Question

真正注意到下面神秘的大师的帮助。

我在html中使用了getJSON()。

只有硬编码数组可以是json_encode（即通过设置$DEBUG = true :)并传递给javascript，然后浏览器显示结果。但是从mysql生成文本时失败（通过设置$DEBUG = false）。

我正在挠头让mysql生成的动态数组工作？我可以在浏览器中运行两个场景，在浏览器中使用JSON格式化文本输出，即http://www.example.com/phpTWLLT/json_encoded_array.php。

如果$DEBUG设置为true，

从localhost / phpTWLLT / json_encode_array.php输出

[{"active":"0","first_name":"Darian","last_name":"Brown","age":"28","email":"darianbr@example.com"},{"active":"1","first_name":"John","last_name":"Doe","age":"47","email":"john_doe@example.com"}]

浏览器中显示的列表。 0 1

如果$DEBUG设置为false，

从localhost / phpTWLLT / json_encode_array.php输出

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

浏览器显示为空白。

html文件：

<!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脚本：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" => "darianbr@example.com"
        ),
        array(
            "active" => "1",
            "first_name" => "John",
            "last_name" => "Doe",
            "age" => "47",
            "email" => "john_doe@example.com"
        ) 
    );
} 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);
?>

Answer 1

您需要在PHP脚本中添加标题以将其输出为json：json_encoded_array.php

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

默认情况下，PHP将返回text/html，这不是$.getJSON()的有效JSON

JSON文本的MIME媒体类型是application / json。默认编码为UTF-8。（来源：RFC 4627）。

Answer 2

$.each使用javascript数组或对象循环遍历not a json string首先需要使用

解析json字符串

jQuery.parseJSON()

所以你的代码看起来像这样

data = jQuery.parseJSON(data);

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

Answer 3

发现Netbeans的默认许可标头（自动生成）会阻止Javascript识别JSON结构。

通过将默认许可证标题修改为空白来删除默认许可证标题后。 PHP脚本的输出仅包含JSON结构。浏览器显示正确。

请比较下面的两个调试输出（无法发布图像） debugFalse4.jpg：http://www.arthurmak.hk/debugFalse4.jpg debugFalseWrking.jpg：http://www.arthurmak.hk/debugFalseWorking.jpg

非常感谢Saquieb展示如何获取调试信息!!!

PHP脚本json_encode mysql请求无法通过getJSON（）

3 个答案: