Reading form array

Question

I have this previous post where I need to change my HTML form to accept an array. This post has the full details of the code.

So I changed all forms accordingly, such as:

<select name="camera_type[]">

But now I'm stuck on how to run the update query. Now that I store everything in an array I obviously can't use my query as is. Here is a post that seems to be exactly what I need, but I'm not sure I understand: Update MySQL table with an Array using Foreach

Any further help on this is appreciated.

What error are you getting when you run the UPDATE query? It looks like you have all of the code on the other post. — Homer6
– Homer6, Commented Nov 14, 2011 at 3:53

footy · Accepted Answer · 2011-11-14 02:06:35Z

1

Use a foreach loop on the submitted value. Then you can iterate over the entire array and construct queries as you please.

For example: You have a $_POST['camera_type'] variable which is an submitted from the browser. In your php script do the following.

foreach($_POST['camera_type'] as $value) {
  //do your processing.
}

answered Nov 14, 2011 at 2:06

footy

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3 Comments

Tom Over a year ago

Sorry, wasn't clear. I've added the foreach loop on the submitted value. Just not sure how to construct the update query. Do I update in each loop? Seems...inefficient?

Luke Stevenson Over a year ago

Agreed, it seems inefficient, but it'd the only way. Unless you want to go down the track of creating a single UPDATE query and using CASE arguments to set the values for the various feeds based on the tuple's ID field.

footy Over a year ago

@TomPepernic Tho may be inefficient its a general way. The POST['camera_type'] holds ONLY those which are submitted, so it wont have redundant values. Hence its better.

Yes Barry · Accepted Answer · 2011-11-14 09:07:34Z

1

To help understand looping through an array and running UPDATE queries check this out:

// first setup your MySQLi connection
$db = new mysqli('localhost', 'user', 'pass', 'dbname');

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit;
}

// disable autocommit for transactions
$db->autocommit(false);

// **EDIT** forgot to escape input data. fixed now. note the mysqli_real_escape_string() and int type casting
foreach ($_POST['camera_type'] as $type) {
    // assuming 'id' is set in the POST array
    $query = sprintf(
        'UPDATE table SET column = "%s" WHERE id = %d', 
        $db->real_escape_string($type), 
        (int) $_POST['id']
    );
    $db->query($query);
}

// commit transactions (or commit the updates that were run) and then close
$db->commit();
$db->close();

edited Nov 14, 2011 at 9:07

answered Nov 14, 2011 at 7:56

Yes Barry

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6 Comments

Yes Barry Over a year ago

hmm oops missed that one. fixed. Thanks! ...but why was this downvoted?

Your Common Sense Over a year ago

because of sql injection which still persists. you have to either cast the id to int instead of escaping OR quote it.

Yes Barry Over a year ago

Ok touche. Thanks for commenting! I fixed it now. :) Did I miss anything else? (by the way, I swear I always validate all my data before running queries lol)

Your Common Sense Over a year ago

data validation and query composing are different matters.

Your Common Sense Over a year ago

I am taking validation as data-specific routine, say, email validation. while query composing should be quite contrary - data-irrelevant, allowing any kind of data into query while making it safe way.

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