ace(array("Blueberry", "Blue"), array("Strawberry", "Red"), $paragraph);
_
Marcus Rasmussen
[EMAIL PROTECTED]
www.marcusr.dk
-
On 31-03-2003 at 13:52 René Fournier wrote:
--
Putting an & sign in front of the $id in the first line should do the trick:
$variable = &$id;
A short example:
$bar = 0;
$foo = &$bar;
$bar = 2;
print $foo; //prints 2
______
Marcus Rasmussen
[EMAIL PROTECTED]
ww
This should work:
parse_str($example_string, $_GET);
___
Marcus Rasmussen
[EMAIL PROTECTED]
www.marcusr.dk
-
On 01-04-2003 at 03:20 Jose wrote:
-
>
Hello.
My question is:
How do I detect if mysql_connect() failed when I'm surpressing the error message with
@ like
$linkid = @mysql_connect("host","user","pass");
And how do I, if it failes, get the error message?
The manuel says that it:
[quote]Returns a positive MySQL link identifier on succ
d surpress and write the error message myself, but I just want the
original one, wich I should be able to get with mysql_error()
PS. It is only mysql_connect that do it.
Marcus R.
*** REPLY SEPARATOR ***
On 20-04-01 at 12:35 David Robley wrote:
>On Fri, 20 Apr 2001
p://www.php.net/manual/en/language.variables.external.php
_____
Marcus Rasmussen
-
On 24-04-02 at 15:19 Liam MacKenzie wrote:
-
>Hi all,
You could do it this way
WHERE column LIKE 's%' ORDER BY column
Now you'll get all rows sorted and where column is staring with an 's' or 'S'
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On 28-03-02 at 17:30 Sebastian A. wrote:
--
You can allso do like this:
$i = 1;
${"test". $i} = "123";
echo $test1;
_______
Marcus Rasmussen
[EMAIL PROTECTED]
On 21-09-01 at 11:21 _lallous wrote:
>you can always use eval to create a variable too!
>
>$i = 1;
>eva
You can allso do this:
//foo.class.php
class Foo{
var $bar;
function Foo($bar){
$this->bar = $bar;
}
}
//foo.php
include("foo.class.php");
$bar = "1234";
$foo = new Foo($bar);
//Now: $foo->bar = "1234"
Regards:
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