HTML form in my PHP

[tom_b]

I have the following code with a small form in it. The form is fine when on its' own but doesn't show up when combined with the PHP code. Suggestions anyone??

Thanks!!

<?php

$username="root ";
$password="woot s";
$database="text ";
$id=$_GET['id'];
mysql_connect(l ocalhost,$usern ame,$password);
@mysql_select_d b($database) or die( "Unable to select database");

$query="SELECT * FROM trc_text where id='$id'";
$result=mysql_q uery($query);
$num=mysql_numr ows($result);
mysql_close();

$i=0;
while ($i < $num) {
$text=mysql_res ult($result,$i, "text");
?>

<form action="updated _text.php" method="post">
<input type="hidden" name="ud_id" value="<?php echo $id; ?>">
Text: <input type="text" name="ud_text" value="<?php echo $text; ?>"><br>
<input type="Submit" value="Update">
</form>

<?php
++$i;
}

?>

[ronverdonk]

Your form states ACTION=POST. But you read the value using a $_GET['id'].

Ronald :cool:

[iam_clint]

there is also a better way to loop through your sql results

while($row=mysql_fetch_array( $result )) {
$text = $row['text'];
}

anyways i like to use this way better.

[bevort]

Do you get something back from your database. A common error is asking for "$ID" as a string instead of $ID as an integer.
Check this using an

Echo($num);

after counting tne number of records.
Another way is to check for the result like

if ($result) {echo ("No data recieved from the database);}
else { show your form }

Using your form in a faked environment dit not give any problems

Vincent