nutstretch

I have solved this. It did not like the name of one of my fields as it was a designated word 'Desc'

where will your database be sitting on the yahoo server or on your own Nuts

I have a table i am trying to insert some data into and I am getteing the following error' "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Desc, addedby ) VALUES( 'test', 'this test', 'angie' )' at line 1" my code is: $result = mysql_query("INSERT INTO scheme( name, Desc, addedby ) VALUES( '$scheme', '$desc', '$user' )", $linkID)or die(mysql_error()); the fields in my table are name, Desc, addedby Please can anyone see an obvious error as I can't Hoping one day to create something useful. Nuts

I use die(mysql_error()); to tell me what the prob is.

Yes I do and i have given $courseid a value earlier. What I need to do is carry that and the value from the drop down box across to my next page which will process both. In HTML I just add that value to a hidden field so that on my next page I can $courseid = $_POST['courseid']; it. But when i have put it into the php it is not bringing the value across. print "<input type='HIDDEN' NAME='bob' VALUE='$courseid'>";

sorry I am not sure what you mean. Usually if I am doing a html form extra values I need to carry over are just added as hidden values in the form. That is what i have done here but is not readying it on the next page. I am not sure why i have to use javascript? sorry if i am appearing thick

I use <a href='mailto:$email' >$email</a> Hope this helps Nuts

I am working on an application where the user can check to see if someone has registered and if so use thier details to make a booking. I want to use a dropdown list in a form which allows the user to check name and email address as these are the fields i am using to make sure it is a unique person. I am populating the form with the list of registrants and want to make a hidden field which holds the bookingid. The problem I am having is that it is bringing through the client id but not the booking id. I have no problems with this if it is in html but for some reason it is not working here. code from form - courseid defined above this print "<form name='Clients' method='post' action='justtesting.php'>"; print "<input type='HIDDEN' NAME='bob' VALUE='$courseid'>"; print "<p>Search by Client Name<p>"; print "<select name='client' size='1' style='WIDTH: 80%' onchange='this.form.submit()' >"; print "<option value='clientname' selected>Choose Client</option>"; while ($row1 = mysql_fetch_assoc($resultID)) { $item = $row1['name']; $value = $row1['clientID']; $desc = $row1['email']; print "<option value='$value'>$item : $desc</option>"; } print" </select>"; print "</form>"; code from process form <?php $courseid = $_POST['bob']; $clientid = $_POST['client']; print "$courseid , $clientid "; ?> Anyone who can shed alight on what i have done wrong would be revered higher than ice cream in my estimations and that is pretty high already Nuts