小编Tas*_* H.的帖子

我想用php代码中的选项填充以下select语句

<select name='friends[]' id='friends' class='selectpicker show-tick form-control'
        data-live- search='true' multiple>
    <!-- options here -->
</select>

我的jQuery代码

$.ajax({
    url:'get_togethers.php', //this returns object data
    data:'user_id='+user_id,
    type:'POST',
    datatype:'json',
    success:function(data) { //data = {"0":{"id":1,"name":"Jason"},"1":{"id":2,"name":"Will"},"length":2 }
        data = JSON.parse(data);
        var options;
        for (var i = 0; i < data['length']; i++) {
            options += "<option value='"+data[i]['id']+"'>"+data[i]['name']+"</option>";
        }
        $("#friends").append(options);
    }
});

select标记内的静态值显示,但从ajax函数添加的值不显示.编辑:如果我从中删除引导程序,值显示,但启动引导程序,它们不会显示.