Am hi guys currently am having problem updating my myqsl database I have a popup window the updates it there is no error but its not updating what the problem.
here is the popup window php code:
<?php
session_start();
include("selectDB.php");
myconnection();
if(isset($_POST['submit']))
{
$myquery="update member set contactno='".$_POST['contactno']."',email='".$_POST['email']."',religion='".$_POST['religion']."',deptid='".$_POST['deptid']."',where memberid='".$_POST['mid']."'";
mysql_query($myquery);
echo "Record has been saved.";
}else{
$myquery="select * from member where memberid='".$_SESSION['memberid']."'";
$results=mysql_query($myquery);
$rs=mysql_fetch_array($results);
?>
<form action="editacct.php" method="post" align="left">
<input type="text" name="mid" size="30" value="<?php echo $rs['memberid'];?>" hidden /><br />
Contact Number:<br /> <input type="text" name="contactno" size="30" value="<?php echo $rs['contactno'];?>" /><br />
E-mail Adress:<br /> <input type="text" name="email" size="30" value="<?php echo $rs['email'];?>"/><br />
Religion:<br /> <input type="text" name="religion" size="30" value="<?php echo $rs['religion'];?>"/><br />
Address:<br /> <input type="text" name="address" size="30" value="<?php echo $rs['address'];?>"/><br />
Department:<br /><select name="deptid" value="<?php echo $rs['deptid'];?>"><br />
<option value="">Select</option>
<option value="CAH001">Development Communication</option>
<option value="CAH002">English</option>
<option value="CAH003">Filipino</option>
<option value="CAH004">Fine Arts</option>
<option value="CAH005">History and Social Sciences</option>
<option value="CAH006">Music</option>
<option value="CAH007">P.E</option>
<option value="CAH008">Psychology</option>
<option value="CAH006">Music</option>
<option value="CAH007">P.E</option>
<option value="CAH008">Psychology</option>
<option value="COB001">Accountancy</option>
<option value="COB002">Commerce</option>
<option value="COB003">Office Administration</option>
<option value="COD001">Dentistry</option>
<option value="COE001">Elementary Education</option>
<option value="COE002">Secondary Education</option>
<option value="COH001">Medical Laboratory</option>
<option value="COH002">Nutrition and Dietetics</option>
<option value="CON001">Nursing</option>
<option value="COT001">Theology</option>
<option value="CST001">Biology</option>
<option value="CST002">Chemistry</option>
<option value="CST003">Computer Science</option>
<option value="CST004">Computer Technology</option>
<option value="CST005">Library Science</option>
<option value="CST006">Math and Physics</option>
<input type="submit" name="submit" value="Update" />
</select>
<br />
</form>
<?php
}
?>
this is the javascript of the popup:
<script type="text/javascript">
function popup(){
cuteLittleWindow = window.open("editacct.php", "", "width=700,height=500");
}
</script>
This is the link to show the popup
<div class="clearfix grpelem" id="pu325-3"><!-- group -->
<a class="nonblock nontext anim_swing clearfix grpelem" id="u325-3" href="javascript:popup()"><!-- content --><p> </p></a>
<a class="nonblock nontext anim_swing clearfix grpelem" id="u333-6" href="javascript:popup()"><!-- content --><p>ACCOUNT</p><p>INFO</p></a>
</div>
Remove the trailing comma just before you start the where clause:
deptid='".$_POST['deptid']."',where
^ right there
Having used mysql_query($myquery) or die(mysql_error());
would have signaled the error.
I would like to note that your present code is open to SQL injection. Use mysqli with prepared statements, or PDO with prepared statements, they're much safer.
mysql_real_escape_string() till you do switch over.Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// rest of your code
Sidenote: Error reporting should only be done in staging, and never production.
Edit: (conversion from mysql_ to mysqli_, a basic method.
Change your present DB connection to the following.
This is an example taken from http://php.net/manual/en/function.mysqli-connect.php
<?php
$link = mysqli_connect("myhost","myuser","mypassw","mybd")
or die("Error " . mysqli_error($link));
// rest of your code
Then change mysql_query($myquery); to mysqli_query($link, $myquery);
$results=mysql_query($myquery); to $results=mysqli_query($link, $myquery);
$rs=mysql_fetch_array($results); to $rs=mysqli_fetch_array($results);
If you have any other functions that start with mysql_, they must be changed to mysqli_.
All mysqli_ functions require DB connection be passed as the first parameter.
This I did to help you out. If this still doesn't work, then I will simply delete this answer.