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我在下拉菜单中设置值但它没有插入。为什么?但是我在另一个例子中做同样的事情,我得到了结果。

 <div data-role="fieldcontain">
                        <label for="text-12" style="text-align:left;margin-left: 0px;"> Font Type:</label>
                        <select name="select-choice-1" id="select-choice-1" class="fontFamily_h">
                            <option>Select Font </option>
                            <option value="AntiquaAntiqua">AntiquaAntiqua</option>
                            <option value="Arial">Arial </option>
                           <option value="Times New Roman">Times New Roman</option>

                        </select>
                    </div>
<div data-role="fieldcontain">
                        <label for="text-12" style="text-align:left;margin-left: 0px;">Font Size:</label>
                        <select name="select-choice-1" id="select-choice-2" class="fontSize_h">Select Size
                            <option>Select Size</option>
                            <option value="9">9 px</option>
                            <option value="10">10 px</option>
                            <option value="11">11 px</option>
                            <option value="12">12 px</option>
                            <option value="13">13 px</option>

                        </select>
                    </div>


    $(document).on('click', '.default_h', function(event) {
       var i =12;
        alert("Hi"+ $('#select-choice-2').val()); getting select Size
       $('#select-choice-1').val('Arial').selectmenu("refresh");
       $('#select-choice-2').val('12 px').selectmenu("refresh");
        alert("Hi"+ $('#select-choice-1').val());  getting Arial
           alert("Hi"+ $('#select-choice-2').val());;getting select Size


});   

我得到 Arrial 但在第二个例子中我没有得到 12 px

4

1 回答 1

2

您需要将val()属性设置为实际值属性,而不是文本

$('#select-choice-1').val('Arial').selectmenu("refresh");
$('#select-choice-2').val('12').selectmenu("refresh");
于 2013-07-17T07:38:16.110 回答