Re: New Problem

Hi Well... If you look at your script, you'll see that you are trying to test on $submit, the 'problem' is that you have not defined $submit, not as a local var and not as a var from eg a URL, in other words: You are trying to test on something that you have not yet defined - IMHO 'sloppy' code. You can however remove the error message by using the error_reporting() or change error_reporting in the php.ini, but the best solution is to either test if the var isset or define and assign it before using it ! eg: if(isset($submit) && $submit === true) { .... } Best Steen "Eugene Wolff" <eugenew@htm.co.za> wrote: > Hi, > > I have the following script. When I open it says: > > Notice: Undefined variable: submit in > c:\inetpub\wwwroot\phpTest9.php on > line 4 > Although the input boxes all show fine. Obviously it does > not do the > first part of the code because ($submit) is not equal to > true. > > <html> > <body> > <?php > if ($submit){ > //Process form > $db = mysql_connect("eugene","root"); > mysql_select_db("mydetailsonline",$db); > $sql="insert into employees > (first_name,last_name,cell_phone,home_phone) values > > ('$first_name','$last_name','$address','$position')"; > $result=mysql_query($sql); > echo "Thank you! information entered.\n"; > }else{ > //display form > ?> > <form method="post" action="<?php echo > $_SERVER['PHP_SELF']?>"> > First name:<input type="Text" name="first_name"><br> > Last name:<input type="Text" name="last_name"><br> > CellPhone:<input type="Text" name="Cell_phone"><br> > Home phone:<input type="Text" name="home_phone"><br> > <input type="Submit" name="submit" value="Enter > information"> > </form> > <?php > } //end if > ?> > </body> > </html> > > Thanks again > > Eugene