我有一个脚本,它读取Json字符串的内容并创建一个表来显示数据。只要我在JQuery函数中包含Json字符串,这工作正常。我需要做的是调用另一个php文件,它返回Json字符串。Json从json读取数据调用表
我写了返回JSON字符串的脚本:
[{"ClientName":"Name","RoomName":"Room 2","RoomFromTime":"06:00:00","RoomToTime":"17:00:00"},{"ClientName":"Name","RoomName":"Room 6","RoomFromTime":"06:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 1","RoomFromTime":"07:00:00","RoomToTime":"17:00:00"},{"ClientName":"Name","RoomName":"Room 14","RoomFromTime":"07:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 12","RoomFromTime":"07:00:00","RoomToTime":"19:00:00"},{"ClientName":"Name","RoomName":"Room 10","RoomFromTime":"07:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 9","RoomFromTime":"07:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 8","RoomFromTime":"07:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 7","RoomFromTime":"07:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 5","RoomFromTime":"07:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 3","RoomFromTime":"07:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 4","RoomFromTime":"08:00:00","RoomToTime":"23:00:00"},{"ClientName":"Name","RoomName":"Room 15","RoomFromTime":"08:00:00","RoomToTime":"19:00:00"}]
我JQuery的功能有以下几点:
$(document).ready(function() {
var json = $.getJSON("get_data.php", function(data){
var tr ;
for (var i = 0; i < json.length; i++) {
tr = $('<tr/>');
tr.append("<td><div class='clientname-text'>" + json[i].ClientName + "</div></td>");
tr.append("<td><div class='roomname-text'>" + json[i].RoomName + "</div></td>");
tr.append("<td><div class='time-text'>" + json[i].RoomFromTime + " - " + json[i].RoomToTime + "</div></td>");
$('table').append(tr);
}
});
});
使用呼叫转移到其他的PHP脚本不显示返回的数据在桌子里。现在我知道我在某个地方出了问题,但是anyoe可以看到我做错了什么。
非常感谢您的时间。
清楚,没有成才。现在我的表格行显示“Undefinded”这个词,为什么? – DCJones
做一个'console.dir(json);'来找出json中的内容。 – Adder
我想你应该从[json] [i] .ClientName中删除[i],然后写上json.ClientName – Zeljka