php sending values to ajax faild error

Question

I Am trying to send value from ajax to php and retrieve it just to test that everything is work, when i click in a button to test i got error and alert('Failed') Appears , how can i fix it in order to get success? thanks

Ajax :

  var a = "test";
    $.ajax({
            url: "search.php",
            dataType: "json",
            data: a ,
            success: function(data) {
               alert('Successfully');
            }, 
            error: function(data) {
               alert('Failed');
           }
        })

PHP :

    <?php
        $pictures = "img1";
        echo json_encode($pictures); 
    ?>

Forgot type:'POST' or 'GET'.

StackSlave
– StackSlave

2017-04-03 23:49:34 +00:00
Commented Apr 3, 2017 at 23:49 — StackSlave
– StackSlave, Commented Apr 3, 2017 at 23:49

cosmoonot · Accepted Answer · 2017-04-03 23:15:07Z

1

I refined your code slightly and it works.

    var a = "test";
    $.ajax({
        type: 'POST',
        url: 'search.php',
        data: 'a=' + a,
        dataType: 'json',
        cache: false,
        success: function (result) {
          alert('Successful');  
        },
        error: function (result) {
          alert('Failed');
        }
    });

answered Apr 3, 2017 at 23:15

cosmoonot

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2 Comments

John_P Over a year ago

thanks, but i got the same error i replaced also type: 'POST', with type: 'GET', still doesn't work

cosmoonot Over a year ago

Are you receiving any other errors? Check your browser console.

VaioWay · Accepted Answer · 2017-04-03 23:44:19Z

0

If you're requesting a JSON, use the $.getJSON from jQuery, it's aready parse the JSON into a JSON object for you.

Seems that you're not return an actual JSON from server, maybe this is what is causing the error.
If you're seeing the 'Failed' message probably the problem is a 500 error which is a server error.

Try this code above.

Javascript:

var a = "test";

$.getJSON("search.php", {
  a: a
}, function (json) {
  console.log(json);
});

PHP:

<?php
$pictures = ["img1"];
echo json_encode($pictures);

The only way to this not work, is if you have a huge mistake on you webserver configuration.

answered Apr 3, 2017 at 23:44

VaioWay

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Comments

Konstantinos · Accepted Answer · 2017-04-03 23:45:40Z

0

Your ajax is wrong, it should be:

var a = "test";
$.ajax({
    type: "POST",
    url: "search.php",
    dataType: "json",
    data: {a:a},
    success: function(data) {
        alert('Successfully');
    },
    error: function(data) {
        alert('Failed');
    }
});

answered Apr 3, 2017 at 23:45

Konstantinos

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