Hi!
Thank you.
But I cannot use that $link1 and/or $link2 to select two databases on
different servers...
If I use $link1 the first time to select the first database on the first
server I got that error. But without that $link1-variable AND only one
connection in the script it works.
And it makes no sense if I use it as shell-prog or with Apache-Builtin-PHP.
eg:
#!/usr/local/bin/php -q
<?php
$link1 = mysql_connect("localhost","a");
echo 'Link 1: '.$link1 .'<BR>';
if (mysql_select_db("test", $link1)) {
echo 'Successful select of test<BR>';
} else {
echo 'Failed to select test<BR>';
}
?>
gives this output:
Warning: Supplied argument is not a valid MySQL-Link resource in
/home/bt/bin/test.php on line 4
I don't know what to do...
David Robley wrote:
> In article <[EMAIL PROTECTED]>, [EMAIL PROTECTED] says...
>
>>Hi!
>>
>>mysql_error returns nothing...
>>
>>The error arrives while selecting the database via:
>>
>>$link = mysql_connect("host","user","passwd");
>>mysql_select_db("bla", $link);
>>
>>echo $link returns always '1'!
>>
>>
>>
>>
>>David Robley wrote:
>> > In article <[EMAIL PROTECTED]>, [EMAIL PROTECTED] says...
>> >
>> >>So,
>> >>
>> >>I've testet all possible I found in my brain...
>> >>CGI or Apache module: same error
>> >>mysql_connect or mysql_pconnect: same error
>> >>Variing settings in php.ini: same error
>> >>
>> >>How can I solve that problem?
>> >>BTW: Here I describe it again:
>> >>Connecting to a database works well: ok
>> >>Working with that database: ok
>> >>Using the additional $link-Parameter: false
>> >>Not a valid resource id ...
>> >>
>> >>And it does not matter how many different links with different servers I
>> >>use! The resource-id is always '1'.
>> >>
>> >>Strange...
>> >>
>> >
>> > Not a valid resource ID usually means a problem with the query - in fact
>> > this might be in the FAQ. Use mysql_error() after the database query to
>> > see if mysql returns a useful error.
>>
>
> Guess I misread your question. However, a litle playing around with this
> script
>
> <?php
> $link1 = mysql_connect("localhost","a");
> $link2 = mysql_connect("localhost");
> echo 'Link 1: '.$link1 .'<BR>';
> echo 'Link 2: '.$link2 .'<BR>';
> if(mysql_select_db("test", $link1)) {
> echo 'Successful select of test<BR>';
> }else{
> echo 'Failed to select test<BR>';
> }
> if(mysql_select_db("test2", $link2)) {
> echo 'Successful select of test2<BR>';
> }else{
> echo 'Failed to select test2<BR>';
> }
>
> ?>
> which gives this output
> Link 1: Resource id #1
> Link 2: Resource id #2
> Successful select of test
> Successful select of test2
>
> seems to indicate that php is smart enough not to open a new link even if
> requested if there is already a link for the user! Removing the "a" in
> the first mysql_connect gives me two links with a resource ID of 1.
>
> I can force errors of course by feeding a wrong username/password which
> eventually leads to a 'Not a valid resource id' error but has other error
> messages as well. You aren't by chance hiding other error responses by
> using @ or 'or die'?
>
>
--
Berthold
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