PHP AJAX Json send object error

Question

var workerEdit = { 
    id: id, 
    name: $('#workers_response input[name="name"][data-id="gear-'+id+'"]').val(),
    photo: $('#workers_response input[name="pas-scan"][data-id="gear-'+id+'"]')[0].files[0],
    bibliography: $('#workers_response textarea[name="bibliography"][data-id="gear-'+id+'"]').val(),
    history: $('#workers_response textarea[name="history"][data-id="gear-'+id+'"]').val(), 
    salary: $('#workers_response input[name="salary"][data-id="gear-'+id+'"]').val()
}

i create this object, then i use ajax to send data throw method POST:

$.ajax({
    type: "POST",
    url: "/admin-workers/edit_worker.php",
    data: 'edit_worker='+JSON.stringify(input_data),

    success: function(data) { 
        var obj = JSON.parse(data); 
        if(obj.error === undefined) { 
            console.log(obj); 
            if(obj.data === null) { 
                alert("Успех"); 
            } 
        } 
        else { 
            alert("Ошибка: "+obj.error); 
            console.log(obj); 
        } 
    }, 
});

and in php i use:

$edit_worker = json_decode($_POST['edit_worker']);
$test = $edit_worker->name;

Why doesn't php get the object?

In your browser's debugging tools, what are the actual details of the AJAX request? Is it a POST request? What does the data look like? In the server-side code, if $_POST['edit_worker'] doesn't contain valid JSON, does it contain anything at all? What are its raw contents? — David
– David, Commented Sep 22, 2017 at 15:55
It's possible data should be a key-value object, not a string. Look at the request. — Halcyon
– Halcyon, Commented Sep 22, 2017 at 15:56
Check the network tab in dev tools and see the status of the request... — ProEvilz
– ProEvilz, Commented Sep 22, 2017 at 15:56

Neil Girardi · Accepted Answer · 2017-09-22 16:43:38Z

1

Your data is not properly serialized. You cannot just stringify a JSON map and POST it as 'edit_worker='+JSON.stringify(input_data) It should look like this:

key1=value1&key2=value2&key3=value3

On the backend, you will extract individual values by their keys:

If ($_SERVER['REQUEST_METHOD'] === 'POST') {
    $val1 = ( isset($_POST['val1']) ) ? trim($_POST['val1']) : '';
    $val2 = ( isset($_POST['val2']) ) ? trim($_POST['val2']) : '';
    // etc. etc. 

}

Your validation probably needs to contain more than just trim, depending upon what the data source is and what you're doing with it, but that's another discussion.

answered Sep 22, 2017 at 16:43

Neil Girardi

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user3720435 · Accepted Answer · 2017-09-22 16:09:14Z

0

You are creating an object with variable workerEdit but your are posting the data from input_data. It appears you are not posting any data because of incorrect variable names.

answered Sep 22, 2017 at 16:09

user3720435

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