在用戶使用Spring MVC提交表單后，如何獲取表單的數據？

Question

我收到一條錯誤消息：org.springframework.web.util.NestedServletException：請求處理失敗; 嵌套異常是java.lang.ClassCastException：java.lang.Object無法強制轉換為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!!!");
    }

}

Answer 1

嘗試此解決方案：

JSP

<form:form action="yourUrl" modelAttribute="login" method="POST">
<% ... %>
</form:form>

調節器

// 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();
    // ...
}

說明

注釋@ModelAttribute與model.addAttribute(String name, Object value)完全相同。 例如，@ @ModelAttribute Login login與model.addAttribute("login", new Login()); 。

也就是說，使用onGet方法，您將這樣的對象傳遞給您的視圖。 由於屬性modelAttribute="login" ，標簽<form:form>將查看模型的屬性列表，以找到名稱為login的屬性。 如果找不到，則拋出異常。

然后，這是神奇的部分：使用標簽<form:input path="userName" /> ，Spring MVC將自動設置bean的屬性userName ，該屬性位於modelAttribute="login"屬性中，即在您的情況下， login 。 如果您<form:input path="wtf" />了類似<form:input path="wtf" /> ，則會拋出異常，因為bean Login沒有這樣的屬性。

所以，最后，在你的onSubmit方法上（再次感謝注釋@ModelAttribute ），你可以訪問以前由Spring MVC自動提取的login bean。

注意

我個人（幾乎）從不使用ModelAndView實例，但按以下步驟操作：

// 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的路徑在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>
...

您應該閱讀文檔以獲取更多信息（參見16.5解析視圖）。

Answer 2

在'onSubmit'方法中，您將command為Login 。

顯然它不是。