我的问题是要用jQuery的ajax检测一个用户名是否存在,但它老是弹出"检测失败,请检查网络!",
不能弹出“aa”,跪求大神解释怎么用jQuery的ajax,以下是代码
页面jsp代码
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
<!DOCTYPE html>
<html>
<head>
<title></title>
<script type="text/javascript" src="${pageContext.servletContext.contextPath}/jquery.min.js"></script>
<script type="text/javascript">
function ajaxCheck(v){
var userName = v;
if(userName.trim()!=""){
$.ajax({
type:"post",
url:"ajaxCheck.action",
data:'user.userName='+userName+'',
dataType:"json",
success:tishi,
error:function(){
alert("检测失败,请检查网络!");
}
})
}else {
alert("请先输入用户名!");
}
}
function tishi(){
alert("aa");
/*var json = eval(jsonResult);
var str = json.jsonResult;
if(str=="ok"){
alert(json.userName+"该用户存在,请输入密码进行登陆!");
}else{
alert(json.userName+"该用户不存在,请先注册!");
}*/
}
</script>
</head>
<body>
<form action="save.action" method="post">
userName:<input type="text" name="user.userName" onblur="ajaxCheck(this.value);"><br/>
userPs:<input type="text" name="user.userPs"> <br/>
<input type="submit">
</form>
</body>
</html>struts.xml代码
<package name="ajax" namespace="/" extends="json-default">
<action name="ajaxCheck" class="userAction" method="ajaxCheck">
<result type="json"></result>
</action>
</package>java代码
package com.pro.action;import java.util.List;import javax.servlet.ServletContext;
import javax.servlet.http.HttpServletRequest;import org.apache.struts2.interceptor.ServletRequestAware;
import org.apache.struts2.util.ServletContextAware;
import org.hibernate.criterion.DetachedCriteria;import com.opensymphony.xwork2.ActionSupport;
import com.pro.domain.User;
import com.pro.manager.UserManager;public class UserAction extends ActionSupport implements ServletRequestAware,ServletContextAware {
private User user;
private UserManager userManager;
private HttpServletRequest request;
private ServletContext application;
private String currentPage;
private String jsonResult;
private String userName;
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getJsonResult() {
return jsonResult;
}
public void setJsonResult(String jsonResult) {
this.jsonResult = jsonResult;
}
public String getCurrentPage() {
return currentPage;
}
public void setCurrentPage(String currentPage) {
this.currentPage = currentPage;
}
public void setServletContext(ServletContext application) {
this.application=application;
}
public void setServletRequest(HttpServletRequest request) {
this.request=request;
}
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
public void setUserManager(UserManager userManager) {
this.userManager = userManager;
}
//ajax
public String ajaxCheck() throws Exception{
return SUCCESS;
}
}
jQueryajaxj2ee
不能弹出“aa”,跪求大神解释怎么用jQuery的ajax,以下是代码
页面jsp代码
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
<!DOCTYPE html>
<html>
<head>
<title></title>
<script type="text/javascript" src="${pageContext.servletContext.contextPath}/jquery.min.js"></script>
<script type="text/javascript">
function ajaxCheck(v){
var userName = v;
if(userName.trim()!=""){
$.ajax({
type:"post",
url:"ajaxCheck.action",
data:'user.userName='+userName+'',
dataType:"json",
success:tishi,
error:function(){
alert("检测失败,请检查网络!");
}
})
}else {
alert("请先输入用户名!");
}
}
function tishi(){
alert("aa");
/*var json = eval(jsonResult);
var str = json.jsonResult;
if(str=="ok"){
alert(json.userName+"该用户存在,请输入密码进行登陆!");
}else{
alert(json.userName+"该用户不存在,请先注册!");
}*/
}
</script>
</head>
<body>
<form action="save.action" method="post">
userName:<input type="text" name="user.userName" onblur="ajaxCheck(this.value);"><br/>
userPs:<input type="text" name="user.userPs"> <br/>
<input type="submit">
</form>
</body>
</html>struts.xml代码
<package name="ajax" namespace="/" extends="json-default">
<action name="ajaxCheck" class="userAction" method="ajaxCheck">
<result type="json"></result>
</action>
</package>java代码
package com.pro.action;import java.util.List;import javax.servlet.ServletContext;
import javax.servlet.http.HttpServletRequest;import org.apache.struts2.interceptor.ServletRequestAware;
import org.apache.struts2.util.ServletContextAware;
import org.hibernate.criterion.DetachedCriteria;import com.opensymphony.xwork2.ActionSupport;
import com.pro.domain.User;
import com.pro.manager.UserManager;public class UserAction extends ActionSupport implements ServletRequestAware,ServletContextAware {
private User user;
private UserManager userManager;
private HttpServletRequest request;
private ServletContext application;
private String currentPage;
private String jsonResult;
private String userName;
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getJsonResult() {
return jsonResult;
}
public void setJsonResult(String jsonResult) {
this.jsonResult = jsonResult;
}
public String getCurrentPage() {
return currentPage;
}
public void setCurrentPage(String currentPage) {
this.currentPage = currentPage;
}
public void setServletContext(ServletContext application) {
this.application=application;
}
public void setServletRequest(HttpServletRequest request) {
this.request=request;
}
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
public void setUserManager(UserManager userManager) {
this.userManager = userManager;
}
//ajax
public String ajaxCheck() throws Exception{
return SUCCESS;
}
}
jQueryajaxj2ee
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success:tishi,
这句什么意思?
alert("aa");
}来alert,测试是否成功执行ajax
data:'user.userName='+userName+'',
dataType:"json"
而且你的框架要支持json
比如Spring mvc的要引入jackson包来支持json数据
error:function(a,b){
alert(b);
}
有可能是你返回的数据没有按照标准的json格式导致解析失败,标准的json格式都是加双引号的哦@!
就是以下这3个jar包,谢谢各位了
json-lib-2.1.jar
struts2-json-plugin-2.1.8.1.jar
xwork-core-2.1.6.jar