Java 在jsp中处理Ajax成功和错误
我有JSP中的表单,它有两个输入框,还有提交和清除按钮,如下所示Java 在jsp中处理Ajax成功和错误,java,ajax,jsp,Java,Ajax,Jsp,我有JSP中的表单,它有两个输入框,还有提交和清除按钮,如下所示 <form name="loginForm" method="GET" action="Ajaxexample" id="loginForm"> <table> <tr> <td>From Date</td><td><input type="text" name="n1" value=""/></td> </tr>
<form name="loginForm" method="GET" action="Ajaxexample" id="loginForm">
<table>
<tr>
<td>From Date</td><td><input type="text" name="n1" value=""/></td>
</tr>
<tr>
<td>End Date</td><td><input type="text" name="n2" value=""/></td>
</tr>
<tr></tr>
<tr>
<td><input type="submit" name="validpro_insert" value="Insert"></td>
<td><input type="reset" name="validpro_clear" value="Clear"></td>
</tr>
</table>
</form>
<script type="text/javascript" src="js/jq.js"></script>
<script type="text/javascript">
var form = $('#loginForm');
form.submit(function () {
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
data: form.serialize(),
error: function (theRequest,textStatus, errorThrown) {
// Success = false;
alert (theRequest.responseText);
alert(errorThrown);
alert('No graph found');//doesnt goes here
},
success: function (data) {
var result=data;
alert(result);
}
});
return false;
});
</script>
因为我必须编写一些代码来从servlet中的DB获取数据,并将该响应返回给ajax,ajax在相同的jsp上处理成功和错误,如下所示
<form name="loginForm" method="GET" action="Ajaxexample" id="loginForm">
<table>
<tr>
<td>From Date</td><td><input type="text" name="n1" value=""/></td>
</tr>
<tr>
<td>End Date</td><td><input type="text" name="n2" value=""/></td>
</tr>
<tr></tr>
<tr>
<td><input type="submit" name="validpro_insert" value="Insert"></td>
<td><input type="reset" name="validpro_clear" value="Clear"></td>
</tr>
</table>
</form>
<script type="text/javascript" src="js/jq.js"></script>
<script type="text/javascript">
var form = $('#loginForm');
form.submit(function () {
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
data: form.serialize(),
error: function (theRequest,textStatus, errorThrown) {
// Success = false;
alert (theRequest.responseText);
alert(errorThrown);
alert('No graph found');//doesnt goes here
},
success: function (data) {
var result=data;
alert(result);
}
});
return false;
});
</script>
变量形式=$(“#loginForm”);
表单提交(函数(){
$.ajax({
类型:form.attr('method'),
url:form.attr('action'),
数据:form.serialize(),
错误:函数(请求、文本状态、错误抛出){
//成功=错误;
警报(theRequest.responseText);
警报(错误抛出);
警报('未找到图形');//此处不显示
},
成功:功能(数据){
var结果=数据;
警报(结果);
}
});
返回false;
});
但问题是,我并没有从ajax中的servlet中获得任何值来处理成功或错误
我认为由于servlet doget()方法代码,我面临这个问题。。如果有任何其他问题,请告诉我。感谢您的帮助我已将代码粘贴在此处,效果良好 尝试更改参数 您的JSP页面
<script src="http://code.jquery.com/jquery-1.10.2.js"
type="text/javascript"></script>
}))
您的servlet文件
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class GetUserServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String userName = request.getParameter("userName").trim();
if(userName == null || "".equals(userName)){
userName = "Guest";
}
String greetings = "Hello " + userName;
response.setContentType("text/plain");
response.getWriter().write(greetings);
}
}
在我的代码中进行这些更改后,它将成功运行
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
try (PrintWriter out = response.getWriter())
String responseSend = "";
String from = request.getParameter("n1");
String to = request.getParameter("n2");
if ((from == null) || (from.equals(""))) {
System.out.println("From null");
responseSend = "error";
}
else if ((to == null) || (to.equals(""))) {
System.out.println("End null");
responseSend = "error";
}
else{
//jdbc code
System.out.println("got it");
int n1 = Integer.parseInt(request.getParameter("n1"));
int n2 = Integer.parseInt(request.getParameter("n2"));
responseSend = "code";
}
out.print(responseSend);
}
}
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
System.out.println("In get");
processRequest(request, response);
}
因为我添加了一个新方法processrequest(),该方法带有请求和响应参数,将文本/HTML返回到同一jsp上的Ajax代码中
error: function (theRequest,textStatus, errorThrown) {
alert (theRequest.responseText);
alert(errorThrown);
},
success: function (data) {
var result=data;
alert(result);
}
当在给定URL上找不到servlet时,将调用该错误;当成功调用具有给定类型和servlet URL的servlet时,将调用success。您还没有从ajax代码发送请求类型,那么如何进入servlet。基本上,我对从servlet返回的数据有一个问题,在返回数据之后,将首先执行成功或错误。没有必要,因为在成功中默认的请求类型是“GET”,不要将参数值分配给另一个变量,直接像这样使用它:函数(数据){alert(数据);}我刚刚在success中分配了这个变量,但我担心的是,我无法使用response.getWriter().write(问候语)从success中的doget()方法获取数据正如我在google chrome open developer模式下的answeropen页面中提到的那样,我使用了PrintWriter,按下ctrl+shift+I和goto network tab,并在提交表单时检查servlet文件返回的内容,一旦您单击/提交表单,它就会显示您的servlet文件名。单击表单并检查返回数据,并检查单击按钮后,您的文件是否显示在网络选项卡中
error: function (theRequest,textStatus, errorThrown) {
alert (theRequest.responseText);
alert(errorThrown);
},
success: function (data) {
var result=data;
alert(result);
}