Make checkbox appear in $_POST array even if unchecked

Question

I have an invoice in a HTML table. You select item name then item lot and the item #. There is a check box at the end to mark if it is sold out.

<input type="checkbox" name="soldout[]">

When I process the form the $_POST array of check boxes obviously don't contain the unchecked boxes. That leads me to problems when matching up the check boxes to the item

foreach($productID as $key=>$id){
    $p = $id;
    if(isset($soldout[$key])){
        $so = true;
    }
}

So if I have three line items and the first and last are marked sold out it processes as the first 2 sold out and the last one not sold out.

I don't think i can do this:

<input type="checkbox" name="soldout[<?php echo $i; ?>]">

because I am adding empty invoice lines with javascript when needed and wouldn't know how to increment the array key.

Any ideas on how to handle this?

Does a newly added row (through JavaScript) have a productID as well? If so, and if they are unique per row, I'd use the productID as a key to soldout[], i.e. soldout[<?php echo $productId; ?>]. Obviously, with JavaScript, you'd have to analyze the newly created row for its productID then, and add is as a key to the checkbox name. — Decent Dabbler
– Decent Dabbler, Commented Mar 1, 2013 at 2:28

user1765062 · Accepted Answer · 2013-03-01 02:23:48Z

1

When you increase the array using JS just change the name to soldout_new and then increase a variable.

Something like this:

var soldout_new = 0;

function add()
{
'<input type="checkbox" name="soldout_new[' + soldout_new + ']">';

soldout_new += 1;
}

or also you can just set the variable to the id from php:

var soldout = <?php echo $i; ?>;

function add()
{
'<input type="checkbox" name="soldout[' + soldout + ']">';

soldout += 1;
}

answered Mar 1, 2013 at 2:23

user1765062

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Comments

Mike · Accepted Answer · 2013-03-01 02:30:00Z

There is no straight HTML solution for this. If Javascript is an option, you could look there, but if you want a straight HTML solution, an unchecked checkbox isn't going to show up in PHP, but a radio button will.

<input type="radio" name="soldout[<?php echo $i; ?>]" value="true"> Yes
<input type="radio" name="soldout[<?php echo $i; ?>]" value="false"> No

Then in PHP you can access it like

foreach($productID as $key=>$id){
    $p = $id;
    if($soldout[$key] == "true"){
        $so = true;
    }
    else {
        $so = false;
    }
}

Just remember the quotes around "true" because it's a string

Then again, if you are populating the checkboxes server-side you should also presumably know which values should exist and if they are not present in $_POST, assume they are unchecked.

foreach ($availableProductIDs as $id) {
    if(isset($_POST['soldout'][$id]){
        $so = true;
    }
    else {
        $so = false;
    }
}

Michael L. · Accepted Answer · 2013-03-01 04:46:48Z

0

Why not do something like this?

<input type="checkbox" name="soldout[]" value="<?= $productID[0] ?>">
<input type="checkbox" name="soldout[]" value="<?= $productID[1] ?>">

Then in PHP

foreach($productID as $key=>$id) {
    $p = $id;
    if(in_array($id, $soldout) {
        $so = true;
    }
}

answered Mar 1, 2013 at 4:46

Michael L.

6183 silver badges17 bronze badges

Comments

Shehary · Accepted Answer · 2017-06-06 06:22:10Z

0

Java free solution:

&lt;input type="hidden" name="hsoldout[]" value="$value"&gt;
&lt;input type="checkbox" name="soldout[]" value="$value"&gt;<br />

After submit, if a difference between soldout and hsoldout appears, it means that the checkbox was engaged. In fact, you need only one hidden key for each form:

if(array_key_exists($_POST[hsoldout],$i)) { <br/>
    $_POST[soldout][$i] = $_POST[soldout][$i];   // makes sure it is set as a 
    $_POST var <br />
    .... <br />
}

edited Jun 6, 2017 at 6:22

Shehary

9,99810 gold badges47 silver badges75 bronze badges

answered Oct 21, 2013 at 13:23

19cool53

11 bronze badge

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