从服务器请求JSON并使用Javascript进行解析
[英]Requesting JSON from server and parsing in Javascript
问题描述
I am trying to pull json down from my server and parse it into a Javascript object. 
我正在尝试从服务器中提取json并将其解析为Javascript对象。 here is what my json looks like: 这是我的json的样子：
{
"tour1": [
{
"title": "building1",
"description": "Tour of building1",
"image": "Icon.png",
"video": "tour.mp4",
"length": "0.00",
"version": "1.0",
"timestamp": "1111111111"
}
]
}
Here is the requst to the server: 这是服务器的要求：
<!DOCTYPE html>
<html>
<body>
<h2>Parse JSON from Server</h2>
<p id="demo"></p>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"> </script>
<script>
$.ajax({
url: "mysite.com/videos/tour.json";
var j = [];
$.ajax({
type: 'GET',
url: url,
dataType: 'json',
});
window.alert(j.tour1[0].title);
</script>
</body>
</html>
I cant understand why its not working. 我不明白为什么它不起作用。 I am new to javascript. 我是javascript新手。 I appreciate any help with this issue. 感谢您提供有关此问题的帮助。
3 个解决方案
解决方案1
3 已采纳 2015-12-03 09:28:22
I think better if you use getJson when you want to load JSON-encoded data from the server using a GET HTTP request, check example bellow : 我想如果要使用GET HTTP请求从服务器加载JSON编码的数据时使用getJson更好，请查看以下示例：
$.getJSON( "mysite.com/videos/tour.json", function( j ) {
alert( j.tour1[0].title );
});
Hope this helps. 希望这可以帮助。
解决方案2
1 2015-12-03 09:29:18
Ajax call is an asynchronous call, j is not populated as soon as your ajax statement ends. Ajax调用是一个异步调用，一旦您的ajax语句结束，就不会填充j。
Use success handler of Ajax to alert the j (didn't tested the code) 使用Ajax的成功处理程序来警告j（未测试代码）
$.ajax({
url: "mysite.com/videos/tour.json",
method: "GET",
dataType: "JSON",
success: function( data ){
window.alert(data.tour1[0].title);
}
});
解决方案3
0 2015-12-03 10:35:02
gurvinder give the best solution, because if you are firing the ajax-request then you must catch the response, but if you make the alert in your code your request is still working and j must be undefined because it is not a global variable. gurvinder提供了最佳解决方案，因为如果您触发ajax请求，那么您必须捕获响应，但是如果您在代码中发出警报，则您的请求仍然有效，并且j必须是未定义的，因为它不是全局变量。 there a lot of mistakes: 有很多错误：
- check your ajax-synthax , because the declaration of "var j.." is wrong, 检查您的ajax-synthax，因为“ var j ..”的声明是错误的，
- never use ";" 永远不要使用“;” in a object declaration, always use "," 在对象声明中，始终使用“，”
- in your alert you try to find a variable "j", but j don't exist, it is undefined. 在您的警报中，您尝试找到变量“ j”，但是j不存在，它是未定义的。
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