2013-10-21 114 views
0

我有两个下拉菜单。一个是主Menu和Other是SubMenu。下拉菜单显示问题jQuery mobile

<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--> 

我已填充Main菜单上有阵列的数据,但不荫能够与相应的主菜单的数据显示子菜单。我曾尝试是

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'); 
} 

这里路过的IAM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iam无法显示子菜单项。

+0

可以张贴[的jsfiddle(http://jsfiddle.net/)? – Omar

回答

0

试试这个

submenuID.trigger('create'); 
0
subMenuId.html(subCatHtml).trigger('create'); 
subMenuId.html(subCatHtml).selectmenu('refresh', true);