JQuery Ajax returns Source code

Question

I'm using JQuery Ajax on my website. I've tested it on our test server it worked perfectly, but when I tried to put it on our productive server the Ajax just returns the source code of the website itself.

JS File:

$(document).ready(function () {
    $('#sshin').keyup(function () {
        var query = $(this).val();
        console.log(query);
        if (query != '') {
            $.ajax({
                url: "search.php",
                method: "POST",
                data: {
                    query: query
                },
                success: function (data) {

                    console.log(data);

                }
            });
        }
    });

    $(document).on('click', 'article', function () {
        var text = $('#sshin').val();
        text = text.substring(0, text.lastIndexOf(',') + 1);
        text += $(this).text();

        var uniq = text.match(/\b\w+\b/g).filter(function (el, idx, a) {
            return idx === a.lastIndexOf(el);
        });

        text = uniq.join(',');

        $('#sshin').val(text);
        $('#sshout').fadeOut();
    });
});

PHP File:

<?php
  if(isset($_POST["query"])){
     $query = $_POST["query"];
     return '<ul><li>1</li><li>2</li></ul>';
  }
?>

Any idea why it returns something different than it should?

What do you mean by "source code of the website itself"? Can you post an example bit? — Pekka
– Pekka, Commented Jan 12, 2017 at 11:53
Nothing wrong with ajax. see the implementation of search.php — Jyothi Babu Araja
– Jyothi Babu Araja, Commented Jan 12, 2017 at 11:53
Is it any possibility that you forget to use a parser of your PHP files? — Chu Ya'an Xiaonan
– Chu Ya'an Xiaonan, Commented Jan 12, 2017 at 11:54

XAF · Accepted Answer · 2017-01-12 12:55:29Z

0

This method once worked for me, hope it might help. Let me know.

 this->output->set_content_type('application/json');
        return $this->output->set_output(json_encode('<ul><li>1</li><li>2</li></ul>'));

answered Jan 12, 2017 at 12:55

XAF

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2 Comments

XAF Over a year ago

@Damon just remove the return '<ul><li>1</li><li>2</li></ul>'; of your code. and then paste this and check what console.log(data) shows.

Damon Over a year ago

removed this and it works, but still returns the source code.. maybe it helps i found this: https://myserver.ch/search.php?query=test,etc when you go into inspect in chrome and then on network..

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