MySQL IN only deleting single row, should delete multiple from variable - PHP

Question

I'm trying to delete multiple records in a database based on selections from check boxes in a form. But if multiple check boxes are selected, the current sql only deletes first id for some reason, and not deleting the rest.

$del_ids = implode(',', array_keys($_POST['check_list'])); 
$query = "DELETE FROM enhancements WHERE id IN('$del_ids')";
$result = mysqli_query($connection, $query);

For example my array looks like this:

Array
(
    [check_list] => Array
        (
            [4] => on
            [6] => on
        )
)

And after running:

$del_ids = implode(',', array_keys($_POST['check_list']));

my $del_ids string looks like this:

4,6

But for some reason only record with id 4 deletes, and record with id 6 does not delete and they should both delete. Any suggestions?

it seems you are quoting the IN values , resulting in something like IN ('4,6') — darioguarascio
– darioguarascio, Commented Mar 26, 2015 at 11:37
Change WHERE id IN('".$del_ids."') to WHERE id IN(".$del_ids.") — Abhik Chakraborty
– Abhik Chakraborty, Commented Mar 26, 2015 at 11:38

darioguarascio · Accepted Answer · 2015-03-26 11:38:51Z

2

you have to use this:

$del_ids = implode("','", array_keys($_POST['check_list']));

answered Mar 26, 2015 at 11:38

darioguarascio

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Comments

zdeniiik · Accepted Answer · 2015-03-26 11:54:37Z

0

$query = "DELETE FROM enhancements WHERE id IN('".$del_ids."')";

sends keys like a string, try replace with

$query = "DELETE FROM enhancements WHERE id IN(".$del_ids.")";

and it should work

edited Mar 26, 2015 at 11:54

answered Mar 26, 2015 at 11:39

zdeniiik

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Comments

Flo draven · Accepted Answer · 2015-03-26 11:38:42Z

-1

either run muliple delete statements, or append OR clauses in your statement.

The array you have with the ids to delete needs to be enumerated, you could go with something like this :

        foreach($_POST['check_list']as $key) {
    $query = "DELETE FROM enhancements WHERE id IN('".$key."')";
$result = mysqli_query($connection, $query);
    }

answered Mar 26, 2015 at 11:38

Flo draven

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1 Comment

Kevin Over a year ago

and the fact that you're using as $key instead of $key => $value, you'll get those on values instead of the actual key, this will yield IN('on')

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