using variables in the query string

[hellonoko]

I have a feeling im formatting this query string wrong and thats why it is not working.

 

Any ideas?

 

Thanks

 

if ( isset($current_category) )
{
	$query = "SELECT count(*) FROM gallery WHERE category == ".$current_category."";
}
else
{
	$query = "SELECT count(*) FROM gallery";
}

$result = mysql_query($query);
$query_data = mysql_fetch_row($result);
$numrows = $query_data[0];

echo $numrows;
exit();

[GingerRobot]

Indeed. Try:

 

$query = "SELECT count(*) FROM gallery WHERE category = '$current_category'";

 

Unlike php, the comparison operator in mysql is a single = sign. Also, whenever you have a problem with a query, try adding an or die statement:

$result = mysql_query($query) or die(mysql_error().'<br />Query was: '.$query);

[localhost1]

u can also write the query this way

 

"SELECT count(*) FROM gallery WHERE category == "$current_category";

[localhost1]

this is also the way of writing the query

 

"SELECT count(*) FROM gallery WHERE category == $current_category";

[d.shankar]

First try to filter the variable $current_category before directly passing to the SQL Query.

[redarrow]

where $_POST['what_ever']

[hemlata]

Hello,

 

You have used '==' in you query for comparison. This might be the cause of your issue. Modify the query

 

$query = "SELECT count(*) FROM gallery WHERE category == ".$current_category."";

 

With the following

 

$query = "SELECT count(*) FROM `gallery` WHERE `category` = '" . $current_category . "'";

 

Hope this will solve your issue.

 

Regards

[localhost1]

hello

this code is correct n work. no need to filter the php variable.

"SELECT count(*) FROM gallery WHERE category == $current_category";

]just try this one

[hemlata]

Hello,

 

hello

this code is correct n work. no need to filter the php variable.

Code:

 

"SELECT count(*) FROM gallery WHERE category == $current_category";

 

]just try this one

 

I just wanted to clarify that my last post was just not to include proper query syntax, but i also pointed out '==' sign in query which is incorrect and will return query error.

I hope i made myself clear.

 

Regards,