如何更改dojox DropDownSelect中的显示值

时间:2011-03-31 18:40:42

标签: javascript dojo

我用标签声明了DropDownSelect:

<select id="someId" dojoType="dojox.form.DropDownSelect" >
<option>Loading...</option>
</select>

然后当调用xhr加载函数时,我想删除上面的选项并创建新的“选择选项”,例如。

load: function(response) {
        //remove option, but it stays as displayedValue, displayed
        dojo.byId("someId").removeChild(dojo.byId("someId").lastChild); 
        //I would like to show the option below
        dijit.byId("someId").addOption({label: "Choose Option", value:"Select"});
        //dijit.byId("someId").setAttribute('displayedValue','Select'); // not working
        dijit.byId("someId").addOption(response); // add ajax options. 
        return response;
        },

请告诉我如何在DropDownSelect中显示不同的选项???使用版本1.3。

1 个答案:

答案 0 :(得分:1)

// 1。 html:定义下拉列表

<select id="ddselect" name="select" dojoType="dojox.form.DropDownSelect">
    <option value="none">
        Loading...
    </option>
</select>

// 2. onLoad handler:

    load: function(response) {
                    // get widget by id
        var dd = dijit.byId("ddselect");
        // remove loading option
                    dd.removeOption(0);
                    // add new option
        dd.addOption({ label: 'newValue1', value: 'nv1'});
                    // or add array of options
                    dd.addOption([{ label: 'newValue2', value: 'nv2' }, { label: 'newValue3', value: 'nv3'}]);
    }