Question

我有两个下拉菜单。一个是主菜单，另一个是SubMenu。

<div id="menu">
    <select data-native-menu="false" id="mainMenu" name="select-choice-0">
        </select> <select data-native-menu="false" id="subMenu" name=
        "select-choice-1">
        </select>
</div><!--End of menu-->

我已使用数组数据填充主菜单，但我无法显示具有相应主菜单数据的子菜单。我试过的是

function setSubMenu(indexPos) {
    var selectedMenuArray = [];
        for (var i = 0; i < sortedArray[menuKeys[indexPos]].length; i++) {
            selectedMenuArray.push(sortedArray[menuKeys[indexPos]][i]);
        }
        //creating submenu
        var subMenuId = $('#subMenu');
        var subCatHtml = '';
        for (var i = 0; i < selectedMenuArray.length; i++) {
            console.log(selectedMenuArray[i]['subcategory']);
            subCatHtml += '<option>' + selectedMenuArray[i]['subcategory'] + '</option>';
        }

    subMenuId.html(subCatHtml);
    subMenuId.selectmenu();
    subMenuId.selectmenu('refresh');
}

这里iam传递主菜单项的索引位置。然后iam将各个项目推送到SelectedMenuArray以在SubMenu中显示它们。这里我的问题是我无法显示子菜单项。

Answer 1

试试这个

submenuID.trigger('create');

Answer 2

 subMenuId.html(subCatHtml).trigger('create');
 subMenuId.html(subCatHtml).selectmenu('refresh', true);

下拉菜单显示问题jQuery mobile

2 个答案: