Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/typescript/9.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Jquery mobile 下拉菜单显示问题jQuery mobile_Jquery Mobile - Fatal编程技术网

Jquery mobile 下拉菜单显示问题jQuery mobile

Jquery mobile 下拉菜单显示问题jQuery mobile,jquery-mobile,Jquery Mobile,我有两个下拉菜单。一个是主菜单,另一个是子菜单 <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"> </s

我有两个下拉菜单。一个是主菜单,另一个是子菜单

<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-->

我已经用数组数据填充了主菜单,但我无法用相应的主菜单数据显示子菜单。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');
}
功能设置子菜单(indexPos){
var selectedMenuArray=[];
对于(变量i=0;i
这里我传递主菜单项的索引位置。然后我将相应的项目推送到SelectedMenuArray以在子菜单中显示它们。 这里我的问题是我无法显示子菜单项

试试这个

submenuID.trigger('create');
你能发一封信吗?
 subMenuId.html(subCatHtml).trigger('create');
 subMenuId.html(subCatHtml).selectmenu('refresh', true);