Problem using PHP var in SQL code

[djwinder]

Hi everyone

I am having problems with the below code. $result contains no value at the end of the query. I have tried to see why this would be and echo some values at the end of the query. $annoid comes back containing the correct number, the referances to tables are correct soIi think the error is in how I am referancing the variable $annoid in the code. Can anyone help me out with this?

 $query = "SELECT 'oldannoid' FROM 'announceedit' where 'annoid' = '" . $annoid . "' ";  

       $result = mysql_query($query) ;  //or die ('could not select the old announcement ID'); 
     echo "annoid = $annoid...   -  result = $result ..." ;

Thanks in advance

[laserlight]

You should have used the or die to debug. The problem is that you are using single quotes to quote identifier names in the SQL statement. With MySQL, you should use backticks to quote identifier names. You should also cast $annoid to int, or as in my example, use the appropriate format specifier with sprintf:

$query = sprintf("SELECT `oldannoid` FROM `announceedit` where `annoid` = '%d'", $annoid);
$result = mysql_query($query);