Re: newbie: Supplied argument is not a valid MySQL result..?

On Tue, 4 Dec 2001 08:29, Lee Philip Reilly wrote: > Hi, > > I wonder if someone could point out what is wrong with the following > piece of code (taken directly from the 'PHP - fast & easy web > development' book) , which gives the following warning : > > Warning: Supplied argument is not a valid MySQL result resource in > c:\program files\apache group\apache\htdocs\cms\pick_modcontact.php on > line 23 > > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= > $result = mysql_query($sql,$connection); <--line 23 Use mysql_error() here to return the error string from mysql, which will tell you more about your problem; it will be in the code before line 23 > > while ($row = mysql_fetch_array($result)) { Using extract($row) here will save you typing :-) > $id = $row['id']; > $fname = $row['fname']; > $lname = $row['lname']; > > $option_block .= "<option value=\"$id\">$lname, > $fname</option>"; > } > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= > > The mySQL query that is executed returns a result set when I use the > mySQL shell, and the connection string is valid. If someone could > advise me as to the error, I would greatly appreciate it. > > - Best regards, > > Lee -- David Robley Techno-JoaT, Web Maintainer, Mail List Admin, etc CENTRE FOR INJURY STUDIES Flinders University, SOUTH AUSTRALIA He who always plows a straight furrow is in a rut.