Re: Problem with string comparison

At 22:31 13.03.2003, Charles Kline said: --------------------[snip]-------------------- >I am trying to get this sql query to work and am getting the error >"Insufficient data supplied". Can anyone help with this? > >$sql = "SELECT id, concat(fname, ' ', lname, ' ', degree1) name > FROM tbl_personnel "WHERE name LIKE \'%"; >$sql .= $attributes['nametofind']; >$sql .= "%\' ORDER BY lname ASC"; > >What I am trying to do is allow the user to find a personnel record >that contains the name to find. --------------------[snip]-------------------- At the first glance I see 3 items where the first one might be the culprit: - if your query is exactly as you sent it here you're missing a comma after the closing bracket of "concat()", just before "name" - before constructing the query make sure $attributes['nametofind'] is formatted correctly, esp. correctly escaping single quotes - when using double quotes to construct a string you don't need to escape single quotes. -- >O Ernest E. Vogelsinger (\) ICQ #13394035 ^ http://www.vogelsinger.at/