Two linkage (dropdown data acquisition)

I. Introduction
Note: (Get values from the database)
to obtain data of a drop-down box directly after loading the page, when the value is selected, the query data value matches the second drop down box into effect as FIG. :

Second, the code shows
1, HTML Code:

<div class="layui-form-item">
    <label class="layui-form-label">公寓</label>
    <div class="layui-input-inline" style="width: 100px">
        <select name="selectBuilding" style="width: 100px;height: 30px;margin-top: 5px;text-align: center">
            <option value="">请选择公寓</option>
        </select>
    </div>
    <label class="layui-form-label" style="width: 40px">宿舍</label>
    <div class="layui-input-inline" style="width: 100px">
        <select name="selectDormitory" style="width: 100px;height: 30px;margin-top: 5px;text-align: center">
            <option value="">请选择宿舍</option>
        </select>
    </div>
</div>

2, js codes:
the first drop box load direct access to the page data, when the first drop-down box is changed, a second drop-down box of Ajax, obtaining data corresponding to

<script>
		$.ajax({
            type: 'post',
            url: 'findAllBuilding',
            dataType: 'json',
            data: {},
            success: function (data) {
                console.log(data);//使用console.log()方法来找到对象的内容
                $.each(data, function (i) {
                    var trString = " <option value='" + data[i].buildingId + "'>" + data[i].buildingName + "</option>";
                    $("[name='selectBuilding']").append(trString);
                });
            }
        });
        
        $("[name='selectBuilding']").change(function () {
            $.ajax({
                type: 'post',
                url: 'findAllDormitory',
                dataType: 'json',
                data: {
                    buildingId: $("[name='selectBuilding']").val()
                },
                success: function (data) {
                    $("[name='selectDormitory']").html("");
                    console.log(data);//使用console.log()方法来找到对象的内容
                    $.each(data, function (i) {
                        var trString = " <option value='" + data[i].dormitoryId + "'>" + data[i].dormitoryName + "</option>";
                        $("[name='selectDormitory']").append(trString);
                    });
                }
            });
        });
</script>

Third, the end of
which is the main code, other codes do not show up! !