[SOLVED] get next id in mysql...

[pfkdesign]

hi everybody,

 

i made get_menu_id function :

<?php
function get_menu_id($menu) {
    $result = mysql_query("SHOW TABLE STATUS LIKE 'menu'");
    $rows = mysql_fetch_assoc($result);
    return $rows['auto_increment'];
}

echo get_menu_id()
?>

 

this is return the next auto_increment id in the mysql table (5.0.5a),

PROBLEM !

when i check it on local server(localhost) i dont have any problem and i 'm getting the next id  , but as soon as i upload in my server i will get the warning.

here are the warning:

 

Warning: Missing argument 1 for get_menu_id(),

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

 

any idea?

 

tnx

[paul2463]

when you wrote the function you stipulated that you would supply it with $menu

 

function get_menu_id($menu) {

 

and when you call it you dont give it that information

 

echo get_menu_id()

 

it fails because it wants to know what $menu is, you dont use it in the function anyway so I am confused as to why you want it

[pfkdesign]

than you  paul2463  for you reply ,

 

the first warning is solved but still i have this warning :

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource...

 

in local server (localhost) is not show up , but in main server is shewing this warning!

 

any idea?

[JasonLewis]

Make this:

$result = mysql_query("SHOW TABLE STATUS LIKE 'menu'");

 

This:

$result = mysql_query("SHOW TABLE STATUS LIKE 'menu'") or die("Error: ".mysql_error());

 

Also, does the 'menu' in the query need to be the variable $menu?

[pfkdesign]

tnx it solved.