Re: Variable conversion problem

On Fri, 21 Sep 2001 11:15, Neil Silvester wrote: > Another PHP problem has kept me up all night, scouring through my > database books and trying everything I could think of. I am still new > to the MySQL and PHP field, but never the lass I will not give up. > $cat is a variable that is passed to the query from the previous page. > The SELECT statement works faultlessly, except for the point at which > no choice is made from the drop down on the page prior to this one. I > have set a value > > of "x" to the drop down for a non selection, but am having problems > converting that "x" value to a "%" value (the MySql wildcard symbol) > withing my SELECT statement. > I am assuming that I will need an IF THEN before the SELECT, but I have > tried several variation to no avail. > > $result = mysql_query("SELECT CompetitorName > FROM competitor, competitorproducts, productcategory > WHERE competitorproducts.CompID=competitor.ID > AND productcategory.ID=\"$cat\" > AND competitorproducts.CatID=productcategory.ID" > ); > > Any help would be mostly appreciated. > Thanks. > > Neil Silvester > Webmaster / Systems Administrator > Heat and Control Inc. There are several possible solutions here, starting with seting a default selection in your dropdown so you avoid the problem. Otherwise you could test whether the dropdown value is empty, and if so, not add the 'AND productcategory.ID' clause to your select. -- David Robley Techno-JoaT, Web Maintainer, Mail List Admin, etc CENTRE FOR INJURY STUDIES Flinders University, SOUTH AUSTRALIA "I've lived through a lot of windstorms," Tom regaled.