PHP脚本json_en code MySQL的请求无法传递到的getJSON（） [英] PHP script json_encode mysql request cannot pass through to getJSON()
问题描述
真的注意，从大师的帮助在下面的神秘面纱。
我用的getJSON（）在我的HTML。
只有硬codeD数组可以json_en code（通过设置即 $ DEBUG = TRUE :)，并传递给JavaScript和浏览器随后显示结果。但是，当失败产生从MySQL文本（通过设置 $ DEBUG = FALSE ）。
我抓我的头获得MySQL生成的动态数组工作？我可以在浏览器中JSON格式化文本输出浏览器同时运行的情况下，即 HTTP ：//www.example.com/phpTWLLT/json_en$c$cd_array.php
如果 $ DEBUG 设置真正，
从本地主机/ phpTWLLT / json_en code_array.php
输出<$p$p><$c$c>[{\"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\"}]在浏览器中显示的列表。
0
1
如果 $ DEBUG 设置假，
从本地主机/ phpTWLLT / json_en code_array.php
输出 [{积极：1}，{积极：1}]
浏览器显示为空白。
HTML文件：
＆LT;！DOCTYPE HTML＆GT;
＆LT;！ -
为了改变这种许可证标头，选择项目属性许可头。
要改变这种模板文件，选择工具|模板
并打开编辑器的模板。
- ＆GT;
＆LT; HTML和GT;
＆LT; HEAD＆GT;
＆LT;！ -
＆LT;脚本类型=文/ JavaScript的SRC =HTTP：//ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js'> ＆LT; / SCRIPT＆GT;
- ＆GT;
＆LT;脚本类型=文/ JavaScript的'SRC ='JS / jquery.min.js'＆GT;＆LT; / SCRIPT＆GT; ＆LT;间的charset =UTF-8＆GT;
＆LT; /头＆GT;
＆LT;身体GT; ＆LT;！ - 这UL将与来自PHP数组中的数据进行填充 - ＆GT;
＆LT; UL＆GT;＆LT; / UL＆GT; ＆LT;脚本类型=文/ JavaScript的'＆GT;
$（文件）。就绪（函数（）{
/ *调用具有PHP数组是json_en codeD *的PHP /
//$.getJSON('json_en$c$cd_array.php'功能（数据）{
$ .getJSON（'json_en coded_array.php'功能（数据）{
/ *数据将持有PHP数组作为一个JavaScript对象* / $。每个（数据，功能（键，VAL）{ $（UL）追加。（'＆LT;李ID =+按键+'＆GT;'+ val.active +'＆LT; /李＆GT;'）;
}）; }）;
}）;
＆LT; / SCRIPT＆GT; ＆LT; /身体GT;
＆LT; / HTML＆GT;
PHP脚本：json_en coded_array.php
＆LT; PHP/ *
*要改变这种许可证标头，选择项目属性许可头。
*要改变这个模板文件，选择工具|模板
*并打开编辑器的模板。
* // *载文档类型为text / JavaScript的，而不是text / html的* /
$ DEBUG = TRUE;如果（$ DEBUG）{
标题（内容类型：text / JavaScript的）;
$ ARR =阵列（
阵列（
活动= GT; 0，
FIRST_NAME=＆GT; 达里安
姓氏=＆GT; 棕色，
时代=＆GT; 28，
电子邮件=＆GT; darianbr@example.com
）
阵列（
活动= GT; 1，
FIRST_NAME=＆GT; 约翰，
姓氏=＆GT; 李四，
时代=＆GT; 47，
电子邮件=＆GT; john_doe@example.com
）
）;
}其他{
require_once（'connection.php'）;
// $ m_id = 8有许多就读课程，并得到了11正好一门课程就读。
$ m_id = 8;
$ P_ID = 1; $ qry1 =SELECT DISTINCT event.active为活动，主题。code作为当然code'，subject.name为课程名，event.event_desc为eventDesc'FROM申请人，事件，主题，部分其中，applicant.applicant_id = $ m_id和applicant.event_id = event.id和event.subject_id = subject.id和part.id = subject.owner_id和part.id = $ P_ID由event.active DESC，event.from_month DESC命令;
mysqli_set_charset（$ BD，'utf-8'）;
$结果= mysqli_query（$ BD，$ qry1）; $ ARR =阵列（）;
$ I = 0;
如果（mysqli_num_rows（$结果）大于0）{
而（$ RS = mysqli_fetch_assoc（$结果））{
$ colhead =主动;
$海峡= $ RS ['主动']; $常用3 [$ i] =阵列（$ colhead =＆GT; $海峡）;
$ I ++; //只是生成两个记录进行测试
如果（$ I === 2）
打破; }
}
}
回声json_en code（$ ARR）;
？＆GT;
您需要一个头在PHP脚本添加到输出为 JSON ：json_en coded_array .PHP
标题（内容类型：应用程序/ JSON）;
默认情况下
PHP将返回的text / html 这是不是 $有效的JSON。的getJSON（）
有关JSON文本的MIME媒体类型是application / JSON。默认的编码是UTF-8。 （来源： RFC 4627 ）
。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":"darianbr@example.com"},{"active":"1","first_name":"John","last_name":"Doe","age":"47","email":"john_doe@example.com"}]
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" => "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);
?>
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).
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