form $_POST error, dynamic fields

Question

I have an error that I can't figure out...

Om my webpage there is a form that the user has the ability to add some new input fields to. If the user is submitting the form, then the optional fields is empty when the php-file is handing them, why?

HTML:

        <form method="post" action="newRequest.php">
            <input type="text" name="title" />
            <input type="hidden" name="fname" value="0" />
            <input type="checkbox" name="fname" value="1"/>
            <input type="hidden" name="ename" value="0" />
            <input type="checkbox" name="ename" value="1" />
            <input type="hidden" name="seat" value="0" />
            <input type="checkbox" name="seat" value="1" />
            <input type="hidden" name="fields" value="0" />
            <input type="text" id="fields" name="fields" />
            <input type="submit" />
        </form>

PHP:

if (strlen($_POST[title]) > 2) {
    $toDb[title] = $_POST[title];
} else {
    error('title');
}

$toDb[fname] = $_POST[fname];
$toDb[ename] = $_POST[ename];
$toDb[seat] = $_POST[seat];

if ($_POST[fields] > 0) {
    $i = 0;
    while ($i < $_POST[fields]) {
        $toDb[optional][$i] = $_POST[optional-$i];
        $i++;
    }
    $toDb[optional] = serialize($toDb[optional]);
} else {
    $toDb[optional] = 0;
}
newEvent($toDb,$dbh);

JQuery that is adding dynamical fields:

$(document).ready(function() {
    $('#fields').focusout(function(){

        var fields = $('#fields').val();
        var i = 0;

        while(i < fields) {
            $('#fields').after("Valfritt fält "+(i+1)+":<input type='text' name='optional"+i+"' />");
            i++;
        }

    })
})

This is how you use hash arrays: $foo["bar"] = $_POST["bar"];, the string index should be quoted. — HamZa
– HamZa, Commented Mar 20, 2013 at 13:33
All your field names have duplicates; the first one gets overwritten by the second one when it gets parsed in PHP. — Ja͢ck
– Ja͢ck, Commented Mar 20, 2013 at 13:34
The duplicates is there because the first one is setting a default value that is sent to PHP if the user leaves the field empty. — rablentain
– rablentain, Commented Mar 20, 2013 at 13:37

ulentini · Accepted Answer · 2013-03-20 13:38:14Z

1

You should quote array indexes. It should be

 $toDb['optional'][$i] = $_POST['optional'.$i];

answered Mar 20, 2013 at 13:38

ulentini

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Comments

Muhammad Raheel · Accepted Answer · 2013-03-20 13:38:18Z

1

You are missing commas in $_POST

$toDb['fname'] = $_POST['fname'];
$toDb['ename'] = $_POST['ename'];
$toDb['seat'] = $_POST['seat'];

Here is your modified code

if (strlen($_POST['title']) > 2) {
    $toDb['title'] = $_POST['title'];
} else {
    error('title');
}

$toDb['fname'] = $_POST['fname'];
$toDb['ename'] = $_POST['ename'];
$toDb['seat'] = $_POST['seat'];

if (count($_POST) > 0) {
    $i = 0;
    while ($i < count($_POST)) {
        $toDb['optional'][$i] = $_POST['optional-'.$i];
        $i++;
    }
    $toDb['optional'] = serialize($toDb['optional']);
} else {
    $toDb['optional'] = 0;
}
newEvent($toDb,$dbh);

Also use count() to check if $_POST has values > 0.

answered Mar 20, 2013 at 13:38

Muhammad Raheel

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Comments

Community · Accepted Answer · 2017-05-23 11:57:26Z

0

I faced the same problem and I solved it using Javascript, like this :

add a new text field every time a button is pressed

edited May 23, 2017 at 11:57

CommunityBot

11 silver badge

answered Mar 20, 2013 at 13:33

A human being

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