【发布时间】：2015-03-09 22:07:48
【问题描述】：
我正在按照一些教程从服务器传递一个 JSON 文本文件，以在对 html 文件进行一些 javascript 处理后显示数据。作为测试，尝试显示一列的 LI，但无法在浏览器中获得任何输出。感谢您的帮助。
我尝试了两种方法。
方法一 xmlhttp:
显然，浏览器抱怨 html 格式： 未捕获的 SyntaxError：意外字符串（15:08:42:080 | 错误，javascript） 在 testJSON3.html:12
我的 xmlhttp 调用格式是否正确？
提前感谢您的帮助。
这是 JSON 文本 myTutorial.txt：
[
{
"active":"1",
"courseCode":"208.01.00",
"courseName":"course name 1",
"eventDesc":"2015 class of course name 1"
},
{
"active":"1",
"courseCode":"208.01.00",
"courseName":"course name21",
"eventDesc":"2015 class of course name "
}
]
并通过下面的html来处理xmlhttp访问服务器localhost目录phpTWLLT上的文件
<!DOCTYPE html>
<html>
<head>
<script type='text/javascript' src='js/jquery.min.js'></script>
<meta charset="UTF-8">
</head>
<body>
<div id="id01"></div>
<script>
var xmlhttp = new XMLHttpRequest();
var url = "http://localhost/phpTWLLT/myTutorial.txt";
xmlhttp.onreadystatechange = function () {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
var myArr = JSON.parse(xmlhttp.responseText);
myFunction(myArr);
}
};
xmlhttp.open("GET", url, true);
xmlhttp.send();
function myFunction(arr) {
var out = "";
var i;
for (i = 0; i < arr.length; i++) {
out += '<li'> + arr[i].courseCode +'</li><br>';
}
document.getElementById("id01").innerHTML = out;
}
</script>
</body>
</html>
方法2 getJSON()：
这个很有趣。如果服务器端是静态数组（$DEBUG = true:)，javascript 能够处理并获取浏览器显示。但是从 mysql ($DEBUG = false) 生成文本时失败。
我正在为 $DEBUG=false 工作而摸不着头脑？显然，这两种情况都生成了有效的 JSON 文本。
如果 $DEBUG 设置为真，
来自 localhost/phpTWLLT/json_encode_array.php 的输出
[{"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"}]
在浏览器中显示的列表。 0 1
如果 $DEBUG 设置为 false，
来自 localhost/phpTWLLT/json_encode_array.php 的输出
[{"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);
?>
【问题讨论】：
-
'<li'>应该是'<li>'，请经常检查您的语法 -
@Patrick，谢谢你的建议。但是，浏览器中仍然没有显示列表。
-
您是否遇到完全相同的错误，可能是不同的错误，或者根本没有？
标签： javascript php mysql json xmlhttprequest