PHP mysql query not producing result

Question

I have a problem with the below query not producing a result. There is nothing wrong with the query. But it is being run after a lot of other time and memory consuming instructions within a PHP script. If I restrict the number of operations performed in the script before this query is executed, it works.

How do I find out what the problem is? Is there a way to be notified if the script exceeds memory or execution time limitations? Or could there be another reason?

$query="SELECT * FROM user" or die(mysql_error());
$result=mysql_query($query);

if($result)
{
    success();
}
else //error
{
    failure();
}

As you said the problem lies in the resource hungry codes preceding it. The issue can be accurately tackled only after seeing those sections of the code. — Shamim Hafiz - MSFT
– Shamim Hafiz - MSFT, Commented Jan 11, 2012 at 12:18

Rene Pot · Accepted Answer · 2012-01-11 12:16:42Z

2

or die(mysql_error()) should be added after the query is executed:

$query="SELECT * FROM user";
$result=mysql_query($query) or die(mysql_error());

Also, the else will never execute because you die() before that.

I suggest this:

$query="SELECT * FROM user";
$result=mysql_query($query);

if($result)
{
    success($result);
}
else //error
{
    failure(mysql_error());
}

answered Jan 11, 2012 at 12:16

Rene Pot

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Comments

Gaurav · Accepted Answer · 2012-01-11 12:17:07Z

0

You are using or with string.

$query="SELECT * FROM user";
$result=mysql_query($query);

if($result)
{
    success();
}
else //error
{
    failure();
}

answered Jan 11, 2012 at 12:17

Gaurav

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Comments

Orentet · Accepted Answer · 2012-01-11 12:18:56Z

0

First of all, you can always check if there is a mysql error by using mysql_error() which returns the mysql error string if it exists.

Second, to see if your query returned any rows, use mysql_num_rows($result) > 0.

answered Jan 11, 2012 at 12:18

Orentet

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