Neil,
Please indicate what you mean by converting the string "x" to a percent
value???
-Jason Garber
www.deltacron.com
At 08:51 AM 9/21/2001 +1000, 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.
>$cat is a variable that is passed to the query from the previous page. The
>SELECT statement works faultlessly, except for the point at which no
>choice is made from the drop down on the page prior to this one. I have
>set a value of "x" to the drop down for a non selection, but am having
>problems converting that "x" value to a "%" value withing my SELECT statement.
>I am assuming that I will need an IF THEN before the SELECT, but I have
>tried several variation to no avail.
>
>$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.
>
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>
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