XmlHttpRequest status = 0

时间:2013-02-03 18:58:33

标签: java javascript ajax servlets xmlhttprequest

我有来自javascript文件的这个ajax调用,我想作为参数传递我想要删除的用户的id:

function eliminaUtente(id,nome){
if (confirm("Sei sicuro di voler eliminare l'utente "
    + nome
    + "?")) {
var xmlHttp2 = new XMLHttpRequest();
xmlHttp2.open("POST", "EliminaUtente", true);
xmlHttp2.setRequestHeader("Content-type",
        "application/x-www-form-urlencoded");
var params2 = "id=" + id;
xmlHttp2.send(params2);
xmlHttp2.onreadystatechange = function() {
    if (xmlHttp2.readyState == 4) 
    {
                    alert(xmlHttp2.status);  <-----------this prints always 0!
        if (xmlHttp2.status == 200) //
        {
            alert("utente eliminato!");
        } else {
            alert("An error occurred while communicating with server.");
        }
    }
};

} }

在名为EliminaUtente的对应Servlet中,我有这个代码:

  protected void doPost(HttpServletRequest request,
        HttpServletResponse response) throws ServletException, IOException {
    // TODO Auto-generated method stub
    String id = request.getParameter("id");
    System.out.println(id);
    String query = "delete from utente where idutente=" + id;
    System.out.println(query);
    try {
        Class.forName("com.mysql.jdbc.Driver").newInstance();

        Connection con = DriverManager
                .getConnection("jdbc:mysql://localhost/Spinning?user=root");
        PreparedStatement prest = con.prepareStatement(query);
        prest.executeUpdate();

        response.setContentType("text/plain");
        PrintWriter ajaxWriter = response.getWriter();
        ajaxWriter.print("ok");
        ajaxWriter.flush();
        ajaxWriter.close();

        con.close();
    } catch (Exception e) {
        e.printStackTrace();
        response.setContentType("text/plain");
        PrintWriter ajaxWriter = response.getWriter();
        ajaxWriter.print("ko");
        ajaxWriter.flush();
        ajaxWriter.close();
    }

}

}

我无法理解问题出在哪里......你能帮帮我吗? ;)

3 个答案:

答案 0 :(得分:1)

我尝试你的代码并改变一点我想解释我和我从中学到了什么。我读了一些来源。首先,我阅读XMLHttpRequest对象和onreadyState事件。 我在PUTGET操作方法中实施了您的示例。

的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    version="2.5">

    <servlet>
        <servlet-name>testServlet</servlet-name>
        <servlet-class>com.test.testServlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>testServlet</servlet-name>
        <url-pattern>/test/*</url-pattern>
    </servlet-mapping>

</web-app>

testServlet.java

package com.test;

import java.io.IOException;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class testServlet extends HttpServlet {
    /**
     * 
     */
    private static final long serialVersionUID = 1L;

    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp)
            throws ServletException, IOException {
        // TODO Auto-generated method stub
        /*super.doPost(req, resp);*/
        String strId = req.getParameter("id");
        System.out.println(strId);
    }

    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp)
            throws ServletException, IOException {
        // TODO Auto-generated method stub
        //super.doGet(req, resp);
        String strId = req.getParameter("id");
        System.out.println(strId);
    }
}

和主要部分NewFile.jsp

<%@ page language="java" contentType="text/html; charset=UTF-8"
    pageEncoding="UTF-8"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Insert title here</title>
</head>
<body>
    <input type="button" value="Submit" onclick="eliminaUtente('1')" width="100%" />
</body>
<script language="javascript" type="text/javascript">
    function eliminaUtente(id) {

        var xmlHttp = new XMLHttpRequest();
        var url = "test/NewFile.jsp?id=" + id;
        xmlHttp.open("POST", url, true);
        xmlHttp.send(url);
        xmlHttp.onreadystatechange = function() {
            if (xmlHttp.readyState == 4 && xmlHttp.status == 200) {
                alert("utente eliminato!");
            } else {
                alert("An error occurred while communicating with server.");
            }
        };
    }
</script>
</html>

用这种方式我写参数1(我在jsp文件方法调用中硬编码)并将其写入控制台,这里的第一件事就是你的代码与我的差异我删除xmlHttp2.setRequestHeader("Content-type","application/x-www-form-urlencoded");因为如果方法type是POST默认加密就是这个。

enctype = content-type [CI]
This attribute specifies the content type used to submit the form to the server (when the value of method is "post"). The default value for this attribute is "application/x-www-form-urlencoded". The value "multipart/form-data" should be used in combination with the INPUT element, type="file".
  • 因此POST无需输入默认值即可。
  • 这是开放方法签名open(method, url, async, user, password)这里async是参数,这意味着如果它是假的,不等待来自服务器的响应实现另一行,当响应来时它将运行。如果确实等到响应来了。实际上我尝试了他们的机器人,它是有效的。
  • 最后,我使用GET进行尝试。如果您想将其与GET一起使用,则应添加xmlHttp2.setRequestHeader("Content-type","application/x-www-form-urlencoded");代码进行加密,并从url方法中删除send()参数。

    function eliminaUtente(id){

        var xmlHttp = new XMLHttpRequest();
        var url = "test/NewFile.jsp?id=" + id;
        xmlHttp.open("GET", url, true);
            xmlHttp2.setRequestHeader("Content-type","application/x-www-form-urlencoded");
        xmlHttp.send();
        xmlHttp.onreadystatechange = function() {
            if (xmlHttp.readyState == 4 && xmlHttp.status == 200) {
                alert("utente eliminato!");
            } else {
                alert("An error occurred while communicating with server.");
            }
        };
    } 
    

注意:我在firefox中尝试此代码,我创建了xmlHttpRequest对象。对于所有浏览器(包括IE6),确定您知道使用:

var xmlhttp;
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }

答案 1 :(得分:0)

错字?

alert(xmlHttp2.status);&lt; -----------这打印始终为0!现在有“2”

答案 2 :(得分:-1)

问题在于我必须在这一行中加错:

xmlHttp2.open("POST", "EliminaUtente", FALSE);