How to perform a like query Typeorm

Question

Hello guys I\'m trying to find all the results that have a in them. I have tried a couple of ways but the problem is nothing works. It just returns an empty array

Answer 1

TypeORM provides out of the box Like function. Example from their docs:

import {Like} from "typeorm";

const loadedPosts = await connection.getRepository(Post).find({
    title: Like("%out #%")
});

in your case:

var data = await getRepository(User).find({
    name: Like(`%${firstName}%`)
});

Answer 2

You can also use the database function for concatenation. In postgres for instance:

 var data = await getRepository(User)
              .createQueryBuilder("user")
              .where("user.firstName like '%' || :name || '%'", {name: firstName })
              .getMany();

Answer 3

var data = await  getRepository(User)
                        .createQueryBuilder("user")
                        .where("user.firstName ILIKE %q, {q:`%${VALUE_HERE}%` })
                .getMany();

This is how I do it. Hope it helps

Answer 4

Correct way is:

 var data = await getRepository(User)
                  .createQueryBuilder("user")
                  .where("user.firstName like :name", { name:`%${firstName}%` })
                  .getMany();

Answer 5

If you have already used .find methods to support your repository needs you might not want to switch to QueryBuilder.

There is an easy way to implement LIKE filter using findConditions:

this.displayRepository.find({ where: "Display.name LIKE '%someString%'" });

OR for case insensitive (in postgres):

this.displayRepository.find({ where: "Display.name ILIKE '%someString%'" });

Keep in mind this is susceptible to Injection attacks, so you must protect the dynamic value explicitly.