从Javaservlet方法doGet将数据返回到ajax
我的函数ajax如下所示:从Javaservlet方法doGet将数据返回到ajax,java,jquery,ajax,jsp,servlets,Java,Jquery,Ajax,Jsp,Servlets,我的函数ajax如下所示: $.ajax({ url: "associer_type_flux", // It's my servlet dataType : "xml", type : "GET", data : { }, success: function(response){ alert("fine"); }, error: function(data
$.ajax({
url: "associer_type_flux", // It's my servlet
dataType : "xml",
type : "GET",
data : { },
success: function(response){
alert("fine");
},
error: function(data, status, er){
alert(data+"_"+status+"_"+er);
}
});
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String forward = "";
try {
String fluxXML = "";
ServicesCodeTypeFluxGlobaux servicesCodeTypeFluxGlobaux = new ServicesCodeTypeFluxGlobauxImpl();
fluxXML = "<lescodetypeflux>";
fluxXML += "</lescodetypeflux>";
PrintWriter printWriter = response.getWriter();
printWriter.println(fluxXML);
}
forward = "/associerCode/accueil_association.jsp";
getServletContext().getRequestDispatcher(forward).forward(request, response);
}
catch (Exception e) {
forward = "/erreur.jsp";
request.setAttribute("msg", e.getMessage());
}
}
我的方法doGet在我的servlet中是这样的:
$.ajax({
url: "associer_type_flux", // It's my servlet
dataType : "xml",
type : "GET",
data : { },
success: function(response){
alert("fine");
},
error: function(data, status, er){
alert(data+"_"+status+"_"+er);
}
});
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String forward = "";
try {
String fluxXML = "";
ServicesCodeTypeFluxGlobaux servicesCodeTypeFluxGlobaux = new ServicesCodeTypeFluxGlobauxImpl();
fluxXML = "<lescodetypeflux>";
fluxXML += "</lescodetypeflux>";
PrintWriter printWriter = response.getWriter();
printWriter.println(fluxXML);
}
forward = "/associerCode/accueil_association.jsp";
getServletContext().getRequestDispatcher(forward).forward(request, response);
}
catch (Exception e) {
forward = "/erreur.jsp";
request.setAttribute("msg", e.getMessage());
}
}
受保护的void doGet(HttpServletRequest请求,HttpServletResponse响应)抛出ServletException,IOException{
向前字符串=”;
试试{
字符串fluxXML=“”;
ServicesCodeTypeFluxGlobaux服务CodeTypeFluxGlobaux=新服务CodeTypeFluxGlobauxImpl();
fluxXML=“”;
fluxXML+=“”;
PrintWriter PrintWriter=response.getWriter();
printWriter.println(fluxXML);
}
forward=“/associerCode/accueil_association.jsp”;
getServletContext().getRequestDispatcher(forward).forward(请求,响应);
}
捕获(例外e){
forward=“/erreur.jsp”;
setAttribute(“msg”,即getMessage());
}
}
所以我的问题是,我无法在jsp中获取数据。。但我不知道如何从doGet方法中获取数据或返回数据。。现在我有一个警报错误
Thx一些建议可以让它发挥作用-
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("application/xml");
String forward = "";
try {
String fluxXML = "";
ServicesCodeTypeFluxGlobaux servicesCodeTypeFluxGlobaux = new ServicesCodeTypeFluxGlobauxImpl();
fluxXML = "<lescodetypeflux>";
fluxXML += "</lescodetypeflux>";
PrintWriter printWriter = response.getWriter();
printWriter.println(fluxXML);
printWriter.close();
}
}
catch (Exception e) {
//print xml with error value
}
}
受保护的void doGet(HttpServletRequest请求,HttpServletResponse响应)抛出ServletException,IOException{
setContentType(“应用程序/xml”);
向前字符串=”;
试试{
字符串fluxXML=“”;
ServicesCodeTypeFluxGlobaux服务CodeTypeFluxGlobaux=新服务CodeTypeFluxGlobauxImpl();
fluxXML=“”;
fluxXML+=“”;
PrintWriter PrintWriter=response.getWriter();
printWriter.println(fluxXML);
printWriter.close();
}
}
捕获(例外e){
//打印带有错误值的xml
}
}
打开浏览器网络控制台并检查发生了什么。我有以下错误:状态:parserror…:'(…无需确认响应。内容是什么?Java或Ajax中的响应..如果在Ajax中,我无法获取它,因为..告诉我状态:ParserError网络控制台将包含每个请求和响应的信息,包括标题和正文。我怀疑解析错误是什么。只需选择响应并查看其中的内容。或者,我们e一个不同的客户端(可能是Java)好的,我明白了。如果我这样做了,我的页面将返回XML。我如何才能看到我的页面JSP和表等等。我得到了这个:-1101你不能在一次单击中获得数据,也不能发送请求。你可以尝试其他替代方法,如-1。首先重定向到页面,在该页面中,你将数据显示到表中,然后通过ajax调用获取数据。2.通过ajax调用获取数据,然后使用javascript在要显示的div中创建表。