RE: [PHP] PHP3: eval() problem AGAIN

> > I'm creating generic single record editing forms from within a > database and > > want to pass the where clause to MySQL, so... > > > > I want to put something like > > > > ID='$SessionParms[\"User_ID\"]' > > > > into the Mysql database and then build my string > > > > $WhereClause = "ID='$SessionParms[\"User_ID\"]'"; // Value > > Taken From > > Database > > > > eval("\$Query =\" SELECT * FROM Table WHERE \$WhereCluse\";"); > Well, I may have spotted the obvious, or it could just be the email > message... > There is no a in the word WhereCluse in the line above. > > The line should be something like > eval("\$Query =\" SELECT * FROM Table WHERE \$WhereClause\";"); > > HINT: echo the staement and then cut and paste to mysql terminal > client.. It saves me a lot of time with queries goining astray... > > Hope that was all the problem was... Unfortunately not. I spotted the error myself after I pushed the send button. I still have the issue. I've been playing eval(), variable variables and an assortment of ereg_replace and preg_replace options. The variable variables looks like my best bet so far although I haven't managed to get them to work with arrays. To restate the problem slightly differently: Put "ID='\$User_ID'" into a field in the database. I want to retrieve it from the database into a variable called WhereClause so that it actually says "ID='12345'" rahter than "ID='$User_ID'". This is driving me nuts as it is crucial to my whole web structure for online data entry. Regards Grant Walters Walters & Associates, P O Box 13-043 Johnsonville, Wellington, NEW ZEALAND Telephone: +64 4 4765175, CellPhone 025488265