PHP script json_encode mysql request cannot pass through to getJSON()

Question

Really note help from guru on the below mystery.

I used getJSON() in my html.

Only hardcoded array can be json_encode (i.e. by setting $DEBUG = true:) and pass to javascript and subsequently browser display the result. But fail when generate the text from mysql (by setting $DEBUG = false).

I am scratching my head to get mysql generated dynamic array to work? I can run both scenario in the browser with JSON formated text output in the browser, i.e http://www.example.com/phpTWLLT/json_encoded_array.php.

If $DEBUG is set true,

output from localhost/phpTWLLT/json_encode_array.php

[{"active":"0","first_name":"Darian","last_name":"Brown","age":"28","email":"[email protected]"},{"active":"1","first_name":"John","last_name":"Doe","age":"47","email":"[email protected]"}]

the list displayed in browser. 0 1

If $DEBUG is set false,

output from localhost/phpTWLLT/json_encode_array.php

[{"active":"1"},{"active":"1"}]

The browser display is blank.

html file:

<!DOCTYPE html>
<!--
To change this license header, choose License Headers in Project Properties.
To change this template file, choose Tools | Templates
and open the template in the editor.
-->
<html>
    <head>
        <!--
        <script type='text/javascript' src='http://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js'>    </script>
        -->
    <script type='text/javascript' src='js/jquery.min.js'></script>

    <meta charset="UTF-8">
    </head>    
    <body>

    <!-- this UL will be populated with the data from the php array -->
    <ul></ul>

    <script type='text/javascript'>
        $(document).ready(function () {
            /* call the php that has the php array which is json_encoded */
            //$.getJSON('json_encoded_array.php', function(data) { 
            $.getJSON('json_encoded_array.php', function (data) {
                /* data will hold the php array as a javascript object */

                 $.each(data, function (key, val) {

                 $('ul').append('<li id="' + key + '">' + val.active  + '</li>');
                 });

            });
        });
        </script>

    </body>
</html>

PHP script: json_encoded_array.php

<?php

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */



/* set out document type to text/javascript instead of text/html */


$DEBUG = true;

if ($DEBUG) {
    header("Content-type: text/javascript");
    $arr = array(
        array(
            "active" => "0",
            "first_name" => "Darian",
            "last_name" => "Brown",
            "age" => "28",
            "email" => "[email protected]"
        ),
        array(
            "active" => "1",
            "first_name" => "John",
            "last_name" => "Doe",
            "age" => "47",
            "email" => "[email protected]"
        ) 
    );
} else {
    require_once('connection.php');
// $m_id= 8 has many enrolled course and 11 got exactly one course enrolled. 
    $m_id = 8;
    $p_id = 1;

    $qry1 = "SELECT distinct event.active as active, subject.code as 'courseCode', subject.name as     'courseName', event.event_desc as 'eventDesc' FROM applicant, event, subject, part where applicant.applicant_id = $m_id and applicant.event_id = event.id and event.subject_id=subject.id and part.id = subject.owner_id and part.id = $p_id order by event.active DESC, event.from_month DESC ";
    mysqli_set_charset($bd, 'utf-8');
    $result = mysqli_query($bd, $qry1);

    $arr = array();
    $i = 0;
    if (mysqli_num_rows($result) > 0) {
        while ( $rs = mysqli_fetch_assoc($result)  ) {
             $colhead = "active";
            $str = $rs['active'];

            $arr[$i] = array($colhead => $str);
            $i++;

                // just generate two record for testing
                if ($i === 2)
                break;

        }
    } 
}
echo json_encode($arr);
?>

Saqueib · Accepted Answer · 2015-01-12 12:47:04Z

3

You need to add a header to output it as json in PHP script: json_encoded_array.php

 header("Content-type: application/json");

By default PHP will return text/html which is not a valid JSON for $.getJSON()

The MIME media type for JSON text is application/json. The default encoding is UTF-8. (Source: RFC 4627).

edited Jan 12, 2015 at 12:47

answered Jan 12, 2015 at 12:41

Saqueib

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13 Comments

klauskpm Over a year ago

He could even use $.ajax setting dataType: 'JSON'. Wouldn't even have to mind setting header on PHP, just returning a JSON string.

Saqueib Over a year ago

Ya that can be done, but its good practice to return valid header json with header("Content-type: application/json");

mak arthur Over a year ago

@Saqueib added the header, but no change to browser output

klauskpm Over a year ago

@makarthur: You need to set the header outside the if condition to work on both cases. And you are just passing "active" trough your secound condition, what do you want it to return? The same as the first?

Saqueib Over a year ago

@makarthur if you can share the console screenshot it will be great help

|

Kuldeep Dangi · Accepted Answer · 2015-01-12 12:47:57Z

1

$.each takes a javascript array or object to loop through not a json string first you need to parse your json string using

jQuery.parseJSON()

so your code will look like this

data = jQuery.parseJSON(data);

    $.each(data, function (key, val) {
        $('ul').append('<li id="' + key + '">' + val.active  + '</li>');
    });

answered Jan 12, 2015 at 12:47

Kuldeep Dangi

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2 Comments

mak arthur Over a year ago

tried the above code, neither $DEBUG=true nor $DEBUG=false produce output.

mak arthur Over a year ago

tried the above code, neither $DEBUG=true nor $DEBUG=false produce output. and output error Uncaught SyntaxError: Unexpected token o (22:06:01:876 | error, javascript) at testGetJson.html:1 >

mak arthur · Accepted Answer · 2015-01-12 17:18:50Z

1

Found that default License Header from Netbeans (which is automatically generated) prevented Javascript from recognizing the JSON structure.

After removing the default License Header by modifying the default License Header to blank. The output from PHP script only contains the JSON structure. The browser display correctly.

Please compare the two debug output below (Cannot post images) debugFalse4.jpg: http://www.arthurmak.hk/debugFalse4.jpg debugFalseWrking.jpg: http://www.arthurmak.hk/debugFalseWorking.jpg

Thanks a lot to Saquieb showing how to obtain the debug information!!!

edited Jan 12, 2015 at 17:18

answered Jan 12, 2015 at 17:02

mak arthur

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