Uncaught SyntaxError: Unexpected token < in JSON at position 0 at Function.parse [as parseJSON] () at Object.success (script.js:14)

Question

i am tying to search user search from database database include varchar and image every time i type in search box this error appears in console Uncaught SyntaxError: Unexpected token < in JSON at position 0

Here is my js code

$(function(){
    $('#q').keyup(function(){
        var q = $(this).val();
        var html = '';
        $('.form-wrapper').find('.result-wrapper').remove();

        if (q !== '' && q !== null) {

            $.ajax({
                url: 'suggestions.php',
                data: {q: q},
                success: function (response){
                    /*var reader = new FileReader();*/
                    response = $.parseJSON(response);

                    html = generateDOM(response);
                    $('.form-wrapper').append(html);

                },
                error: function(response){
                    console.log(response);
                }
            });

        }

    });
});

function generateDOM(response){
    var html = '<div class="result-wrapper"/>';
    html += '<ul>';
    $.each(response, function(index, value){
        html += '<li><img src="'+value.q_links+'"/><a href="'+value.q_links+'">'+value.q_title+'</a></li>';
    });

    html += '</ul>';
    html += '</div>';

    return html;
}

and connection to db code

<?php

function connect(){
    $connection = mysqli_connect('localhost', 'root', '', 'img');
    if (!$connection){
        die('Error: Failed to connect DB!!');
    }
    return $connection;
}

function get_suggestions( $q ){
    $connection = connect();
    $sql = "SELECT `id`, `q_image`, `q_title`, `q_links` FROM `users` WHERE (`q_title` LIKE '%$q%')";
    $result = mysqli_query($connection, $sql);
    if ( mysqli_num_rows($result) ){
        return mysqli_fetch_all($result, MYSQLI_ASSOC);
    }
    return false;
}

function debug($arg){
    echo '<pre>';
    print_r($arg);
    echo '</pre>';
    exit;
}

?>

please help me to solve this problem

error screenshot

https://drive.google.com/file/d/1S72_AmxILPZHA7nZZ0g4ZZhx0OxD9wjp/view?usp=sharing

This will work `return json_encode(mysqli_fetch_all($result, MYSQLI_ASSOC));` — Rohit Rasela, Dec 16 '18 at 17:28
If you are calling `debug()`, this will put some HTML tags out and will cause the error. — Nigel Ren, Dec 16 '18 at 17:35
If that PHP script is `suggestions.php` then you need some main line code (not a function) to call one of these functions. They will not call themselves — RiggsFolly, Dec 16 '18 at 17:45
@ Rohit Rasela bro a new error occurred Uncaught SyntaxError: Unexpected token < in JSON at position 25 — Kamran Siddique, Dec 16 '18 at 17:46
Add new key `dataType:"json"` in your ajax function object. And remove `response = $.parseJSON(response);`. also write `return json_encode(mysqli_fetch_all($result, MYSQLI_ASSOC));` in php then try — Rohit Rasela, Dec 16 '18 at 17:57
@RohitRasela this now this error came https://drive.google.com/file/d/1jyDWFMQ9rGeYtHswNp19wELdcdZAzkvf/view?usp=sharing look like i am a error man — Kamran Siddique, Dec 16 '18 at 18:04
Check answer and try to change your function according this. — Rohit Rasela, Dec 16 '18 at 18:08
Also check `console.log(response);` what the out of this inside success. — Rohit Rasela, Dec 16 '18 at 18:10
If you are using you debug function somewhere to print data you will not be able to get valid response. — Rohit Rasela, Dec 16 '18 at 18:13
@RohitRasela actually i'm making an instant search result search box which load title along with images and show to the user — Kamran Siddique, Dec 16 '18 at 18:14
Possible duplicate of ["SyntaxError: Unexpected token < in JSON at position 0"](https://stackoverflow.com/questions/37280274/syntaxerror-unexpected-token-in-json-at-position-0-in-react-app) — miken32, Dec 18 '18 at 18:26

score 1 · Accepted Answer · answered Dec 16 '18 at 18:08

Change your php function to

function get_suggestions( $q ){
    $connection = connect();
    $sql = "SELECT `id`, `q_image`, `q_title`, `q_links` FROM `users` WHERE (`q_title` LIKE '%$q%')";
    $result = mysqli_query($connection, $sql);
    if ( mysqli_num_rows($result) ){
        return json_encode(mysqli_fetch_all($result, MYSQLI_ASSOC));
    }
    return false;
}

And you ajax code to

$(function(){
    $('#q').keyup(function(){
        var q = $(this).val();
        var html = '';
        $('.form-wrapper').find('.result-wrapper').remove();

        if (q !== '' && q !== null) {

            $.ajax({
                url: 'suggestions.php',
                data: {q: q},
                dataType: 'json',
                success: function (response){
                    /*var reader = new FileReader();*/
                    /*response = $.parseJSON(response);*/

                    html = generateDOM(response);
                    $('.form-wrapper').append(html);

                },
                error: function(response){
                    console.log(response);
                }
            });

        }

    });

Uncaught SyntaxError: Unexpected token < in JSON at position 0 at Function.parse [as parseJSON] () at Object.success (script.js:14)

1 Answers1