javascript - 使用 j 查询在一个下拉列表中选择选项时如何填充 json？

Question

当我在第一个选择框中选择城市时，显示城市的相关数据

如何在第一个下拉列表中选择城市时填充 json，以及如何在选择不同城市时显示重新分配的数据？

这是html

$(document).ready(function() {
  var cityData = [{
      cityName: 'Bengaluru',
      value: "Bengaluru",
      data: [{
          movieName: 'ABC',
          theaterName: 'Tulsi Theatre'
        },
        {
          movieName: 'DEF',
          theaterName: 'PVR'
        },
        {
          movieName: 'GHI',
          theaterName: 'Srinivasa Theatre'
        }
      ]
    },


    {
      cityName: 'Hyderabad',
      value: "Hyderabad",
      data: [{
          movieName: '123',
          theaterName: 'Theatre1'
        },
        {
          movieName: '456',
          theaterName: 'PVR2'
        },
        {
          movieName: '789',
          theaterName: 'Theatre3'
        }
      ]
    },

    {
      cityName: 'Guntur',
      value: "Guntur",
      data: [{
          movieName: 'ABC1',
          theaterName: 'Theatre4'
        },
        {
          movieName: 'DEF2',
          theaterName: 'PVR3'
        },
        {
          movieName: 'GHI3',
          theaterName: 'Theatre5'
        }
      ]
    },

    {
      cityName: 'Ongole',
      value: "Ongole",
      data: 'currently not available'
    }
  ];
  $("#selectCity").on('change', function() {
    var locations = cityData[$(this).val()];
    var locationString = 'locations';
    console.log(locations)
    $.each(locations, function(i, item) {

      console.log(JSON.stringify(item));

      // OUCH!!! 

      locationString += '<option value="' + item.id + '">' + item.name + '</option>';
    });
    //console.log(locationString)
    $('#secondselectbox').html(locationString);
  });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="UserData">
  <h1>MyMovie-Ticket-Booking</h1>
  <select class="selectCity" id="selectCity">
    				<option value="City">Select City</option>
    				<option value="Bengaluru">Bengaluru</option>
    				<option value="Hyderabad">Hyderabad</option>
    				<option value="Guntur">Guntur</option>
    				<option value="Ongole">Ongole</option>
    			</select>
  <span id="welcome"> </span>
</div>
<div>
  <select id="secondselectbox"></select>
  <select id="secondselectbox"></select>
</div>

当我选择城市类加罗尔时，会显示类加罗尔剧院列表和电影列表，但它显示的是未定义的....

Answer 1

根据您的 json 结构，您没有正确检索“位置”数据。

此外，选项在 json 中没有“id”、“item”。

一定要将最后一个选择的 ID 更改为唯一的东西(比如“thirdselectbox”...)。

将您选择的更改监听器修改为:

$("#selectCity").on('change', function() {
    var locations = cityData.filter(c => c.cityName === $(this).val())[0].data;
    var locationString = '';
    var locationString2 = '';
    console.log(locations)
    $.each(locations, function(i, item) {

        console.log(JSON.stringify(item));
        locationString += '<option value="' + item.theaterName + '">' + item.theaterName + '</option>';
        locationString2 += '<option value="' + item.movieName + '">' + item.movieName + '</option>';
    });
    $('#secondselectbox').html(locationString);
    $('#thirdselectbox').html(locationString2);
});

参见 working fiddle .

希望这对您有所帮助。