JSON_decode problem with array posted to CodeIgniter

Question

I am certain I am making a simple mistake... but simply can't find it.

Ultimately I am posting a JSON array from an Android app (that part is working), but for the time being I am simply testing between two PHP pages (1: test PHP page with basic form, and 2: the CodeIgniter final destination) Here is what I have:

At the form page:

<form action="bambooinvoice/index.php/api2/newinvoice/4/0/0" method="post">
    <?php 
        $array = array("items"=>array(
            "taxable"=>1, 
            "quantity"=>1, 
            "amount"=>123.99, 
            "work_description"=>"this is a test"));
        $json = json_encode($array);
    ?>
    <input type="hidden" name=json value=<?php $json ?> />
    <input type="submit" name="btnSendForm" value="Send" />
</form>

This creates (which looks good to me):

{"items":{"taxable":1,"Quantity":1,"amount":123.99,"work_description":"this is a test"}}

On the codeIgniter side, I have:

$input = $this->input->post('json');
$items = json_decode($input, TRUE);

$amount = 0;
foreach ($items as $item) // In case there are multiple 'items'
{
    $taxable = (isset($item['taxable']) && $item['taxable'] == 1) ? 1 : 0;

    $invoice_items = array(
        'quantity' => $item['quantity'],
        'amount' => $item['amount'],
        'work_description' => $item['work_description'],
        'taxable' => $taxable
    );

    $this->_addInvoiceItem($invoice_items); //simply adding contents to DB
}

In the end I receive the error: (i have received numerous errors actually in all my tweaking, but this is the one I can't seem to shake)

Message: Invalid argument supplied for foreach()

Edited - to correct a typo.

have you tried var_dump()ing $items and $json to ensure they are what you expect them to be? — Shad
– Shad, Commented Feb 28, 2011 at 22:50
You're dumping unescaped JSON data to an attribute. I'm assuming this is causing some brokenness in the DOM, and the POST data is also broken. — Dan Lugg
– Dan Lugg, Commented Feb 28, 2011 at 22:51
If is not a typo here, this line ` <input type="hidden" name=json value=<?php $json ?> />` looks wrong to me. you are not echoing the data. it should be <?=$json?> or <?php echo $json?> — Francesco Laurita
– Francesco Laurita, Commented Feb 28, 2011 at 22:53
Yes, thank you for reminding me, I have output'd the various variables. var_dump($input); = 1 I had seen this previously but thought it meant there was '1' associative array inside the posted JSONObject (ie - 1 'item')... but perhaps I have handled this incorrectly and it is being treated as a bool? — ctgScott
– ctgScott, Commented Feb 28, 2011 at 23:27
...and var_dump($items)=NULL. This tells me my mistake is between the post on the form side and decode on the service side. — ctgScott
– ctgScott, Commented Feb 28, 2011 at 23:32

bschaeffer · Accepted Answer · 2011-03-01 04:35:46Z

3

You are using $this->input->post('items') when your form is posting a hidden value named json.

If you var_dump($this->input->post('items')), it should be FALSE or NULL.

Try this in your CI script instead:

$input = $this->input->post('json'); // not 'items'
$items = json_decode($input, TRUE);

// Rest of your code...

That should fix that problem, but you also need to make sure your json data is being sent correctly to begin with! var_dump($_POST) should show you if it's making it to your script in one piece.

answered Mar 1, 2011 at 4:35

bschaeffer

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2 Comments

ctgScott Over a year ago

Aaaaaarrghh!!! (sighh...) It is working just as it should. Thank you bschaeffer, I can't tell you how much that stupid mistake was irritating me.

bschaeffer Over a year ago

Yeah... those kind of things can be hard to notice because you never think you messed up the "easy part". Glad to help!

azat · Accepted Answer · 2011-02-28 22:50:54Z

0

Try this
<input type="hidden" name=json value='<?=$json?>' /> Or this
<input type="hidden" name=json value='<?=str_replace('\'', ''', $json)?>' />
See htmlspecialchars

answered Feb 28, 2011 at 22:50

azat

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1 Comment

ctgScott Over a year ago

Thanks, azat. You are right... I needed to correct this line. It is now: <input type="hidden" name=json value=<?php echo $json; ?> />

icc · Accepted Answer · 2011-02-28 22:52:48Z

0

When you encode an assoc. array as JSON it becomes an object and is no longer an array. Now when you decode your JSON php creates objects of your JSON and not an assoc. array.

answered Feb 28, 2011 at 22:52

icc

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1 Comment

azat Over a year ago

No correct.. see this ru2.php.net/manual/en/function.json-decode.php, second parameter

Elias Rodrigues · Accepted Answer · 2011-03-01 05:00:12Z

You can change and not use json_encode and use serialize in you array.

Like this in your form:

<form action="bambooinvoice/index.php/api2/newinvoice/4/0/0" method="post">
    <?php 
        $array = array("items"=>array(
            "taxable"=>1, 
            "quantity"=>1, 
            "amount"=>123.99, 
            "work_description"=>"this is a test"));
        $json = serialize($array);
    ?>
    <input type="hidden" name=json value="<?php echo $json ?>" />
    <input type="submit" name="btnSendForm" value="Send" />
</form>

This serialize function will create this output:

a:1:{s:5:"items";a:4:{s:7:"taxable";i:1;s:8:"quantity";i:1;s:6:"amount";d:123.9899999999999948840923025272786617279052734375;s:16:"work_description";s:14:"this is a test";}}

Then in the codeigniter side you can do this:

<?php
$input = $this->input->post('json');
$items = unserialize($input);

$amount = 0;
foreach ($items as $item) // In case there are multiple 'items'
{
    $taxable = (isset($item['taxable']) && $item['taxable'] == 1) ? 1 : 0;

    $invoice_items = array(
        'quantity' => $item['quantity'],
        'amount' => $item['amount'],
        'work_description' => $item['work_description'],
        'taxable' => $taxable
    );

    $this->_addInvoiceItem($invoice_items); //simply adding contents to DB
}
?>

but be careful, because when you use a foreach with an array who contain more the 1 level. I hope that this help you.

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