While (mysql_fetch_array) Question

cmpennington

code is returning an error:
Parse error: parse error in /home/bpafuner/public_html/gate/select.php on line 16

Here's the code: I know the mistake must be something simple.

<?php

$db = mysql_connect("localhost", "bpafuner_cmpenni","unclemel");

mysql_select_db("bpafuner_test",$db);

foreach($HTTP_POST_VARS as $value) {
echo "<BR>You clicked checkbox number $value ";

}

$result = mysql_query("SELECT * FROM employees WHERE id=$id");

$myrow = mysql_fetch_array($result)){
$fullname = $myrow["Fullname"];
}
?>
<html>
<form action=edit_obit.php method=post>
<center>
<input type=hidden name="id" value="<?php echo $id ?>">
<table>
<tr align=right><td> Deceased's First Name:</td><td><input type="Text" size=40 name="first" value="<?php echo $first ?>"><br>
<tr align=right><td> Deceased's Last Name:</td><td><input type="Text" size=40 name="last" value="<?php echo $last ?>"><br>
<tr align=right><td> Deceased's Full Name:</td><td><input type="Text" size=40 name="Fullname" value="<?php echo $Fullname ?>"><br>
</td></tr>
<tr align=right><td> Dateline:</td><td><input type="Text" size=40 name="address" value="<?php echo $address ?>"><br>
</td></tr>
<tr align=right><td> Date Of Death:</td><td align=left><input type="Text" size=8 name="date" value="<?php echo $date ?>"><br>
</td></tr>
<tr align=right><td> Photo Location:</td><td align=left><input type="Text" size=40 name="photo" value="<?php echo $photo ?>"><br>
</td></tr>
</table>
Obituary:<br></td></tr><tr><td><TEXTAREA NAME="position" cols=50 rows=15 TYPE="TEXT" ALIGN="ABSMIDDLE" WRAP="virtual"><?php echo $position ?></textarea><br>

</form>
</center>
</html>

Thanks, cmpennington

yahelb

here is the problem:

your code:

$myrow = mysql_fetch_array($result)){  

      $fullname = $myrow["Fullname"]; 
}

i think you ment for somthing like

if($myrow = mysql_fetch_array($result)){  

      $fullname = $myrow["Fullname"]; 
}

olaf

or just like this:

$myrow = mysql_fetch_array($result);
$fullname = $myrow["Fullname"];

qo2o

if youre trying to do a while loop with it, it would be

$result = mysql_query("SELECT * FROM employees WHERE id=$id"); 

while($myrow = mysql_fetch_array($result)){ 
$fullname = $myrow["Fullname"]; 
}

looks like you just need to do what olaf said though, since it seems that youre only searching through one row anyway, right?

olaf

three answers for such a small spell-mistake...

cmpennington

Actually I solved the problem, I used the mysql_fetch_assoc and it worked just fine. I thank all of you for sharing you knowledge.

Here is the final code:

<?php

foreach($HTTP_POST_VARS as $value) { }

$db = mysql_connect
connection........
mysql_select_db....................
$result = mysql_query("SELECT * FROM employees WHERE id=$id");
While ($myrow = mysql_fetch_assoc($result)){

$last= $myrow["last"];
$position = $myrow["position"];
$address = $myrow["address"];
$date = $myrow["date"];
$first = $myrow["first"];
$photo = $myrow["photo"];
$Fullname = $myrow["Fullname"];

}

olaf

he didn't noticed the problem...we see him again...