Question

我正在做struts2 + spring + hibernate + liferay集成。在这种情况下，不会在struts2 Action类中调用setter方法。我通过jsp发送的属性的名称在Action类和表单中保持相同。然后，setter方法也没有执行。我需要将属性插入数据库，所以我需要在我的Java类中的那些属性。请告诉我我哪里出错了。任何帮助将不胜感激。 Action类的代码如下： -

public class AddUserAction extends ActionSupport   implements ModelDriven<User> {

    /**
     * 
     */
    private static final long serialVersionUID = 4587665276501838677L;

    private String name="";
    private String place=null;
    private String state=null;
    private User user = new User();
    private List<User> userList = new ArrayList<User>();

    private List<String> validateErrors;
    private UserDAO userDao;



    public AddUserAction(UserDAO userDao) {
        this.userDao = userDao;
    }
    @Override
    public User getModel() {
    return user;
    }

    public User getUser() {
        return user;
    }
    public void setUser(User user) {
        this.user = user;
        System.out.println("hello");
    }
    @Override
    public String execute() throws Exception {
        //validateBookmark();

        userDao.saveUser(user);
        setUserList(userDao.listUser());
        System.out.println("List users"+userDao.listUser());
        return SUCCESS;
    }


    /*public void validateBookmark() throws Exception {
        Student studentObject = getNewStudent();
        validateErrors = new ArrayList<String>();
        StudentValidator validator = new StudentValidator();

        if(validator.validateStudent(studentObject, validateErrors)) {
            StudentLocalServiceUtil.addStudent(studentObject);
            addActionMessage(MessageStore.STUDENT_ADDED_SUCCESSFUL);

        }
        else {
            //handle the error massage 
        addActionError(MessageStore.BOOKMARK_ADDED_FAILED);
            addFieldError("failed", MessageStore.BOOKMARK_ADDED_FAILED);
            Iterator<String> errorIter = validateErrors.iterator();
            int count = 0;
            while (errorIter.hasNext()) {
                String error = errorIter.next();
                addFieldError("error" + count, error);
                count++;
            }
        }
    }*/

    /**
     * Create a new Bookmark Object from the view
     * @return 
     */
    /*private User getNewUSer() {
        User user =  new User();
        user.setName(getName());
        user.setPlace(getPlace());
        user.setState(getState());

        return user;
    }*/

    /*
     * Getters and Setters start here
     */
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
        System.out.println("name is hfdj");
    }

    public List<String> getValidateErrors() {
        return validateErrors;
    }

    public void setValidateErrors(List<String> validateErrors) {
        this.validateErrors = validateErrors;
    }

    public String getPlace() {
        return place;
    }

    public void setPlace(String place) {
        this.place = place;
    }

    public String getState() {
        return state;
    }

    public void setState(String state) {
        this.state = state;
    }
    public void setUserList(List<User> userList) {
        this.userList = userDao.listUser();
    }
    public List<User> getUserList() {
        return userList;
    }

}

JSP的代码如下： -

<s:form action="adduserrecord">
      <s:textfield name="name" label="Student Name" value="abhi"/>
      <s:textfield name="place" label="Student Place" value=""/>
      <s:textfield name="state" label="Student State" value=""/>
      <s:submit value="Add User" align="center"/>
</s:form>

我想澄清一下hibernate部分没有问题，因为插入的值为null。但问题是我想实际存储通过jsp传递的值。所以请帮助我..提前做好准备。

User.java

@Entity(name = "user")
@Table(name="USER")
public class User {

    private Long id;
    private String name;
    private String place;
    private String state;

    @Id
    @GeneratedValue
    @Column(name="USER_ID")
    public Long getId() {
        return id;
    }
    public void setId(Long id) {
        this.id = id;
    }

    @Column(name="USER_NAME")
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }

    @Column(name="USER_PLACE")
    public String getPlace() {
        return place;
    }
    public void setPlace(String place) {
        this.place = place;
    }

    @Column(name="USER_STATE")
    public String getState() {
        return state;
    }
    public void setState(String state) {
        this.state = state;
    }

}

Answer 1

简单地说，您的观点和操作中的更改应该有效，如给定参考中所示：

查看：

<s:form action="adduserrecord"> <s:textfield name="user.name" label="Student Name" value="abhi" /> <s:textfield name="user.place" label="Student Place" value="" /> <s:textfield name="user.state" label="Student State" value="" /> <s:submit value="Add User" align="center" /> </s:form>

<强>动作：

public class AddUserAction extends ActionSupport implements ModelDriven<User> { private static final long serialVersionUID = 4587665276501838677L; private User user; @Override public String execute() throws Exception { // Here you will find your user object populated with form fields userDao.saveUser(user); return SUCCESS; } public User getUser() { return user; } public void setUser(User user) { this.user = user; } }

Reference

在struts 2中没有调用setter方法

1 个答案: