Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • Home
  • SEARCH
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 6609397
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 25, 20262026-05-25T19:42:48+00:00 2026-05-25T19:42:48+00:00

public User findUser(String email) { User user = null; user = (User) sessionFactory.getCurrentSession().createCriteria(User.class).add(Restrictions.eq(email, email)).uniqueResult();

  • 0
 public User findUser(String email) {
                User user = null;
                user = (User) sessionFactory.getCurrentSession().createCriteria(User.class).add(Restrictions.eq("email", email)).uniqueResult();

                return user;

        }

Now I want to add one more restriction as if active is 1 how can I do it. I am not finding any way that I could check for two restrictions
Please suggest
Romi.

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-25T19:42:48+00:00Added an answer on May 25, 2026 at 7:42 pm

    The add(Criteria) method returns itself (this) to allow for chaining.

    public User findUser(String email) {
        User user = null;
        Criteria c = sessionFactory.getCurrentSession().createCriteria(User.class);
    
        //You can chain the add method, because it returns `this`.
        c = c.add(Restrictions.eq("email", email)).add(Restrictions.eq("active", 1));
        return (User) c.uniqueResult();
    }
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have the following piece of code in my DataModel.cs class: public User ValidateUser(string
I have this Class: public class User { public string id{ get; set; }
Im using DataContractJsonSerializer to serealize this class: public class User { public string id
Which class design is better and why? public class User { public String UserName;
public class User { private final String _first_name; private final String _last_name; private final
class User { public: int v() { return min_pass_len; } static const int min_pass_len
I have a class like public User class { public string Name{get;set;} public string
The code is listed below: @Document @XmlRootElement public class User { @Indexed(unique=true) private String
I have this class: class user { private: string userid; string password; public: user(){};
I have a class: public class User { public virtual int Id { get;

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.