On Wed, 3 Mar 2010 08:21:06 -0600, halip...@gmail.com (haliphax) wrote:
>> On Fri, Feb 26, 2010 at 11:01 PM, wrote:
>> > while ($i < $j) { $b[$i] = $a[$i++]; } B.
>> >
>> > You get $b[0] = $a[1], and so on (as you would expect).
>>
>
>Wouldn't that be $b[0] = $a[0], with the value of $i
> On Fri, Feb 26, 2010 at 11:01 PM, wrote:
> > while ($i < $j) { $b[$i] = $a[$i++]; } B.
> >
> > You get $b[0] = $a[1], and so on (as you would expect).
>
Wouldn't that be $b[0] = $a[0], with the value of $i being 1 *after* the
statement was finished executing? You used a post-decrement
Thanks for taking time to provide the examples, Clancy, I'll know what
potential pitfalls to wary of now :)
On Fri, Feb 26, 2010 at 11:01 PM, wrote:
> A week ago Dasn asked a question about converting arrays, and I quoted one
> possible way of
> achieving his task, using the operation:
>
> $i =
Mess
Dne 27.2.2010 5:01, clanc...@cybec.com.au napsal(a):
A week ago Dasn asked a question about converting arrays, and I quoted one
possible way of
achieving his task, using the operation:
$i = 0; while ($i< $k) { $b[$a[$i++]] = $a[$i++]; }
I added the comment that "I have always been war
A week ago Dasn asked a question about converting arrays, and I quoted one
possible way of
achieving his task, using the operation:
$i = 0; while ($i < $k) { $b[$a[$i++]] = $a[$i++]; }
I added the comment that "I have always been wary of using statements like this
because I
was unsure when the
> >> $Count = $Count + 1; is *exactly(?)* same as $Count++; Â or ++$Count
> >> But not exactly same. Â PostFix notation adds the value after assigning
> .
> >> PreFix notation adds the value right away.
> >> But optimized programming argues about how machine is coded nowadays.
>
Thanks Mike for cl
Ford, Mike wrote:
> On 27 April 2009 14:21, PJ advised:
>
>
>> Ford, Mike wrote:
>>
>>> On 26 April 2009 22:59, PJ advised:
>>>
>>>
>>>
kranthi wrote:
> if $Count1 is never referenced after this, then certainly this
> assignment operation is redundent.
Richard Quadling wrote:
> 2009/4/27 9el :
>
>>> Thanks for the clarification, Mike. In my ignorance, I was under the
>>> impression that the right side of the equation was only for the use of
>>> the left part. How stupid of me. So what I should have been doing was
>>> $Count1 = $Count + 1; righ
On 27 April 2009 14:21, PJ advised:
> Ford, Mike wrote:
>> On 26 April 2009 22:59, PJ advised:
>>
>>
>>> kranthi wrote:
>>>
if $Count1 is never referenced after this, then certainly this
assignment operation is redundent. but assignment is not the ONLY
operation of this statement
Thats why I used a (?) after exactly. PJ didn't have a need for the value.
;)
2009/4/27 9el :
>>
>> Thanks for the clarification, Mike. In my ignorance, I was under the
>> impression that the right side of the equation was only for the use of
>> the left part. How stupid of me. So what I should have been doing was
>> $Count1 = $Count + 1; right?
>>
$Count1 = $Count++;
is n
>
> Thanks for the clarification, Mike. In my ignorance, I was under the
> impression that the right side of the equation was only for the use of
> the left part. How stupid of me. So what I should have been doing was
> $Count1 = $Count + 1; right?
>
$Count = $Count + 1; is exactly(?) same as $Coun
Ford, Mike wrote:
> On 26 April 2009 22:59, PJ advised:
>
>
>> kranthi wrote:
>>
>>> if $Count1 is never referenced after this, then certainly this
>>> assignment operation is redundent. but assignment is not the ONLY
>>> operation of this statement. if u hav not noticed a post increment
>>
If you want to increment $Count by one, you can use any one of the 3
following lines to do this:
$Count = $Count + 1;
$Count += 1;
$Count++;
The 3 lines are equivalent, but have varying degrees of elegance (suit
yourself). However, make sure that your mysql function returned a
number (and not an er
On 26 April 2009 22:59, PJ advised:
> kranthi wrote:
>> if $Count1 is never referenced after this, then certainly this
>> assignment operation is redundent. but assignment is not the ONLY
>> operation of this statement. if u hav not noticed a post increment
>> operator has been used which will aff
Phpster wrote:
> What parameters are you pasing in the link? That will be the telling
> point of what you are doing wrong. You could pass the search params (
> though these are best kept in a session or cookie ) and the offset
> counter to get the next block of results.
Actually, I am trying to use
> Look for "Pagination with PHP + MySQL" and watchit
>
>
> http://www.google.com/url?sa=t&source=web&ct=res&cd=10&url=http%3A%2F%2Fwww.bestechvideos.com%2F2008%2F07%2F02%2Fsampsonvideos-php-pagination-part-2&ei=ohn0SdnlL8KLkAWQrNTbCg&usg=AFQjCNEGKOIG3791BpgeVqCiiq5-cikbRA
>
> Dont miss this Sampso
9el wrote:
> I have pagination set up and the number for pages "next" has a link but
> the "next" does not. I have experimented with all sorts of
> configurations of the code but the only thing that works (and this is
> totally "off the wall") is to do this
> *$Count* = *mysql_num_rows($results);*
kranthi wrote:
> if $Count1 is never referenced after this, then certainly this
> assignment operation is redundent. but assignment is not the ONLY
> operation of this statement. if u hav not noticed a post increment
> operator has been used which will affect the value of $Count as well,
> and this
What parameters are you pasing in the link? That will be the telling
point of what you are doing wrong. You could pass the search params
( though these are best kept in a session or cookie ) and the offset
counter to get the next block of results.
Sorry for top posting.
Bastien
Sent from
I have pagination set up and the number for pages "next" has a link but
the "next" does not. I have experimented with all sorts of
configurations of the code but the only thing that works (and this is
totally "off the wall") is to do this
*$Count* = *mysql_num_rows($results);*
$Count1=*$Count++;* /
if $Count1 is never referenced after this, then certainly this
assignment operation is redundent. but assignment is not the ONLY
operation of this statement. if u hav not noticed a post increment
operator has been used which will affect the value of $Count as well,
and this operation is required fo
otron.ca]
> Enviada em: sexta-feira, 24 de abril de 2009 21:14
> Para: php-general@lists.php.net
> Assunto: [PHP] inexplicable behaviour
>
> Frankly, I don't know what to look for or why something so weird would
> happen:
> I have pagination set up and the number for page
Is the $Count++..
-Mensagem original-
De: PJ [mailto:af.gour...@videotron.ca]
Enviada em: sexta-feira, 24 de abril de 2009 21:14
Para: php-general@lists.php.net
Assunto: [PHP] inexplicable behaviour
Frankly, I don't know what to look for or why something so weird would
happen:
I
Frankly, I don't know what to look for or why something so weird would
happen:
I have pagination set up and the number for pages "next" has a link but
the "next" does not. I have experimented with all sorts of
configurations of the code but the only thing that works (and this is
totally "off the wa
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