
This tip is an example of how you can create a trivial login form in JSF 2.0 . The credentials are hardcoded as final variables – in real application, you need will compare the entered values agains a database.
. . .
@ManagedBean
@SessionScoped
public class User implements Serializable {
private static final long serialVersionUID = 1L;
private final String userName = "User";
private final String userPassword = "12345";
private String name;
private String password;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String login() {
if((userName.equals(name))&&(userPassword.equals(password))){
return "welcome";
} else {
return "index";
}
}
}
The login page should look something as below:
. . .
<h:form>
User : <h:inputText value="#{user.name}" />
Password : <h:inputSecret value="#{user.password}" />
<h:commandButton action="#{user.login}" value="Submit" />
</h:form>
. . .
If the
userName and userPassword corresponds with the name and password entered, you will be redirected to the welcome.xhtml page. Otherwise, you will remain on the login form page.
3 comments:
Huh. Please step back and read how security in java webapp works. Go on with http://download.oracle.com/javaee/6/tutorial/doc/bncbx.html or check out Spring Security.
Remember: This is not security! If some other beginner read those articles they will never get the point of how it is done correctly.
Stop playing smart, because you are not! Do you know what "trivial" means? Is just a simple example, like RoseIndia one (http://www.roseindia.net/jsf/JSFLoginApplication.shtml). I did not say anything about security, so please forget it!
It's very great post. This is really helpful for me.Thanks for sharing it.
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