furqan219 -9 Junior Poster in Training
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Jump to Posthello
you can get this value like this in update.php file
$c=$_post['siteid']; echo "value".$c;
I hope your problem may be solve
Thanks.
Jump to Post$c= $_POST
echo "value".$c
gives this errorParse error: parse error in C:\xampp\htdocs\update.php on line 24
hi
you have no pu ';' semicolon after the
this code$c= $_POST['siteid']; echo "value".$c;
parse error means you miss ; in script ok.
Jump to Posthi
this is your code but you can also do easy way in php
so i just ediit it like<?php while($row = mysql_fetch_array($result)) { ?> <table cellpadding=2 cellspacing=2 width=100%> <tr> </tr> <tr> <th bgcolor=#5D9BCC >SiteID</th> <td bgcolor=#FEE9A9><?= $row['SiteId'] ?>/td> </tr> </table> <? $b =$row['SiteId']; // store …
Jump to Posthi
if you want to fetch only one row so that time no need of while loop ok
$row=mysql_fetch_array($result);
this is enough code.
now first of all you need to put name address and city value in text box ok<form action="update.php" method="post"> …
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furqan219 -9 Junior Poster in Training
Tulsa 1 Junior Poster in Training
furqan219 -9 Junior Poster in Training
furqan219 -9 Junior Poster in Training
Tulsa 1 Junior Poster in Training
Tulsa 1 Junior Poster in Training
furqan219 -9 Junior Poster in Training
Tulsa 1 Junior Poster in Training
furqan219 -9 Junior Poster in Training
furqan219 -9 Junior Poster in Training
Tulsa 1 Junior Poster in Training
furqan219 -9 Junior Poster in Training
Tulsa 1 Junior Poster in Training
furqan219 -9 Junior Poster in Training
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