[英]Display submitted form data in Java Spring MVC bootstrap modal
[英]How to get a form's data after the user submitted it with Spring MVC?
我收到一条错误消息:org.springframework.web.util.NestedServletException:请求处理失败; 嵌套异常是java.lang.ClassCastException:java.lang.Object无法强制转换为com.crimetrack.business.Login
Login.java
public class Login {
private String userName;
private String password;
private boolean loggedin;
public Login(){};
/**
* @return the loggedin
*/
public boolean isLoggedin() {
return loggedin;
}
/**
* @param loggedin the loggedin to set
*/
public void setLoggedin(boolean loggedin) {
this.loggedin = loggedin;
}
/**
* @param userName
* @param password
*/
public Login(String userName, String password) {
this.userName = userName;
this.password = password;
}
/**
* @return the userName
*/
public String getUserName() {
return userName;
}
/**
* @param userName the userName to set
*/
public void setUserName(String userName) {
this.userName = userName;
}
/**
* @return the password
*/
public String getPassword() {
return password;
}
/**
* @param password the password to set
*/
public void setPassword(String password) {
this.password = password;
}
}
@Controller
public class AuthenticationController {
private final Logger logger = Logger.getLogger(getClass());
private AuthenticationManager authenticationManager;
private Login login = new Login();
String message = "Congrulations You Have Sucessfully Login";
String errorMsg = "Login Unsucessful";
@RequestMapping(value="login.htm")
public ModelAndView onSubmit(Object command) throws ServletException {
String userName = ((Login)command).getUserName();
String password = ((Login)command).getPassword();
login.setUserName(userName);
login.setPassword(password);
logger.info("Login was set");
logger.info("the username was set to " + login.getUserName());
logger.info("the password was set to " + login.getPassword());
if (authenticationManager.Authenticate(login) == true){
return new ModelAndView("main","welcomeMessage", message);
}
//return new ModelAndView("main","welcomeMessage", message);
return new ModelAndView("login","errorMsg", "Error!!!");
}
}
尝试此解决方案:
JSP
<form:form action="yourUrl" modelAttribute="login" method="POST">
<% ... %>
</form:form>
调节器
// your method that prints the form
public ModelAndView onGet(@ModelAttribute Login login) {
// return ...
}
@RequestMapping(value="login.htm")
public ModelAndView onSubmit(@ModelAttribute Login login) {
String userName = login.getUserName();
String password = login.getPassword();
// ...
}
说明
注释@ModelAttribute
与model.addAttribute(String name, Object value)
完全相同。 例如,@ @ModelAttribute Login login
与model.addAttribute("login", new Login());
。
也就是说,使用onGet
方法,您将这样的对象传递给您的视图。 由于属性modelAttribute="login"
,标签<form:form>
将查看模型的属性列表,以找到名称为login
的属性。 如果找不到,则抛出异常。
然后,这是神奇的部分:使用标签<form:input path="userName" />
,Spring MVC将自动设置bean的属性userName
,该属性位于modelAttribute="login"
属性中,即在您的情况下, login
。 如果您<form:input path="wtf" />
了类似<form:input path="wtf" />
,则会抛出异常,因为bean Login
没有这样的属性。
所以,最后,在你的onSubmit
方法上(再次感谢注释@ModelAttribute
),你可以访问以前由Spring MVC自动提取的login
bean。
注意
我个人(几乎)从不使用ModelAndView
实例,但按以下步骤操作:
// the methods can have the name you want
// not only onGet, onPost, etc. as in servlets
@RequestMapping("url1.htm")
public String loadAnyJsp(@ModelAttribute Login login) {
return "path/to/my/views/login";
}
@RequestMapping("url2.htm")
public String redirectToAnotherController(@ModelAttribute Login login) {
return "redirect:url1.htm";
}
JSP的路径在web.xml文件中指定,例如:
...
<bean class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver" p:favorPathExtension="true" p:favorParameter="true" p:ignoreAcceptHeader="true" p:defaultContentType="text/html">
<description>Depending on extension, return html with no decoration (.html), json (.json) or xml (.xml), remaining pages are decoracted</description>
<property name="mediaTypes">
<map>
<entry key="xml" value="application/xml" />
<entry key="json" value="application/json" />
<entry key="html" value="text/html" />
<entry key="action" value="text/html" />
</map>
</property>
<property name="defaultViews">
<list>
<bean class="org.springframework.web.servlet.view.xml.MarshallingView" p:marshaller-ref="xstreamMarshaller" />
<bean class="org.springframework.web.servlet.view.json.MappingJacksonJsonView" />
</list>
</property>
<property name="viewResolvers">
<list>
<bean id="nameViewResolver" class="org.springframework.web.servlet.view.BeanNameViewResolver">
<description>Maps a logical view name to a View instance configured as a Spring bean</description>
</bean>
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver" p:prefix="/WEB-INF/views/" p:suffix=".jsp" />
</list>
</property>
</bean>
...
您应该阅读文档以获取更多信息(参见16.5解析视图)。
在'onSubmit'方法中,您将command
为Login
。
显然它不是。
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