【Layui__下拉框】layui下拉框默认选中其中的一项

话不多说，上代码

    window.onload = function () {
        var HallNameId = $("#HallNameId").val();
        //console.log(HallNameId);
        var select = \'dd[lay-value=\' + HallNameId + \']\';
        $(\'#hallName\').siblings("div.layui-form-select").find(\'dl\').find(select).click();
    }

layui在select标签中，会被渲染为dd列表，此时，我们赋值select标签已无太大的意义，需要对dd列表中的值设置选中

    <div class="layui-form-item">
        <label class="layui-form-label">选项</label>
        <div class="layui-inline layui-show-xs-block">
            <select id="hallName" name="hallName" lay-filter="hallName">
                <option value="1">请选择</option>
                <option value="1">测试1</option>
                <option value="1">测试2</option>
                <option value="1">测试3</option>

            </select>
            <div class="layui-unselect layui-form-select">
                <div class="layui-select-title">
                    <input type="text" placeholder="请选择" value="请选择" readonly="" class="layui-input layui-unselect">
                    <i class="layui-edge" />
                </div>
                <dl class="layui-anim layui-anim-upbit">
                    <dd lay-value="1" class="layui-this">请选择</dd>
                    <dd lay-value="1" class="layui-this">测试1</dd>
                    <dd lay-value="1" class="layui-this">测试2</dd>
                    <dd lay-value="1" class="layui-this">测试3</dd>
                </dl>
            </div>
        </div>
    </div>

第二种方法

直接给select赋值，然后重新渲染选择框

    <div id="divHide" style="display:none">
        <input type="hidden"  id="exhibitorId" value=""/>
    </div>

            var topId = $(\'#exhibitorId\', top.document);
            console.log("exhibitorId ", topId.val());
            layui.use([\'form\'], function () {
                var form = layui.form;
                $("#ExhibitorId").val(topId.val());
                layui.form.render(\'select\');
                console.log("设置为上次的完毕 ", topId.val());
            });