What you said wasn't wrong - you'd just produce a couple of warnings.
By taking away the assignment of the $link variable you're unsetting
it, meaning it will have a value equal to NULL. All of the following
should work.
[old code]
if (!($link = mysql_connect($a, $b, $c))) return;
if (!mysql_select_db($dbname)) return;
$res = mysql_query("SELECT * FROM manual;", $link);
[/old code]
[new code]
if (!mysql_connect($a, $b, $c)) return;
if (!mysql_select_db($dbname)) return;
$res = mysql_query("SELECT * FROM manual;", $link);
[/new code]
OR, optionally, to surpress the warnings:
[new code]
if (!mysql_connect($a, $b, $c)) return;
$link = null;
if (!mysql_select_db($dbname)) return;
$res = mysql_query("SELECT * FROM manual;", $link);
[/new code]
Evert
On Thu, Jun 19, 2008 at 9:12 AM, Isaak Malik <isooik@xxxxxxxxx> wrote:
> My mistake then, it's been a while that I used it that way so some things do
> fade away from my mind.
>
> On Thu, Jun 19, 2008 at 1:57 AM, Chris <dmagick@xxxxxxxxx> wrote:
>
>> Isaak Malik wrote:
>> > Because then the connection resource isn't stored in the $link variable
>> > and you will be able to use the mysql functions without passing that
>> > variable to each function.
>>
>> RTM again.
>>
>> The link parameter is completely optional.
>>
>> If you don't specify it, it uses the last connection created.
>>
>> I can do this and it's perfectly valid:
>>
>> $link = mysql_connect($server, $user, $pass);
>> if (!$link) {
>> die ("Unable to connect to the database server");
>> }
>>
>> $db_selected = mysql_select_db($databasename);
>> if (!$db_selected) {
>> die("Unable to connect to the database $databasename");
>> }
>>
>> $result = mysql_query("select 1");
>>
>> --
>> Postgresql & php tutorials
>> http://www.designmagick.com/
>>
>
>
>
> --
> Isaak Malik
> Web Developer
>
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