Question

我有一个表单，我收集数据，然后我将它们插入url字符串，因此我得到像http://terminal2.expedia.com/x/mhotels/search?city=hersbruck&checkInDate=2016-12-01&checkOutDate=2016-12-03&room1=2&apikey=RZjKNligNc0QZJvV9HP1nbFm1jrghgLK这样的网址此url直接在页面上返回有效的json

{
  numberOfRoomsRequested: 1,
  filterUnavailableHotelsRequested: false,
  totalHotelCount: 94,
  availableHotelCount: 60,
  searchRegionId: "6124915",
  searchRegionCity: "Hersbruck",
  deepLinkUrl: "https://www.expedia.com/Hersbruck-Hotels.d6124915.Travel-Guide-Hotels",
     hotelList: [
       {
     sortIndex: "0",
     hotelId: "2813073",
     name: "Md Hotel Restaurant Bauer",
     localizedName: "Md Hotel Restaurant Bauer",
     nonLocalizedName: "Md Hotel Restaurant Bauer",
     address: "Martin Luther Str 16",
     city: "Hersbruck",
     ........

我尝试从中获取一些数据来创建表，使用这样的代码：

$(document).ready(function(){

   $('#submit-btn').click(function(s) {
    $("form").submit(function (event) {
        var place = (document.querySelectorAll('#place')[0].value);
        var check_in = (document.querySelectorAll('#check-in')[0].value);
        var check_out = (document.querySelectorAll('#check-out')[0].value);
        var api_key = "RZjKNligNc0QZJvV9HP1nbFm1jrghgLK";
        var url = "http://terminal2.expedia.com/x/mhotels/search?city=" + place + "&checkInDate=" + check_in + "&checkOutDate=" + check_out + "&room1=2&apikey=" + api_key + ""

       $.getJSON(url, function (data){
          var div = $("#result");
          var table = '<table><tbody>';
            for (i in data) {
                var tr = '<tr><td>'+data[i].name+'</td><td>'+data[i].city+'</td></tr>'
                table = table+tr;
           }
            table = table + "</tbody></table>";
            div.html(table);
            });
        event.preventDefault();
    });
  });
});

但我收到了undefined的表格我做错了，请指出我。

Answer 1

我猜你想要显示API返回的hotelList酒店。那么你的循环不会更像：

for (hotel in data.hotelList) {
  var tr = '<tr><td>'+ hotel.name+'</td><td>'+ hotel.city+'</td></tr>'
  table = table+tr;
}

为了清楚起见，我改变了我的酒店。

但是因为我无法看到整个JSON，所以我不知道我的猜测是否正确。

通过网页上的json迭代（jquery）

1 个答案: