[Resolved] DB listing under MysqL 4.1.6 and PHP 5.0.2

bsachnoff

Since using PHP 5.0.2 and MySQL 4.1.6 I'm having basic issues.

$db_link=@mysqli_connect($host,$username,$password);
$db_list=mysql_list_dbs($db_link);
die;

Error: PHP Warning: mysql_list_dbs(): supplied argument is not a valid MySQL-Link resource in c:\Inetpub\wwwroot\view_sql_databases.php on line 13

All I would like to do is list the databases that are currently there. I even tried using a standard mysqli_query to the $db_link. Not sure if this function is just plain missing or if they have changed the name, etc...

Any suggestions would be great!

Thanks,

Brent

Kudose

<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
$db_list = mysql_list_dbs($link);

while ($row = mysql_fetch_object($db_list)) {
   echo $row->Database . "\n";
}
?>

If that doesnt work, than I would assume that you do not have the function. Althought the function has been valid since PHP3 and is still active in PHP5.

bsachnoff

If I call $link with mysql then I get a server error. If I change the call to mysqli then it gets past the $link and then I get on the DB listing call:

PHP Warning: mysql_list_dbs(): supplied argument is not a valid MySQL-Link resource in c:\Inetpub\wwwroot\view_sql_databases.php on line 3 PHP Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in c:\Inetpub\wwwroot\view_sql_databases.php on line 5

Apparently some important features are missing from the mysqli.dll file in php5.

I do have both mysqli and mysql active in the php.ini.

Brent

Weedpacket

You could try writing the query yourself and seeing what you get back from that

http://dev.mysql.com/doc/mysql/en/SHOW_DATABASES.html

bsachnoff

Weedpacket, that is what I original did but must have had a typo there since it did not work. I redid it this AM and now its fine:

<?
$dbconnect=@mysqli_connect($host,$username,$password);
$db_query="show databases";
$getdbs=mysqli_query($dbconnect,$db_query);
while ($getrow = mysqli_fetch_array($getdbs)) {
echo "$getrow[Database]<br>\n";
}
?>

Thanks for the input all!

Brent