java - XmlHttpRequest 状态=0

Question

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

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，我有以下代码：

  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();
    }

}

}

我不明白问题出在哪里……你能帮帮我吗？;)

score 1 · Accepted Answer

我尝试了您的代码并进行了一些更改，我想解释一下我做了什么以及从中学到了什么。我阅读了一些资料。首先我阅读了XMLHttpRequest对象和onreadyState事件。我实现了您的示例PUT和GET操作方法。

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>

通过这种方式，我编写参数 1（我在 jsp 文件方法调用中对其进行硬编码）并将其写入控制台，这里的第一件事是您的代码和我的代码的区别，我删除了xmlHttp2.setRequestHeader("Content-type","application/x-www-form-urlencoded");因为如果方法类型是POST默认加密就是这样。

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".

所以不需要 enctype 默认值对于POST.
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，则不要等待来自服务器的响应，当响应到来时，它会运行另一行。如果是真的，请等到响应到来。实际上我尝试了他们的机器人并且它有效。

最后我用GET. 如果你想使用它，GET你应该添加xmlHttp2.setRequestHeader("Content-type","application/x-www-form-urlencoded");加密代码并从方法中删除url参数。send()

功能消除（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 xmlHttpRequest 对象。对于所有浏览器（包括 IE6），请确保您知道使用：

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");
  }

score 0 · Accepted Answer

0

错字？

alert(xmlHttp2.status); <-----------这总是打印0！现在有“2”

于 2013-02-03T19:06:32.360 回答

score -1 · Accepted Answer

-1

问题是我必须在这一行加上 false ：

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

于 2013-02-08T11:26:34.027 回答

java - XmlHttpRequest 状态=0

3 回答 3

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