我转换为json并回显我想使用jquery进行检索的php数组。回显的json的格式如下:尝试解析json数组时未返回undefined
[{"id":"1","title":"Test Event 1","description":"This is the first test event","location":"Acme Hall, Room 101","contact":"John Smith","url":"http:\/\/www.example.com","start":"2016-02-29 13:00:00","end":"2016-02-29 14:00:00"},
{"id":"2","title":"Test Event 2","description":"This is the second test event","location":"Venable Hall, Room 101","contact":"Jane Smith","url":"http:\/\/www.example.com","start":"2016-03-08 09:00:00","end":"2016-03-10 10:45:00"},
{"id":"3","title":"Test Event 3","description":"This is the third test event","location":"Sitterson Hall, Room 200","contact":"Jane Smith","url":"http:\/\/www.example.com","start":"2016-03-18 15:00:00","end":"2016-02-22 16:30:00"}]
当我尝试使用Ajax GET来解析并显示JSON时,会返回'undefined'。我的Ajax代码是:
$.ajax({
type: 'GET',
url: 'get-events.php',
dataType: "json",
cache: false,
success: function (result) {
alert(result[start]);
},
});
UPDATE:
感谢您的答复。我终于得到它使用排序
for (var key in result) {
if (result.hasOwnProperty(key)) {
result[key].id
}
}
什么是变量'start'? – epascarello