SQL, PHP: when submit plus 1 in table_field specified from dropdown - php

Question

UPDATED (*the code is working and I updated the full code :) *)
UPDATED (**edit the suggestions still dont work :) **)
Hi guys I have been looking around in many of the post in here regarding using php to plus 1 in a table_field specifed by a dropdown on the php page. I am very new to this so I just tried some things out to see if I could get it to work but for me this is not easy :)
So here we go.
as you can see in the code there is a drow down that fetch data from a member table. In this table there is a field "isMcount" when a person have chosen a member from the drop down and then press submit I need it to +1 in the "isMcount" column for that specified member which is active in the drop down
Hope you will help.
this is the table setup:
isMcount Type=int(100) Null=No Default=(as defiend)=1
<?php
require('includes/config.php'); ?>
<?php
if(isset($_POST['submit'])){
$row1 = $_POST['username'];
$row1 = mysql_real_escape_string($row1);
mysql_query("UPDATE members SET isMcount = isMcount + 1 WHERE username = '".$row1."' ") or die(mysql_error());
$_SESSION['success'] = 'Page Updated';
header('Location: '.DIRADMIN);
exit();
}
?>
<form method="post">
<select name="username">
<?php
$sql = mysql_query("SELECT username FROM members");
while ($row1 = mysql_fetch_array($sql)){
echo "<option value=\"" . $row1['username'] . "\">" . $row1['username'] . "</option>";
}
?>
</select>
<input type="submit" name="submit" value="Submit" class="button" />
</form>

Answer 1

You have wrong HTML part. Try option below. "username" is field name for select and will be available on $_POST['username'] with a value of username for your user.
<form method="post">
<select name="username">
<?php
$sql = mysql_query("SELECT username FROM members");
while ($row1 = mysql_fetch_array($sql)){
echo "<option value=\"" . $row1['username'] . "\">" . $row1['username'] . "</option>";
}
?>
</select>
<input type="submit" name="owner" value="submit"/>
</form>
And you can remove isMCount from $_POST in PHP as you don't need it.
And then you update query would be:
mysql_query("UPDATE members SET isMcount = isMcount + 1 WHERE username = '" . $row1 . "'");

Answer 2

Please check the statements within if condition:
$row1 = $row['username']; // It should be $_POST['owner']
$mc = $row['isMcount']; // You don't even need this
mysql_query("UPDATE members SET isMcount = isMcount + 1 WHERE $row1 = username");
You simply need to get the username from the POST data submitted through the form.
So, you query should be something like this:
mysql_query("UPDATE members SET isMcount = isMcount + 1 WHERE username = '".$row1."' ") or die(mysql_error());
Also, you should change your HTML to this:
echo "<option value=\"" . $row1['username'] . "\">" . $row1['username'] . "</option>";
Hope this helps.
Peace! xD

Answer 3

Some suspected areas are in your code
<form action="#" method="POST">
<?php
if(isset($_POST['submit'])){
$username = $row['username'];
$mc = $row['isMcount'];
$row1 = mysql_real_escape_string($row1);
$mc = mysql_real_escape_string($mc);
mysql_query("UPDATE members SET isMcount = isMcount + 1 WHERE username = '$username'");
$_SESSION['success'] = 'Page Updated';
header('Location: '.DIRADMIN);
exit();
}
?>
<select name="username">
<?php
$sql = mysql_query("SELECT username FROM members");
while ($row1 = mysql_fetch_array($sql)){
echo "<option value='".$row1['username']."'>".$row1['username']."</option>";
}
?>
</select>
<input type="submit" name="submit" value="submit"/>
</form>
Suggestion : Use mysqli or pdo instead of mysql