Ajax call not returning JSON error

Question

I have a client view that calls a PHP file via AJAX to update the database behind a form. The view is basic CRUD operations, and consists mostly of a Form and a set of inputs:

HTML

<div class="item-details">
    <form id="item-info-form">
        <input type="hidden" name="id" value="1" />
                <input type="text" name="name" size="50" value="Item 1" />
                <input type="text" id="start_date" name="start_date" value="2012-01-13" />
                <input type="text" id="end_date" name="end_date" value="2014-02-03" />
                <input type="text" name="location" size="50" value="Home" />
                <select name="item_type">
                    <option value="1" selected="selected"> Type 1 </option>
                    <option value="2"> Type 2 </option>
                    <option value="3"> Type 3 </option>
                </select>
     <button id="update-item-info">Save Changes</button>
    </form>

Upon Clicking of the button, I call a php script to update the Item in question via AJAX.

Javascript

$('#update-item-info').on('click', function (e)
{
    e.preventDefault();
    console.log("Updating Item Info...");
    $.ajax({
        type: "POST",
        url: "AJAX/save-item-info.php",
        data: $('#item-info-form').serialize(),
        dataType: "application/JSON"
    }).done(function (data)
    {
        console.log(data.status);
    });

});

The Server PHP that this code gets posted to is:

<?php

include("../db-connection.php"); 
$mysqli = new mysqli($mysqli_host, $mysqli_user, $mysqli_pass, $db_name);

if ($mysqli->connect_errno) 
{
    $response_array['status']  = "Connect failed: %s\n" . $mysqli->connect_error;
}

if(!($stmt = $mysqli->prepare("UPDATE items SET name = ?, start_date = ?, end_date = ?, location = ?, exhibit_type = ? WHERE id = ?")))
{
    $response_array['status']  = "Prepare Statement failed: " . $mysqli->error . "\n";
}

if (!($stmt->bind_param('sssssi', $_POST['name'], $_POST['start_date'], $_POST['end_date'], $_POST['location'], $_POST['exhibit_type'], $_POST['id'])))
{
    $response_array['status']  = "Binding parameters failed: " . $stmt->error . "\n";
}

if (!($stmt->execute())) 
{
    $response_array['status']  = "Execute failed: " . $stmt->error . "\n";
}
else
{
    $response_array['status']  = "Success";
}

$stmt->close();

header('Content-type: application/json');
die(json_encode($response_array));

?>

This does update the database, and if everything goes correctly, firebug tells me that data.status is set to "Success".

BUT if I comment out the database include (to simulate something going horribly wrong), the server code doesn't return anything at all, much less data.status, and I just get a generic 500 error.

Strangest of all, no matter whether the database is updated or not, ajax.done() isn't firing, so I can't see if data.status is ever being set.

Krish R · Accepted Answer · 2013-12-17 08:27:11Z

1

dataType (default: Intelligent Guess (xml, json, script, or html))

Use:

dataType: 'json',

http://api.jquery.com/jQuery.ajax/

Also, use

echo json_encode($response_array);

instead of

 die(json_encode($response_array));

answered Dec 17, 2013 at 8:27

Krish R

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3 Comments

Kevin Whiteside Over a year ago

That did it precisely, though .done still isn't returning anything if the update script fails. I tried adding a .fail method, but .fail(data) { console.log(data.status) } only outputs the Header of the error of the message (500).

Kevin Whiteside Over a year ago

Removing the header didn't work, but I managed to make the .fail method work with .fail(function(xhr) { console.log(xhr.responseJSON.status); });

Krish R Over a year ago

@Kevin Whiteside Great!!

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