【发布时间】:2015-07-29 20:15:00
【问题描述】:
我想在 tomcat 中建立一个 Jquery 函数和一个 servlet 之间的通信。
Servlet 代码:
import java.io.BufferedReader;
import java.io.File;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class Test extends HttpServlet {
public static String getBody(HttpServletRequest request) throws IOException {
String body = null;
StringBuilder stringBuilder = new StringBuilder();
BufferedReader bufferedReader = null;
try {
InputStream inputStream = request.getInputStream();
if (inputStream != null) {
bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
char[] charBuffer = new char[128];
int bytesRead = -1;
while ((bytesRead = bufferedReader.read(charBuffer)) > 0) {
stringBuilder.append(charBuffer, 0, bytesRead);
}
} else {
stringBuilder.append("");
}
} catch (IOException ex) {
throw ex;
} finally {
if (bufferedReader != null) {
try {
bufferedReader.close();
} catch (IOException ex) {
throw ex;
}
}
}
body = stringBuilder.toString();
return body;
}
public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException, ServletException {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
System.out.println(getBody(request));
out.println("Success Call Ajax POST");
}
public void doGet( HttpServletRequest request, HttpServletResponse
response ) throws ServletException, IOException{
response.setContentType("text/html");
response.setCharacterEncoding( "UTF-8" );
PrintWriter out = response.getWriter();
out.println("Get Method");
}
}
servlet 标识在 web.xml 中定义
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"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<servlet>
<servlet-name>Test</servlet-name>
<servlet-class>com.servlets.Test</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Test</servlet-name>
<url-pattern>/toto</url-pattern>
</servlet-mapping>
</web-app>
以下 HTML 包含 JQuery 函数:
jQuery 代码:
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
function login(){
$.ajax({
type: "POST",
url: "http://localhost:8080/test/toto",
data: "POST Call",
success: function(result){
alert("success call"+result);
},
error: function(XMLHttpRequest, textStatus, errorThrown) {
alert("Status: " + textStatus); alert("Error: " + errorThrown);
}
});
}
</script>
<title>My AJAX</title>
</head>
<body>
<button type="button" onclick="login()">Click Me!</button>
</body>
我用浏览器测试了servlet,没问题。当我尝试 HTML/js 应用程序并单击按钮时,我在 Eclipse 控制台中显示了指令 System.out.println(getBody(request)); 显示的消息。 “POST Call”,但我在浏览器中收到了错误警报。所以ajax函数成功调用了servlet中的post方法,但是servlet不能成功地向浏览器返回响应。 servlet 似乎有问题。有人可以帮帮我吗?
【问题讨论】:
-
如果显示 'POST Call' ,则 ajax 中的
data: "POST Call",可能存在问题。如果您使用的是 chrome 浏览器,请检查网络选项卡的响应和标题。 -
我尝试使用 firefox 和 chrome 并收到相同的错误警报。但是在 chrome 中,当我单击按钮时,我也收到了此错误消息:XMLHttpRequest cannot load localhost:8080/test/toto。请求的资源上不存在“Access-Control-Allow-Origin”标头。因此,不允许访问 Origin 'null'。
-
您是否使用 iframe 加载 servlet?\
-
不,我不使用 iframe 加载
-
在
doPost()中尝试添加请求标头:response.setContentType("text/html"); response.setHeader("Access-Control-Allow-Origin", "*");有什么不同吗?
标签: java javascript jquery ajax servlets