No data receive in Jquery from php json_encode

Question

I need help for my code as i have been browsing the internet looking for the answer for my problem but still can get the answer that can solve my problem. I am kind of new using AJAX. I want to display data from json_encode in php file to my AJAX so that the AJAX can pass it to the textbox in the HTML.

My problem is Json_encode in php file have data from the query in json format but when i pass it to ajax success, function(users) is empty. Console.log also empty array. I have tried use JSON.parse but still i got something wrong in my code as the users itself is empty. Please any help would be much appreciated. Thank you.

car_detail.js

    $(document).ready(function() {


        function $_GET(q,s) {
            s = (s) ? s : window.location.search;
            var re = new RegExp('&amp;'+q+'=([^&amp;]*)','i');
            return (s=s.replace(/^\?/,'&amp;').match(re)) ?s=s[1] :s='';
        }
            var car_rent_id1 = $_GET('car_rent_id');
            car_rent_id.value = car_rent_id1;

                    $.ajax({  
        type: 'POST',
        url: "http://localhost/ProjekCordova/mobile_Rentacar/www/php/car_detail.php",
        dataType: "json",
        cache: false, 
        data: { car_rent_id: this.car_rent_id1 },
        success:  function(users) {

        console.log(users);
        $('#car_name').val(users.car_name); 

        }
    });
    });

car_detail.php

    $car_rent_id = $_GET['car_rent_id'];   
    $query = mysql_query("SELECT c.car_name, c.car_type, c.car_colour, 
    c.plate_no, c.rate_car_hour, c.rate_car_day, c.car_status, 
    r.pickup_location
    FROM car_rent c
    JOIN rental r ON c.car_rent_id=r.car_rent_id
    WHERE c.car_rent_id = $car_rent_id");

    $users = array();
      while($r = mysql_fetch_array($query)){
    $user = array(
        "car_name" => $r['car_name'], 
        "car_type" => $r['car_type'], 
        "car_colour" => $r['car_colour'], 
        "plate_no" => $r['plate_no'], 
        "rate_car_hour" => $r['rate_car_hour'], 
        "rate_car_day" => $r['rate_car_day'], 
        "car_status" => $r['car_status'], 
        "pickup_location" => $r['pickup_location']
        );
         $users[] = $user;
    // print_r($r);die;
    }
    print_r(json_encode($users)); //[{"car_name":"Saga","car_type":"Proton","car_colour":"Merah","plate_no":"WA2920C","rate_car_hour":"8","rate_car_day":"0","car_status":"","pickup_location":""}]

car_detail.html

  <label>ID:</label>
                <input type="text" name="car_rent_id" id="car_rent_id"><br>

                <label>Car Name:</label>
                <div class = "input-group input-group-sm">
                    <span class = "input-group-addon" id="sizing-addon3"></span>
                    <input type = "text" name="car_name" id="car_name" class = "form-control" placeholder = "Car Name" aria-describedby = "sizing-addon3">
                </div></br>

                <label>Car Type:</label>
                <div class = "input-group input-group-sm">
                    <span class = "input-group-addon" id="sizing-addon3"></span>
                    <input type = "text" name="car_type" id="car_type" class = "form-control" placeholder = "Car Type" aria-describedby = "sizing-addon3">
                </div></br>

Duplicating this line data: { car_rent_id: this.car_rent_id1 }, wont help! — RiggsFolly
– RiggsFolly, Commented May 16, 2017 at 16:16
Also, you've got ); after your success() method that doesn't look like it should be there. — Rwd
– Rwd, Commented May 16, 2017 at 16:19
You forgot to close the $.ajax(), your code is full of syntax errors — Junius L
– Junius L, Commented May 16, 2017 at 16:22
@RiggsFolly. Thank you for the reply. Actually i have use that line to get an id from the url, pass to the php to make a query. I am using GET method for that purpose. Then the query value will pass back to the json to be display in the textbox in html. Is my way wrong? — Luqman305
– Luqman305, Commented May 16, 2017 at 16:24
@Luqman305 learn to accept answers, I have seen your passed questions you don't accept answers, when you feel that an answer has solved your problem, accept it as it will help other people in the future. — Junius L
– Junius L, Commented May 16, 2017 at 17:29

Junius L · Accepted Answer · 2017-05-16 17:21:52Z

Remove this in this.car_rent_id1 and cache: false this works with HEAD and GET, in your AJAX you are using POST but in your PHP you use $_GET. And car_rent_id is not defined, your function $_GET(q,s) requires two parameters and only one is passed.

$(document).ready(function() {
    function $_GET(q,s) {
        s = (s) ? s : window.location.search;
        var re = new RegExp('&amp;'+q+'=([^&amp;]*)','i');
        return (s=s.replace(/^\?/,'&amp;').match(re)) ?s=s[1] :s='';
    }
    var car_rent_id1 = $_GET('car_rent_id'); // missing parameter
    car_rent_id.value = car_rent_id1; // where was this declared?

    $.ajax({
        type: 'POST',
        url: "http://localhost/ProjekCordova/mobile_Rentacar/www/php/car_detail.php",
        dataType: "json",
        data: { car_rent_id: car_rent_id1 },
        success:  function(users) {
            console.log(users);
            $('#car_name').val(users.car_name);
        }
    });
});

You can also use $.post(), post is just a shorthand for $.ajax()

$(document).ready(function() {
    function $_GET(q,s) {
        s = (s) ? s : window.location.search;
        var re = new RegExp('&amp;'+q+'=([^&amp;]*)','i');
        return (s=s.replace(/^\?/,'&amp;').match(re)) ?s=s[1] :s='';
    }
    var car_rent_id1 = $_GET('car_rent_id');
    car_rent_id.value = car_rent_id1;

    $.post('http://localhost/ProjekCordova/mobile_Rentacar/www/php/car_detail.php', { car_rent_id: car_rent_id1 }, function (users) {
        console.log(users);
        $('#car_name').val(users.car_name);
    });
});

and in your PHP change

$car_rent_id = $_GET['car_rent_id'];

to

$car_rent_id = $_POST['car_rent_id'];

thankssss!!! I have change according to your suggestion and it works now.

pendo · Accepted Answer · 2017-05-16 17:15:38Z

0

Here is a code skeleton using .done/.fail/.always

<script
  src="https://code.jquery.com/jquery-1.12.4.min.js"
  integrity="sha256-ZosEbRLbNQzLpnKIkEdrPv7lOy9C27hHQ+Xp8a4MxAQ="
  crossorigin="anonymous"></script>

<script>
$(function(){
  $.ajax({
    url: 'theurl',
    dataType: 'json',
    cache: false
  }).done(function(data){
    console.log(data);
  }).fail(function(data){
    console.log(data);
  }).always(function(data){
    console.log(data);
  });
});
</script>

I've adapted your code, so you can see the error, replace the ajax call with this one

<script>
    $.ajax({
      url: "theurl",
      dataType: "json",
      data: { car_rent_id: car_rent_id1 },
      success: function(users) {
        console.log(users);
        $('#car_name').val(users.car_name);
      },
      error: function(data) {
        console.log(data);
        alert("I failed, even though the server is giving a 200 response header, I can't read your json.");
      }
    });
</script>

edited May 16, 2017 at 17:15

answered May 16, 2017 at 16:40

pendo

8522 gold badges8 silver badges28 bronze badges

3 Comments

Junius L Over a year ago

There's difference between $.ajax({}).success(function(){...}); and $.ajax({success: fun...})

pendo Over a year ago

Thank you. I'm aware of that. I'm giving him the non-deprecated code skeleton to build from.

pendo Over a year ago

I just googled this and you're correct. Thank you. stackoverflow.com/questions/15821141/…

Arturo P · Accepted Answer · 2017-05-17 01:05:24Z

0

A couple of recommendations on this, I would follow jQuery API to try an see where the request is failing http://api.jquery.com/jquery.ajax/. Also, I would access the ids for the input fileds with jQuery. e.g.: $("#theID").val().

edited May 17, 2017 at 1:05

user4447799

answered May 16, 2017 at 16:45

Arturo P

161 bronze badge

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