CODEWITHSUNDEEP
phphtml
Tom
Tom

Reputation: 2634

mysql update query with form array

Multiple posts but I'm still stuck...I'm missing something fundamental here. I have a form with a select:

<select name="camera_status[]">
  <option <?php echo $enabled_option; ?>>Enabled</option>
  <option <?php echo $disabled_option; ?>>Disabled</option>
</select>

This form is built with a loop to give a list of all camera settings. So you would have multiple cameras and their corresponding camera_status. Also I have a hidden input field with the camera_id:

The camera_id is processed with some javascript. Then I process that with:

$camera_id = $_POST['camera_id'];

if (is_array($_POST['camera_status']))
{
  foreach ($_POST['camera_status'] as $camera_status) {
    $query_status = 'UPDATE `#__cameras` SET `camera_status` ="'.$camera_status.'" WHERE `camera_id`='.$camera_id;
    $db->setQuery($query_status);
    $db->query();
  }
}

If I echo the camera_id it is correct. But my foreach runs an update query for the full list of cameras instead of just the one selected. So it updates only the last camera in the list. Let me know if it makes sense to update the full code. Obviously I'm going about this all wrong...

Upvotes: 1

Views: 2221

Answers (3)

pho
pho

Reputation: 25489

Easy enough. Add value attributes to the <option> tags so that each of them will have a value

<select name="camera_status">
  <option value="1" <?php echo $enabled_option; ?>>Enabled</option>
  <option value="0" <?php echo $disabled_option; ?>>Disabled</option>
</select>

Then, in your php, look for that value

foreach ($_POST['camera_status'] as $camera_status) {
    if($camera_status == 1) {
        $query_status = 'UPDATE `#__cameras` SET `camera_status` ="'.$camera_status.'" WHERE `camera_id`='.$camera_id;
        $db->setQuery($query_status);
        $db->query();
    }
}

Upvotes: 0

nickb
nickb

Reputation: 59699

You have a fundamental flaw in your thought process for the page. When you output more than one <select name="camera_status[]"> elements, you're going to get that many results back. With two of them, you'll get two values in the array, and so on.

What it sounds like you're doing is outputting a list of cameras, having the user select a camera to modify, and then, from then on, all of the camera settings now only apply to that one specific camera. If this is the case, then you don't need to use arrays for the camera settings, including camera_status. Just remove the array portion and stop outputting more than one HTML element for each camera setting (since you know that once a camera is selected, those values apply to that specific camera).

However, if your page that displays the multiple cameras allows the user to modify every camera and its settings, you'll need to accommodate for the user's input.

If the latter is the case, here's a neat trick - Modify your <select> so it looks like this when you're outputting your camera form:

<select name="camera_status[<?php echo $row['camera_id']; ?>]">
    <option value="1">Enabled</option>
    <option value="0">Disabled</option>
</select>

Now, when you grab $_POST['camera_status'], it'll be an array with the camera IDs as the keys and their selected value as the value. So now, you can do this:

if( is_array($_POST['camera_status']))
{
    foreach ($_POST['camera_status'] as $camera_id => $camera_status) {
        $camera_status = intval( $camera_status); // Be wary of SQL injection
        $camera_id = intval( $camera_id);
        $query_status = 'UPDATE `#__cameras` SET `camera_status` ="'.$camera_status.'" WHERE `camera_id`='.$camera_id;
        $db->setQuery($query_status);
        $db->query();
    }
}

Now this will update every camera with the correct value chosen.

Upvotes: 0

dimme
dimme

Reputation: 4424

EDIT: Well if you have single selection then it is simpler than that:

HTML:

<select name="camera_status">
  <option value="Enabled">Enabled</option>
  <option value="Disabled">Disabled</option>
</select>

And PHP:

$camera_id = (int) $_POST['camera_id']; //Here you had SQL injection.
$camera_status = mysql_real_escape_string($_POST['camera_status']); //Neither that was protected.

$query_status = 'UPDATE `#__cameras` SET `camera_status` ="'.$camera_status.'" WHERE `camera_id`='.$camera_id;
$db->setQuery($query_status);
$db->query();

Upvotes: 1

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