * 100 / $total;
echo "[ ", $row[product]. " = ".
number_format($perc,2), "% ]";
}
?>
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Re: [PHP] PHP querying mysql db for data limited to the last month
Sorry, I took that stuff out because it was making the page not load and so
I figured it was wrong. I guess I should have posted the whole thing, but
since it didn't work I left it out.
Micah, thank you very much for the idea, it is working great!!!
"Micah Gersten" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
1. To get last months date, you can use strtotime("1 month ago")
instead of mktime.
2. I don't see anywhere in the code where you are limiting by date.
Try using > and <. Between is tricky on dates.
Thank you,
Micah Gersten
onShore Networks
Internal Developer
http://www.onshore.com
Vinny Gullotta wrote:
So I have this code I'm working with (pasted below) that queries a
mysql db table called timetracking. The goal of the page is to search
the db for all data based on a certain engineer, sorted by product and
it takes pre-defined values based on actions performed, sums them
based on product and display's the percentage of time an engineer has
spent on each product. Everything works great except I need to limit
the results to the last months data only, but everything I try seems
to just break it. Can anyone push me in the right direction a little?
I have tried using BETWEEN in the SELECT statement, some while
statements and if statements, and all I do is keep breaking it. If
anyone has any ideas, it would be exceptionally helpful.
Thanks in advance,
Vinny
One Month Ago = ", $monthago, "";
$query = "SELECT *, SUM(timespent) FROM timetracking WHERE engineer =
'$engineer' GROUP BY product";
$result = mysql_query($query) or die(mysql_error());
$result2 = mysql_query($query) or die(mysql_error());
echo "";
while($row = mysql_fetch_array($result)){
$total = $row['SUM(timespent)'] + $total;
}
while($row = mysql_fetch_array($result2)){
$perc = $row['SUM(timespent)'] * 100 / $total;
echo "[ ", $row[product]. " = ".
number_format($perc,2), "% ]";
}
?>
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[PHP] Incrementing variables based on database data
below is the output I'm seeing along with my code. There are 11425 items in
the database, many of each of these possible values for $i[4], however
you'll see from my Output that the $iiscount variable is the only one being
incremented and it is getting incremented for every row in the table. Can
anyone see anything I'm doing incorrectly? The data in this column in the
table is formatted as TEXT. Thanks in advance for any help you can provide.
=)
11425
0
0
0
0
0
0
0
$query = "SELECT * FROM table";
$result = mysql_query($query) or die(mysql_error());
$iiscount = 0;
$cfcount = 0;
$nlbcount = 0;
$srcount = 0;
$sacount = 0;
$rebootcount = 0;
$apccount = 0;
$othercount = 0;
while($i = mysql_fetch_row($result)) {
if ($i[4] = "IISRESET") {
$iiscount = $iiscount + 1;
} elseif ($i[4] = "CF Restart") {
$cfcount = $cfcount + 1;
} elseif ($i[4] = "NLB STOP IISRESET") {
$nlbcount = $nlbcount +1;
} elseif ($i[4] = "Service Restart") {
$srcount = $srcount + 1;
} elseif ($i[4] = "Restart System Attendant") {
$sacount = $sacount + 1;
} elseif ($i[4] = "Reboot") {
$rebootcount = $rebootcount + 1;
} elseif ($i[4] = "APC Reboot") {
$apccount = $apccount + 1;
} elseif ($i[4] = "Other") {
$othercount = $othercount + 1;
}
}
echo $iiscount, "";
echo $cfcount, "";
echo $nlbcount, "";
echo $srcount, "";
echo $sacount, "";
echo $rebootcount, "";
echo $apccount, "";
echo $othercount, "";
?>
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[PHP] Re: Incrementing variables based on database data
Nevermind, I figured it out. I needed to make the if statements use ==
instead of = like this:
if ($i[4] == "IISRESET") {
$iiscount = $iiscount + 1;
}
etc. =)
""Vinny Gullotta"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
below is the output I'm seeing along with my code. There are 11425 items
in the database, many of each of these possible values for $i[4], however
you'll see from my Output that the $iiscount variable is the only one
being incremented and it is getting incremented for every row in the
table. Can anyone see anything I'm doing incorrectly? The data in this
column in the table is formatted as TEXT. Thanks in advance for any help
you can provide. =)
11425
0
0
0
0
0
0
0
$query = "SELECT * FROM table";
$result = mysql_query($query) or die(mysql_error());
$iiscount = 0;
$cfcount = 0;
$nlbcount = 0;
$srcount = 0;
$sacount = 0;
$rebootcount = 0;
$apccount = 0;
$othercount = 0;
while($i = mysql_fetch_row($result)) {
if ($i[4] = "IISRESET") {
$iiscount = $iiscount + 1;
} elseif ($i[4] = "CF Restart") {
$cfcount = $cfcount + 1;
} elseif ($i[4] = "NLB STOP IISRESET") {
$nlbcount = $nlbcount +1;
} elseif ($i[4] = "Service Restart") {
$srcount = $srcount + 1;
} elseif ($i[4] = "Restart System Attendant") {
$sacount = $sacount + 1;
} elseif ($i[4] = "Reboot") {
$rebootcount = $rebootcount + 1;
} elseif ($i[4] = "APC Reboot") {
$apccount = $apccount + 1;
} elseif ($i[4] = "Other") {
$othercount = $othercount + 1;
}
}
echo $iiscount, "";
echo $cfcount, "";
echo $nlbcount, "";
echo $srcount, "";
echo $sacount, "";
echo $rebootcount, "";
echo $apccount, "";
echo $othercount, "";
?>
--
Vinny Gullotta
Intermedia System Administrator
E [EMAIL PROTECTED]
T (650) 641-4034
F (650) 424-9936
According to Einstein's Theory of Relativity, Chuck Norris can actually
roundhouse kick you in the face yesterday!
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Re: [PHP] Re: Incrementing variables based on database data
Well, what I need to be able to do is then take the numbers of each count
and figure out which is used the most. We use this database to log actions
taken on servers in our network. What my boss wants me to come up with is a
list of which commands are issued the most, which servers get the most
attention, and for each server, which command is issued the most, and so I
was going to kind of do it messy-like as I'm not really 100% sure of the
best way to do it, but my goal was to create the count variables and then
compare them using if statements to see which one is used the most. If you
have a way of doing this more efficiently, I'd love to hear it. I'm not the
best programmer in the world and I'm kind of just getting back into this
stuff, so I'm all ears for any suggestions you may have =)
"Micah Gersten" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
Why not let the DB do this for you? You can group by whatever column
that is and select count(*), column_your_looking_for.
Thank you,
Micah Gersten
onShore Networks
Internal Developer
http://www.onshore.com
Vinny Gullotta wrote:
Nevermind, I figured it out. I needed to make the if statements use ==
instead of = like this:
if ($i[4] == "IISRESET") {
$iiscount = $iiscount + 1;
}
etc. =)
""Vinny Gullotta"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
below is the output I'm seeing along with my code. There are 11425
items in the database, many of each of these possible values for
$i[4], however you'll see from my Output that the $iiscount variable
is the only one being incremented and it is getting incremented for
every row in the table. Can anyone see anything I'm doing
incorrectly? The data in this column in the table is formatted as
TEXT. Thanks in advance for any help you can provide. =)
11425
0
0
0
0
0
0
0
$query = "SELECT * FROM table";
$result = mysql_query($query) or die(mysql_error());
$iiscount = 0;
$cfcount = 0;
$nlbcount = 0;
$srcount = 0;
$sacount = 0;
$rebootcount = 0;
$apccount = 0;
$othercount = 0;
while($i = mysql_fetch_row($result)) {
if ($i[4] = "IISRESET") {
$iiscount = $iiscount + 1;
} elseif ($i[4] = "CF Restart") {
$cfcount = $cfcount + 1;
} elseif ($i[4] = "NLB STOP IISRESET") {
$nlbcount = $nlbcount +1;
} elseif ($i[4] = "Service Restart") {
$srcount = $srcount + 1;
} elseif ($i[4] = "Restart System Attendant") {
$sacount = $sacount + 1;
} elseif ($i[4] = "Reboot") {
$rebootcount = $rebootcount + 1;
} elseif ($i[4] = "APC Reboot") {
$apccount = $apccount + 1;
} elseif ($i[4] = "Other") {
$othercount = $othercount + 1;
}
}
echo $iiscount, "";
echo $cfcount, "";
echo $nlbcount, "";
echo $srcount, "";
echo $sacount, "";
echo $rebootcount, "";
echo $apccount, "";
echo $othercount, "";
?>
--
Vinny Gullotta
Intermedia System Administrator
E [EMAIL PROTECTED]
T (650) 641-4034
F (650) 424-9936
According to Einstein's Theory of Relativity, Chuck Norris can actually
roundhouse kick you in the face yesterday!
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[PHP] PHP and MySQL SELECT COUNT (*)
What I want to do is find the top 10 servers where the column steps =
iisreset. The following code works great except that the page is not
displaying the servername in the 'Server Name' column of my results (nothing
appears, the column is just blank).
servername and steps are the important columns in the database table.
$_POST[time1] and $_POST[time2] come from a form submitted.
When I copy and paste the entire select statement into the SQL tab in
phpmyadmin (and replace the time variables with actual times corresponding
to the timestamp column), it displays the correct results including
servername. Everything works in the php page's results except for the
servername. I feel like it's right in front of my face and that's why I
can't see it lol. Any help would be greatly appreciated. Thanks in advance
=)
My code...
$query = "SELECT servername, COUNT(steps) FROM monitoring WHERE steps LIKE
'iisreset' AND timestamp <= '$_POST[time2]' AND timestamp >= '$_POST[time1]'
GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10";
$result = mysql_query($query) or die(mysql_error());
# display column titles
echo "";
echo "Count";
echo "Server
Name";
echo "";
#display results
while($i = mysql_fetch_row($result))
{
echo "", $i[COUNT('steps')],
"";
echo "", $i[servername] ,"";
}
echo "";
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Re: [PHP] PHP and MySQL SELECT COUNT (*)
""Dan Joseph"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
On Wed, Sep 17, 2008 at 2:17 PM, Vinny Gullotta
<[EMAIL PROTECTED]>wrote:
What I want to do is find the top 10 servers where the column steps =
iisreset. The following code works great except that the page is not
displaying the servername in the 'Server Name' column of my results
(nothing
appears, the column is just blank).
servername and steps are the important columns in the database table.
$_POST[time1] and $_POST[time2] come from a form submitted.
When I copy and paste the entire select statement into the SQL tab in
phpmyadmin (and replace the time variables with actual times
corresponding
to the timestamp column), it displays the correct results including
servername. Everything works in the php page's results except for the
servername. I feel like it's right in front of my face and that's why I
can't see it lol. Any help would be greatly appreciated. Thanks in
advance
=)
My code...
$query = "SELECT servername, COUNT(steps) FROM monitoring WHERE steps
LIKE
'iisreset' AND timestamp <= '$_POST[time2]' AND timestamp >=
'$_POST[time1]'
GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10";
$result = mysql_query($query) or die(mysql_error());
# display column titles
echo "";
echo "Count";
echo "Server
Name";
echo "";
#display results
while($i = mysql_fetch_row($result))
{
echo "", $i[COUNT('steps')],
"";
echo "", $i[servername]
,"";
}
echo "";
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Change that "COUNT(steps)" to "COUNT(steps) AS CountSteps", that might be
the issue. Then you're using $i['CountSteps']. That seems a bit more
normal looking to me atleast.
Also, try echoing out your query on the screen to see that its formating
properly in the PHP code. You may have something wrong in there, although
I
don't see any off hand.
--
-Dan Joseph
www.canishosting.com - Plans start @ $1.99/month.
"Build a man a fire, and he will be warm for the rest of the day.
Light a man on fire, and will be warm for the rest of his life."
Adding as CountSteps $i['CountSteps'] still leaves the column blank.
echo $result; gives me an output of:
Resource id #3
and
echo $query;
just gives me an error.
One thing I don't understand is why echo $result; gives me Resource id #3 as
an output. What does that mean?
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Re: [PHP] PHP and MySQL SELECT COUNT (*)
""Dan Joseph"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
On Wed, Sep 17, 2008 at 2:30 PM, Vinny Gullotta
<[EMAIL PROTECTED]>wrote:
""Dan Joseph"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
On Wed, Sep 17, 2008 at 2:17 PM, Vinny Gullotta
<[EMAIL PROTECTED]
>wrote:
What I want to do is find the top 10 servers where the column steps =
iisreset. The following code works great except that the page is not
displaying the servername in the 'Server Name' column of my results
(nothing
appears, the column is just blank).
servername and steps are the important columns in the database table.
$_POST[time1] and $_POST[time2] come from a form submitted.
When I copy and paste the entire select statement into the SQL tab in
phpmyadmin (and replace the time variables with actual times
corresponding
to the timestamp column), it displays the correct results including
servername. Everything works in the php page's results except for the
servername. I feel like it's right in front of my face and that's why I
can't see it lol. Any help would be greatly appreciated. Thanks in
advance
=)
My code...
$query = "SELECT servername, COUNT(steps) FROM monitoring WHERE steps
LIKE
'iisreset' AND timestamp <= '$_POST[time2]' AND timestamp >=
'$_POST[time1]'
GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10";
$result = mysql_query($query) or die(mysql_error());
# display column titles
echo "";
echo "class='tableHeader'>Count";
echo "Server
Name";
echo "";
#display results
while($i = mysql_fetch_row($result))
{
echo "", $i[COUNT('steps')],
"";
echo "", $i[servername]
,"";
}
echo "";
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Change that "COUNT(steps)" to "COUNT(steps) AS CountSteps", that might
be
the issue. Then you're using $i['CountSteps']. That seems a bit more
normal looking to me atleast.
Also, try echoing out your query on the screen to see that its formating
properly in the PHP code. You may have something wrong in there,
although
I
don't see any off hand.
--
-Dan Joseph
www.canishosting.com - Plans start @ $1.99/month.
"Build a man a fire, and he will be warm for the rest of the day.
Light a man on fire, and will be warm for the rest of his life."
Adding as CountSteps $i['CountSteps'] still leaves the column blank.
echo $result; gives me an output of:
Resource id #3
and
echo $query;
just gives me an error.
One thing I don't understand is why echo $result; gives me Resource id #3
as an output. What does that mean?
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That's basically your result set ID number inside PHP.
as for $query, what error are you getting? Does this $query echo out:
$query = "SELECT servername, COUNT(steps) AS CountSteps FROM monitoring
WHERE steps LIKE 'iisreset' AND timestamp <= '" . $_POST['time2'] . "' AND
timestamp >= '" . $_POST['time1'] . "' GROUP BY servername ORDER BY
COUNT(*)
DESC LIMIT 10";
--
-Dan Joseph
www.canishosting.com - Plans start @ $1.99/month.
"Build a man a fire, and he will be warm for the rest of the day.
Light a man on fire, and will be warm for the rest of his life."
It's actually not an error, it's the select statement that is echo'd
echo $query;
yields
SELECT servername, COUNT(steps) as CountSteps FROM monitoring WHERE steps =
'IISRESET' AND timestamp <= '2008-09-17 11:40:34' AND timestamp >=
'2008-08-17' GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10
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Re: [PHP] PHP and MySQL SELECT COUNT (*)
Still no luck displaying the stupid servername. Any other things I can try? "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] You'll want to change your Order By statement to 'ORDER BY CountSteps DESC'. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: echo $query; yields SELECT servername, COUNT(steps) as CountSteps FROM monitoring WHERE steps = 'IISRESET' AND timestamp <= '2008-09-17 11:40:34' AND timestamp >= '2008-08-17' GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP and MySQL SELECT COUNT (*)
""Dan Joseph"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
On Wed, Sep 17, 2008 at 2:17 PM, Vinny Gullotta
<[EMAIL PROTECTED]>wrote:
What I want to do is find the top 10 servers where the column steps =
iisreset. The following code works great except that the page is not
displaying the servername in the 'Server Name' column of my results
(nothing
appears, the column is just blank).
servername and steps are the important columns in the database table.
$_POST[time1] and $_POST[time2] come from a form submitted.
When I copy and paste the entire select statement into the SQL tab in
phpmyadmin (and replace the time variables with actual times
corresponding
to the timestamp column), it displays the correct results including
servername. Everything works in the php page's results except for the
servername. I feel like it's right in front of my face and that's why I
can't see it lol. Any help would be greatly appreciated. Thanks in
advance
=)
My code...
$query = "SELECT servername, COUNT(steps) FROM monitoring WHERE steps
LIKE
'iisreset' AND timestamp <= '$_POST[time2]' AND timestamp >=
'$_POST[time1]'
GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10";
$result = mysql_query($query) or die(mysql_error());
# display column titles
echo "";
echo "Count";
echo "Server
Name";
echo "";
#display results
while($i = mysql_fetch_row($result))
{
echo "", $i[COUNT('steps')],
"";
echo "", $i[servername]
,"";
}
echo "";
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$i[servername]
Try: $i['servername']
notice the ' and ' around the name. I've heard you can do w/o those, but
I've had issues in the past where it didn't work. ITs also good practice
to
use 'em.
--
-Dan Joseph
www.canishosting.com - Plans start @ $1.99/month.
"Build a man a fire, and he will be warm for the rest of the day.
Light a man on fire, and will be warm for the rest of his life."
yeah, I've tried that combination before, but just for grins I tried it
again, and same result. It displays the counts but not the servernames.
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Re: [PHP] PHP and MySQL SELECT COUNT (*)
var_dump($i); looks messy, but I can see the server names in there and they are the correct names. "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] Do var_dump($i) in the loop to see if you're getting the data you want. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: Still no luck displaying the stupid servername. Any other things I can try? "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] You'll want to change your Order By statement to 'ORDER BY CountSteps DESC'. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: echo $query; yields SELECT servername, COUNT(steps) as CountSteps FROM monitoring WHERE steps = 'IISRESET' AND timestamp <= '2008-09-17 11:40:34' AND timestamp >= '2008-08-17' GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP and MySQL SELECT COUNT (*)
var_dump looks like this:
array(2) { [0]=> string(9) "wehost006" [1]=> string(2) "72" } array(2) {
[0]=> string(8) "H7848-49" [1]=> string(2) "71" } array(2) { [0]=> string(7)
"H7853-2" [1]=> string(2) "70" } array(2) { [0]=> string(7) "H7842-2" [1]=>
string(2) "64" } array(2) { [0]=> string(9) "WEHOST005" [1]=> string(2)
"64" } array(2) { [0]=> string(7) "h7835-2" [1]=> string(2) "57" } array(2)
{ [0]=> string(9) "wehost007" [1]=> string(2) "56" } array(2) { [0]=>
string(7) "H7814-1" [1]=> string(2) "55" } array(2) { [0]=> string(5)
"H0542" [1]=> string(2) "54" } array(2) { [0]=> string(8) "H7811-12" [1]=>
string(2) "54" }
""Dan Joseph"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
On Wed, Sep 17, 2008 at 4:21 PM, Vinny Gullotta
<[EMAIL PROTECTED]>wrote:
var_dump($i); looks messy, but I can see the server names in there and
they
are the correct names.
"Micah Gersten" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
Do var_dump($i) in the loop to see if you're getting the data you want.
Thank you,
Micah Gersten
onShore Networks
Internal Developer
http://www.onshore.com
Vinny Gullotta wrote:
Still no luck displaying the stupid servername. Any other things I can
try?
"Micah Gersten" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
You'll want to change your Order By statement to 'ORDER BY CountSteps
DESC'.
Thank you,
Micah Gersten
onShore Networks
Internal Developer
http://www.onshore.com
Vinny Gullotta wrote:
echo $query;
yields
SELECT servername, COUNT(steps) as CountSteps FROM monitoring WHERE
steps = 'IISRESET' AND timestamp <= '2008-09-17 11:40:34' AND
timestamp >= '2008-08-17' GROUP BY servername ORDER BY COUNT(*) DESC
LIMIT 10
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Can you show us one of the var_dumps?
--
-Dan Joseph
www.canishosting.com - Plans start @ $1.99/month.
"Build a man a fire, and he will be warm for the rest of the day.
Light a man on fire, and will be warm for the rest of his life."
--
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Re: [PHP] PHP and MySQL SELECT COUNT (*)
If by key you mean the column in the database, it's called: servername "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] What is the key for the server name? That's what you need when you output it. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: var_dump($i); looks messy, but I can see the server names in there and they are the correct names. "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] Do var_dump($i) in the loop to see if you're getting the data you want. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: Still no luck displaying the stupid servername. Any other things I can try? "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] You'll want to change your Order By statement to 'ORDER BY CountSteps DESC'. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: echo $query; yields SELECT servername, COUNT(steps) as CountSteps FROM monitoring WHERE steps = 'IISRESET' AND timestamp <= '2008-09-17 11:40:34' AND timestamp >= '2008-08-17' GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP and MySQL SELECT COUNT (*)
That was it!!! Thank you all so much for your help!!! =D ""Dan Joseph"" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] On Wed, Sep 17, 2008 at 4:30 PM, Vinny Gullotta <[EMAIL PROTECTED]>wrote: If by key you mean the column in the database, it's called: servername "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] What is the key for the server name? That's what you need when you output it. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: var_dump($i); looks messy, but I can see the server names in there and they are the correct names. "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] Do var_dump($i) in the loop to see if you're getting the data you want. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: Still no luck displaying the stupid servername. Any other things I can try? "Micah Gersten" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] You'll want to change your Order By statement to 'ORDER BY CountSteps DESC'. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Vinny Gullotta wrote: echo $query; yields SELECT servername, COUNT(steps) as CountSteps FROM monitoring WHERE steps = 'IISRESET' AND timestamp <= '2008-09-17 11:40:34' AND timestamp >= '2008-08-17' GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Thanks, I know your problem, sorry for not seeing this sooner. mysql_fetch_row returns the numerical array types. $array[0]... You want to use mysql_fetch_array(). Change to that, and you should be seeing the servername's. -- -Dan Joseph www.canishosting.com - Plans start @ $1.99/month. "Build a man a fire, and he will be warm for the rest of the day. Light a man on fire, and will be warm for the rest of his life." -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: PHP and MySQL SELECT COUNT (*)
Thanks all, I appreciate the follow ups and the help with the code. I'm still relatively new with this stuff, and never had any formal training, it's all just been learn as I go, and I have to learn fast as this project is relatively urgent to get completed. I plan on going through all of my code on all of these pages and cleaning it up at the end to make it more efficient, so I will use these tips to help do that. Thanks again to all who helped troubleshoot this. It is working great now and I think my bosses will be happy. =D "Nathan Rixham" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] learn something new every day! cheers Micah :) Micah Gersten wrote: While it's true that '.' concatenates and ',' is a list separator, The comma is actually more appropriate in this instance since you are just outputting each piece. It saves the overhead of concatenation before output. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Nathan Rixham wrote: 6: " vs ' when you use " php will parse the enclosed string for variables, when you use ' it won't; so ' leads for faster code, and also encourages you to code strongly by closing strings and concatenating variables. Further it allows you to use valid html " around attributes rather than the invalid ' 7: , vs . there is no vs :) to concatenate we use . (period) not , (comma) so for 6 & 7.. echo '' . $i['servername'] . ''; I'm going to stop there, hope it helps a little bit; and I won't go any further as half the fun is learning; so you finding out how to save time on queries and write your own db handlers etc is not my domain I reckons Regards nathan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
