我有一个jsp,其中有选择标签,我想获取
 以及从我的Servlet中的jsp中的select中选择的值

<select id="listoffood" name="dropdown" onchange="foodname();">
<option value="bg">Burger</option>
<option value="pas">pasta</option>
<option value="pi">pizza</option>
</select>
<div id='content'></div>


这是JavaScript代码

function foodname()
{

  var xmlHttpReq = false;
    var self = this;
    document.getElementById('content').innerHtml='';
    // Mozilla/Safari
    if (window.XMLHttpRequest) {
        self.xmlHttpReq = new XMLHttpRequest();
    }
    // IE
    else if (window.ActiveXObject) {
        self.xmlHttpReq = new ActiveXObject("Microsoft.XMLHTTP");
    }

    self.xmlHttpReq.open('GET', "InformationServlet", true);
    self.xmlHttpReq.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
    self.xmlHttpReq.send(null);

    self.xmlHttpReq.onreadystatechange= function ()
    {
        //alert(document.getElementById('content'));
        if (self.xmlHttpReq.readyState==4)
        {
        if (self.xmlHttpReq.status == 200)
        {

        document.getElementById('content').innerHTML=self.xmlHttpReq.responseText;
        }
        }
    };

}


我所做的是使用像这样的get Attribute,但它不能正常工作,显示为null

protected void doGet(HttpServletRequest request, HttpServletResponse response)
                throws ServletException, IOException

    {
    // TODO Auto-generated method stub

    String coun = request.getParameter("dropdown");
    PrintWriter out=response.getWriter();
    System.out.println("here : "+coun);
}


在此先感谢您,任何代码都将受到高度赞赏。

最佳答案

只需将您的AJAX open()请求更改为

var select = document.getElementById("listoffood");
self.xmlHttpReq.open('GET', "InformationServlet?dropdown=" + select.options[select.selectedIndex].value, true);

关于java - 将值从jsp发送到Servlet,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/16357111/

10-12 22:26
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