Need help updating database row with values from HTML form

Question

I've got an admin area where the admins can set the level of repair and it shows on a progress bar in the users area. I have it all working apart from updating the mySQL database to the value submitted.

My database has a table called 'users' and fields 'UserID', 'Username', 'Password', 'progress', 'admin'.

Here is the code I'm using to try and make the magic happen:

<?php
$query="SELECT * FROM users";
$result=mysql_query($query);
$num=mysql_numrows($result);

?>
<form id="chooseuseredit" method="post" action="<?php echo $PHP_SELF;?>">
<select name="ChooseUser">
<?php
$i=0;
while ($i < $num) {

$f1=mysql_result($result,$i,"UserID");
$f2=mysql_result($result,$i,"Username");
$f3=mysql_result($result,$i,"progress");
$f4=mysql_result($result,$i,"admin");
?>

<option value="<?php echo $f1; ?>"><?php echo $f2; ?></option>

<?php
$i++;
}
?>
</select>
<input type="submit" name="chooseSubmit" id="chooseSubmit" value="Choose User" />
</form>
<?php
if(isset($_POST['chooseSubmit']) )
{
  $varID = $_POST['ChooseUser'];
  $errorMessage = "Jesus Christ Benton, Choose a User!!";

?>
<br>
<p><strong>Editing UserID: <?php echo "$varID"; ?></strong></p>
<p>Progress:<br>
<form name="edituserform" method="post" action="<?php echo $PHP_SELF;?>">
<select name="editinguser">
<option value="0">Phone Not Recieved</option>
<option value="20">Phone Recieved</option>
<option value="40">Parts Recieved</option>
<option value="60">Repair Started</option>
<option value="80">Repair Finished</option>
<option value="100">Posted Back</option>
</select>
<input type="hidden" name="edituserid" id="edituserid" value="<?php echo "$varID"; ?>" />
<input type="submit" name="edituser" id="edituser" value="Edit" />
</form>
    <?php
    if(isset($_POST['edituser'])){

    $add = $_POST['edituser'];
    $varIDe = $_POST['edituserid'];
    $errorMessage = "Jesus Christ Benton, Choose a User!!";
    $query1 = mysql_query("UPDATE users SET progress = $add WHERE UserID = $varIDe");
    mysql_query($query1) or die("Cannot update");

echo $add;
echo $varIDe;

    }
    ?>

<?php
}
?>

I'm not sure if the variables are working or not, or if it's the way I've used the submit button before? Its got me a little stumped.

@Oliver $_POST['edituser'] is the submit button, I think it should be $_POST['editinguser'], just saying – nine7ySix 10 mins ago — nine7ySix
– nine7ySix, Commented Dec 7, 2011 at 13:44

Community · Accepted Answer · 2020-06-20 09:12:55Z

2

You're query should be

$query1 = mysql_query("UPDATE users SET progress = '$add' WHERE UserID = $varIDe");

Don't forget the quotes

and it would be best to change your

mysql_query($query1) or die("Cannot update");

to mysql_query($query1) or die("MySQL ERROR: ".mysql_error());

to get it to display errors

edit

Found a few errors

mysql_numrows should be mysql_num_rows

and major error

$query1 = mysql_query("UPDATE users SET progress = $add WHERE UserID = $varIDe");

is running a query, change it to

$query1 = "UPDATE users SET progress = '".$add."' WHERE UserID = '".$varIDe."'";

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Dec 7, 2011 at 12:43

nine7ySix

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12 Comments

Oliver Whysall Over a year ago

im still not getting errors and cant get it to work

Oliver Whysall Over a year ago

hmm, still cant get it to update the database. I've tried everything everyone has suggested and it doesn't seem to want to play ball!

nine7ySix Over a year ago

@Oliver Whysall Change mysql_query($query1) or die("Cannot update"); to mysql_query($query1) or die(mysql_error()); and see post the error

Oliver Whysall Over a year ago

changed all of that and heres the code again, still doesn't want to do anything!

Oliver Whysall Over a year ago

i dont get an error, it just goes back to the first part asking to choose a user, try logging in at youriphonerepair.x10.mx/postinprogress with username as demo and password as 7575

|

Manse · Accepted Answer · 2011-12-07 12:41:48Z

0

I think your getting the wrong variable

if(isset($_POST['edituser'])){
  $add = $_POST['edituser']; // this is a button

should be :

if(isset($_POST['editinguser'])){
  $add = $_POST['editinguser']; // this is a select list

But please read the following about SQL Injection

answered Dec 7, 2011 at 12:41

Manse

38.1k11 gold badges88 silver badges112 bronze badges

2 Comments

Oliver Whysall Over a year ago

Thank you, I hadn't noticed that!

Manse Over a year ago

This is still incorrect -> if(isset($_POST['edituser'])){ this is a button ?!?!

user862010 · Accepted Answer · 2011-12-07 12:41:49Z

0

When something's going wrong, with respect to query, you better debugging, adding one: or die ( mysql_error ( ) ) ; and then the error message is displayed.

answered Dec 7, 2011 at 12:41

user862010

Comments

Mohit Bumb · Accepted Answer · 2011-12-07 12:48:12Z

0

$query1 = mysql_query("UPDATE `users` SET `progress` = '".$add."' WHERE UserID = '".$varIDe."'");
if(mysql_query($query1))
{
//DO SOME ACTION
}
else
{
die(mysql_error());
}

answered Dec 7, 2011 at 12:48

Mohit Bumb

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But please read the following about SQL Injection

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But please read the following about SQL Injection

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