jquery post .fail even when php succeed

Question

I have a problem with my jQuery post request :

$.post(
  'http://localhost/***/ajax_bdd-change.php',
    {'id': _id, 'id_key': id_key, 'table': table, 'data': data})
    .fail(function(jqXHR, textStatus, errorThrown){
      alert('Erreur: '+jqXHR.responseText);
    })
    .done(function(data){
      alert($(data).text());
    });

And my PHP :

<?php
$id     = json_decode($_POST['id']);
$id_key = json_decode($_POST['id_key']);
$table  = json_decode($_POST['table']);
$data   = json_decode($_POST['data']);

foreach ($_POST as $k=>$v) {
  unset($_POST[$k]);
}
$rlt = array(
  'erreur' => false,
  'request' => 'none'
  ); 
$tmp = 0;
$request = 'UPDATE '.$table.' SET';

foreach ($data as $target => $value) {
  if ($tmp++>0)
    $request = $request.',';
  $request = $request.' '.$target.' = "'.$value.'"';
}
$request = $request.' WHERE '.$id_key.' LIKE "'.$id.'"';

$rlt['request'] = $request;

require('BDD_connexion.php');
if (!$rlt_bdd = mysqli_query($link, $request)){
  $rlt['erreur'] = 'Erreur: Update not done';
}
$link->close();

echo json_encode($rlt);
exit();

?>

Everytime I run my code, it follow the same path :

• PHP is correctly executed
• jQuery run .fail()
  • jqXHR.responseText is empty

I have try to force php to fail and at that time, the jQuery correctly run the done(function).

• PHP have some error
• jQuery run .done()
  • the alert show the php error

I have try many thing like force an UTF8 encode to each php string variable. I even try to impose a simple string like json_encode('hello world');

After many test, it seem my previous informations :

Maybe it is useful to explain that:

• my javascript is inside a laod() php page.

So it must have a structure like:

• main.php --jQuery-->load(second.php into a div)
  • second.php --jQuery-->$.post(ajax_bdd-change.php)
  • ajax_bdd-change.php --return $rlt -->second.php(jQuery part)

I do not mention it because I do not find it pertinent.

Is the cause of this problem. I have try a call of my php by post from a new html page without a .load and it is working perfectly.

Answer 1

The response code, if nothing bad occurred on the server, should be 200.

It's highly likely, based off the observations you've made, that the response code is something other than 200. Also note, jQuery, or any other framework, doesn't know whether custom code, written on the server, was executed coherently. Usually, the only indication to the client is the response code.

jQuery source

Answer 2

I have find a resolution. In fact, my code is more complex, and the jQuery .post is done in a .submit of a <form> inside the script part of an jQuery DialogBox. So the form was finished to submit before the php response when this on have a Database Query.

In resolution I change this submit with just a .click, and I manualy close when I recive a response from the server.

Answer 3

First you have to decide that you want JSON in you AJAX Query like so:

    $.ajax({
        url:        'http://localhost/***/ajax_bdd-change.php',
        dataType:   'json',     //EXPLICITLY SET THIS TO JSON
        cache:      false,
        type:       "POST",
        data:       {
            id          : _id,
            id_key      : id_key,
            table       : table,
            data        : data
        },

        success: function (data, textStatus, jqXHR) {
            if(data){
                alert("DATA CAME BACK AS JSON...");
                console.log(data);
                // YOU REALLY DON'T WANT TO DO THIS: $(data).text() 
                // THE .text() METHOD IS FOR THE DOM AND YOU ARE IN READING MODE TOO
                // data SHOULD CONTAIN THESE: request AND erreur
                // SO YOU CAN EITHER CHECK THE SQL OR THE ERROR LIKE SO:
                // I WOULD PREFER USING console.log(data.request)
                alert(data.request);
                alert(data.erreur);
            }
        },

        error: function (jqXHR, textStatus, errorThrown) {
            alert('Erreur: '+jqXHR.responseText);
        },

        complete: function (jqXHR, textStatus) {
            //DO SOMETHING HERE IF YOU WISH TO
        } 
    });

And this is your PHP Code:

    <?php
        $id     = isset($_POST['id'])       ? htmlspecialchars(trim($_POST['id']))      : null;
        $id_key = isset($_POST['id_key'])   ? htmlspecialchars(trim($_POST['id_key']))  : null;
        $table  = isset($_POST['table'])    ? htmlspecialchars(trim($_POST['table']))   : null;
        $data   = isset($_POST['data'])     ? htmlspecialchars(trim($_POST['data']))    : null;


        foreach ($_POST as $k=>$v) {
            unset($_POST[$k]);
        }

        $rlt    = array(
            'erreur'    => false,
            'request'   => 'none'
        );

        $updateSQL  = 'UPDATE '. $table . ' SET ';
        if($data){
            foreach ($data as $target => $value) {
                $updateSQL .= ' ' . $target . ' = "' . $value . '", ';
            }
            $updateSQL  = rtrim($updateSQL, ", ");
            // LIKE CLAUSE IS NOT SO FUNKY HERE: DO YOU WANT TO DO "EQUALS"?
            // ' WHERE ' . $id_key . ' = "' . $id. '"';
            $condition  = ' WHERE ' . $id_key . ' LIKE "' . $id. '"';
            $request    = $updateSQL . $condition;

            $rlt['request'] = $request; 
            require('BDD_connexion.php');
            if (!$rlt_bdd = mysqli_query($link, $request)){
                $rlt['erreur'] = 'Erreur: Update not done';
            }
            $link->close();
        }

        die( json_encode($rlt) );

Hope this is a little close...

Answer 4

Try this:

// Javascript

var params = {
    id: _id,
    id_key: id_key,
    table: table,
    data: data
};

// explicitely tells javascript you're expecting json response
// with this $.post shortcut

$.post('url', params, function(data) {
    //console.log(data)
    if (data.erreur) {
        // Error occured
    } else {
        // no error
    }
}, 'json');

// PHP

<?php

// You don't need to json_decode the $_POST data, but you can sanitize or
// perform any validation check if you want

$id = $_POST['id'];
$id_key = $_POST['id_key'];
$table = $_POST['table'];
$data = $_POST['data'];

// Your process here

echo json_encode($rlt);
exit;

Hope this helps.