I'm trying to get a value for a single cell out of a table and into a variable. I 
receive the following error message: Warning: Supplied argument is not a valid MySQL 
result resource in /home/etc....

My code (abbreviated) is as follows:

$sql2 = "SELECT ServiceDes FROM ServiceTypes WHERE ServiceID = $serviceid";

$service_type = mysql_result($sql2, 1);

The value in $serviceid is valid and displays fine. I'm simply trying to retrieve the 
value of ServiceDes from a different table which also has a column named serviceid, 
which is unique to the table.  

I'm very new to PHP, but I do have lots of SQL experience. I'm used to defining 
variables using indicators (ie. SELECT column INTO varname FROM tablename WHERE....), 
but this didn't seem to work using PHP/MYSQL.



Reply via email to