Updating multiple records using one submit query

[blackbox]

Hi all,

After trying to get this to work for what seams like hours I am now asking for help.

I am trying to write a script that will update multiple records using one submit.

This is what I have so far;

<form id="form2" name="form2" method="post" action="insert_image_order.php" target="_self">
<select name="ImageOrder" class="sys_bodytext_left" id="ImageOrder">
                  <option value="1">1</option>
                  <option value="2">2</option>
                  <option value="3">3</option>
                  <option value="4">4</option>
                  <option value="5">5</option>
                  <option value="6">6</option>
                  <option value="7">7</option>
                  <option value="8">8</option>
                  <option value="9">9</option>
                  <option value="10">10</option>
                </select>
<input type="submit" name="button2" id="button2" value="Submit" /></td>
<input type="hidden" name="MM_update" value="form2" />
<input type="hidden" name="RecordID" value="<?php echo $row_ImageLookUp['RecordID']; ?>" />
</form>

if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "form2")) {

foreach ($_POST['RecordID'] as $RecordID) {	

$updateSQL = sprintf("UPDATE Images SET ImageOrder=%s WHERE RecordID=%s",
                   GetSQLValueString($_POST['ImageOrder']= $_POST['ImageOrder'][$ImageOrder], "int"),
                   GetSQLValueString($_POST['RecordID'] = $RecordID, "int"));

mysql_select_db($database_Swift, $Swift);
$Result1 = mysql_query($updateSQL, $Swift) or die(mysql_error());
}
}

Prior to the data been submitted it is validated and sanitized.

I have tested to see how many record are in the $_POST array so that part is what I expect to see.
Can anyone see where I am going wrong.

Many thanks in advance.

[laserlight]

What exactly is the problem?

[blackbox]

Hi Laserlight,

Many thanks for looking at this issue. The problem is the updated data in $_POST['ImageOrder'] (in my test there are 4 records to update) is not getting written to the table "Images".

What I am hoping is the foreach would loop through until all 4 rows are updated.

Cheers again.

Blackbox

[blackbox]

Hi all again,

I am finding this very hard to solve. To try and provide more information.

The form which gets submiited only has one field, a drop down list [1,2,3,4,5] for each row. the user selects one of the options in the list and submits the form.

The form contains dynamic rows and it could have 1 or more drop down list, one for each record the form displays.

Can any one point me in the right direction so I can get this working.

Example form

Row1: image1. dropdown list1
Row2: image2. dropdown list2
Row3: image3. dropdown list3

[NogDog]

If you name your form elements using the square-bracket array convention, then you can loop through the array in your PHP code -- but all the elements need to be within the same <form> element. In a very simple example:

<form action="foo.php" method="post">
<p>
<input type="text" name="data[1][name]" />
<input type="text" name="data[1][email]" />
</p>
<p>
<input type="text" name="data[2][name]" />
<input type="text" name="data[2][email]" />
</p>
</form>

Then in your form-handler

$stmt = $pdo->prepare("INSERT INTO table_name (name, email) VALUES(:name, :email);
foreach($_POST['data'] as $input) {
    $stmt->bindParam(':name', $input['name'];
    $stmt->bindParam(':email', $input['email'];
    $stmt->execute();
}

Of course, I'm also recommending you use PDO instead of the deprecated mysql extension. :p