我的JSON列表中有多个项目。我想循环遍历它并在我的页面上显示它。我似乎无法进入下一个对象。
{
"room":[
{"campusName":"A",
"buildingCode":"B",
"roomNumber":"208",
"times":["7-8", "9-10"]
}],
"room2":[
{"campusName":"C",
"buildingCode":"D",
"roomNumber":"208",
"times":["7-8", "9-10"
]}
]}
$(document).ready(function(){
$.getJSON("data.json", function(data){
$.each(data.room, function(){
for(var i = 0; i < data.length; i++){
$("ul").append("<li>campus: "+this['campusName']+"</li><li>building: "+this['buildingCode']+"</li><li>times: "+this.times+"</li>");
}
});
});
});
答案 0 :(得分:3)
试试这个
var list = '';
$.each(data, function (i, root) {
$.each(root, function (i, el) {
list += "<li>campus: " + this.campusName + "</li><li>building: " + this.buildingCode + "</li><li>times: " + this.times.join(' ') + "</li>";
});
});
$('ul').html(list);
如果root's
在数组中只有一个元素
var list = '';
$.each(data, function (i, root) {
list += "<li>campus: " + root[0].campusName + "</li><li>building: " + root[0].buildingCode + "</li><li>times: " + root[0].times.join(' ') + "</li>";
});
$('ul').html(list);
答案 1 :(得分:2)
$.each(data, ..)
- &gt;每个元素都是:
"room":[
{"campusName":"A",
"buildingCode":"B",
"roomNumber":"208",
"times":["7-8", "9-10"]
}]
然后,this[0]
将提供构建li
:
$.each(data, function(){
$("ul").append("<li>campus: "+this[0]['campusName']+"</li><li>building: "+this[0]['buildingCode']+"</li><li>times: "+this[0].times+"</li>");
});