You are not actually posting the form to a page with your PHP on... so its
trying to insert that form into your DB before u actually fill it out.
either check if the form has been posted before running your PHP, or split
this into 2 pages
Andrew
----- Original Message -----
From: "Denis L. Menezes" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, April 01, 2002 5:08 PM
Subject: [PHP] Please help with code
> Hello friends,
>
> can someone tell me what is wrong with this code :
>
>
> <form name="form1" method="post" action="">
> <p>
> <input type="text" name="newsid">
> </p>
> <p>
> <input type="text" name="title">
> </p>
> <p>
> <input type="text" name="author">
> </p>
> <p>
> <input type="text" name="body">
> </p>
> <p>
> <input type="submit" name="Submit" value="Submit">
> </p>
> </form>
> </body>
> <?php
> $link=mysql_connect("localhost","myid","mypassword");
> if ($link){
> print "link id is $link";
> } else {
> print "error connecting to database";
> }
> $posted=time();
> $query="INSERT INTO news (newsid,title, author, body, posted)
> VALUES($newsid,$title, $author, $body, UNIX_TIMESTAMP())";
> IF (mysql_query($query)){
> print "Row added to table";
> } else {
> print "error adding row";
> }
> ?>
>
>
> It gives the link ID. But then gives "error adding row"
>
> Thanks in advance
>
> Denis
>
>
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