Question

我正在使用JSP，并且选择了医生姓名和返回值选项，并且返回值是医生的对象，具有诸如姓名，专长，地址和电话之类的属性。这是选择代码。

<select class="form-control" name="selectedPhysician"  id="selectedDoctor" onchange="populatePhysicianInfo()"> 
    <option value="" >Select Doctor in the List</option>
    <c:forEach items="${listAllPhysicians }" var="thePhysician" >
        <option value= "${thePhysician}">${thePhysician.physicianName}</option>
    </c:forEach>
</select>

populatePhysicianInfo（）是jQuery函数，我想提取通过列表选项传递的对象的属性。这些属性将弹出到适当的表单元素。如何实现财产的提取？

功能如下：

$(document).ready(function() {
     var doctorObj = new Object
    $("#selectedDoctor option").filter(function(){

        return $(this).val() == $("#name").val();
    }).attr('selected', true);

     $('#selectedDoctor').change(function(){

         doctorObj = $(this).val();
         alert(doctorObj);
         /*
         $('#name').doctorObj['physicianName'];
         $('#specialty').doctorObj['physicianSpecialty'];*/

     });


});

警报用于检查目的，其输出看起来像这样。

Physician [id=22, physicianName=Joe Nguyen, physicianPhone=512-244-9024, physicianAddress=ARC Round Rock, 940 Hesters Crossing Road, Round Rock, TX 78681, physicianSpecialty=Primary Doctor]

如何提取传递给jQuery的对象的属性

0 个答案: