[SOLVED] Not getting proper error message

[psychohagis]

I have the following query/code thing.

 

$check = @mysql_query("SELECT id, username FROM users WHERE rank='$rank'");
if (!$result)  {
$mysql_error = mysql_error();
   exit('<p>Error performing query: ' . $mysql_error . '</p>');

 

When i run it it tells me theres an "error performing query:" but then does print out the mysql_error

 

I dont understans this cos I use this sam bit of code on other pages (except with different querys)  and if theres a problem with me query it tells me what it is.

[papaface]

if (!$result) 

Dont you mean:

if (!$check) 

?

[psychohagis]

Ok Ive changed that but now when I run the following:

 

$check = @mysql_query("SELECT id, username FROM users WHERE rank=$rank");
if (!$check)  {
$mysql_error = mysql_error();
   exit('<p>~ Error performing query: ' . $mysql_error . '</p>');
}

 

it says "Error performing query: Unknown column 'chairman' in 'where clause'"

 

which makes no sense cos i dont ask for column 'chairman'

 

This query is asking it to find $rank (which happens to be 'chairman' here) in column rank

[papaface]

Well for a start you should never use the suppression sign @ - its bad coding.

i dont understand why you are making this so long winded.

Try this though:

$check = mysql_query("SELECT id, username FROM users WHERE rank='{$rank}'") or die("<p>Error performing query: " . mysql_error() . "</p>");

^^ does the same as

$check = @mysql_query("SELECT id, username FROM users WHERE rank='$rank'");
if (!$result)  {
$mysql_error = mysql_error();
   exit('<p>Error performing query: ' . $mysql_error . '</p>');

[psychohagis]

oh ok ill try that

 

im only doing it like that because thats the way i learnt to do it