(PHP) Inserted data appears in MySQL as "Array"

Question

This feels like a common problem, but it's a difficult one to Google!

As explained in the comment in the code, inserting the value was previously not a problem but stopped working when I added the additional INSERT queries below it.

$user = $_POST['name'];
$night = $_POST['club'];
$query = mysql_query("SELECT day FROM nights WHERE name = '$night'");
$email = $_POST['email'];

while ($row = mysql_fetch_assoc($query)) {

    $date = getFullDateString($row['day']);
    $date2 = getDateString($row['day']);

}

// this one previously worked but now enters into the database with $user as "Array"
mysql_query("INSERT INTO guestlists (guest, night, date) VALUES('$user', '$night', '$date') ") or die(mysql_error());

$guest1 = $_POST['name1'];
$guest2 = $_POST['name2'];

// these were added later and work fine but seem to have had an effect on the query above
mysql_query("INSERT INTO guestlists (guest, night, date) VALUES('$guest1', '$night', '$date') ") or die(mysql_error());
mysql_query("INSERT INTO guestlists (guest, night, date) VALUES('$guest2', '$night', '$date') ") or die(mysql_error());

foreach ($_POST as $key){
    if (is_array($key)){
        foreach ($key as $key2 => $value){
            mysql_query("INSERT INTO guestlists (guest, night, date) VALUES('$value', '$night', '$date') ") or die(mysql_error());

        }
    }
}

$value than is proabably an array check it with is_array. And your code asks for mysql injection all over. — Elzo Valugi
– Elzo Valugi, Commented Jun 13, 2012 at 9:57
Oh no, SQL Injection! Your code is very vulnerable to attacks. Please use mysql_real_escape_string or even better, PDO. — kapa
– kapa, Commented Jun 13, 2012 at 9:58
Yeah, I haven't looked at SQL injections yet. Any pointers in the right direction? — Sebastian
– Sebastian, Commented Jun 13, 2012 at 9:58
use the print_r($variable) to check all your variables before adding it in mysql query, and if it is array than get the required value with $variable['key']. And BTW Please stop writing new code with the ancient mysql_* functions. They are no longer maintained. — Sanjay
– Sanjay, Commented Jun 13, 2012 at 10:01

Gustav Bertram · Accepted Answer · 2012-06-13 10:11:47Z

1

You should rename the guest HTML inputs to guests[] instead of using name again. You have a naming conflict.

You need to fix this in your javascript code:

var name = $("<p><input class='input' type='text' name='guests[]' value='' /></p>");

And in your HTML code:

<input class="input" type="text" name="guests[]" />

After that, your PHP code should handle the guest variable as an array:

$guests = $_POST['guests'];

foreach ($guests as $guest)
{
    mysql_query("INSERT INTO guestlists (guest, night, date) VALUES('$guest', '$night', '$date') ") 
    or die(mysql_error());
}

Note that you don't need to go guests[1], guests[2], guests[3] etc.

answered Jun 13, 2012 at 10:11

Gustav Bertram

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This is the tidiest solution.

matchdav
– matchdav

06/13/2012 10:19:09
Commented Jun 13, 2012 at 10:19

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mvds · Accepted Answer · 2012-06-13 09:58:05Z

1

in your website you have:

<input class='input' type='text' name='name["+currentArrayNum+"]' value='' />

to generate the form. The name[...] arguments end up in php as $_POST['name'] which will be an array. If you force this to be a string, the string will be Array. So take your $user and iterate over it using foreach, to handle each name.

Also, please read up on sql injection!

answered Jun 13, 2012 at 9:58

mvds

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Dan Smith · Accepted Answer · 2012-06-13 09:58:07Z

0

You have some Javascript on your page that is inserting form fields that look like this:

<input class="input" type="text" name="name[5]" value="">

The name="name[5]" part is whats causing your problems here - it turns $_POST['name'] in to an array when submitted. You'll need to refactor that code.

answered Jun 13, 2012 at 9:58

Dan Smith

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