Optional parameter in TypeORM queryBuilder

Question

Is it possible to use optional parameters in createQueryBuilder ?

for example, I have this code:

let users = await this.conn
  .getRepository(UserEntity)
  .createQueryBuilder("user")
  .where("user.firstName LIKE :search", { search: dto.search })
  .getMany();

my optional parameter is search , I want to launch clauses .where only when dto.search is not null, but when he is null then should skip this function( where ) and goes to getMany .

Have anybody any idea how to do this?

Answer 1

try this:

let users = await this.conn.getRepository(UserEntity)
.createQueryBuilder('user')
.where(search !== null ? 
    'user.firstName LIKE :search': 'TRUE', {search: dto.search})
.getMany();

Or another way to do whithout the "Where TRUE" is this:

let users = this.conn.getRepository(UserEntity)
.createQueryBuilder('user');
users = search !== null ? users.where('user.firstName LIKE :search',{search: dto.search}) : users
users = await users.getMany();

Answer 2

const query = this.conn
  .getRepository(UserEntity)
  .createQueryBuilder("user");

// Optionally add where condition
if(dto.search){
  query.where("user.firstName LIKE :search", { search: dto.search })
}
// run query
let users = await query.getMany();

Do be mindful of falsy values that would trip up the if statement.