Reputation: 181
I am new here recently joined and New in Type ORM My code that I am trying
FIRST QUERY: I would like to use this approach but not sure how I can group by with count on the column and then order by desc on that column
const result = await this.jobViewsRepository.find({
relations: ["jobs"],
loadEagerRelations: true,
order: { id: "DESC" },
skip: offset,
take: limit,
}
);
I am trying if I can use this in my above query
SECOND QUERY: IT'S WORKING FOR ME PERFECTLY THE RESULT I AM LOOKING
const res = await this.jobViewsRepository.createQueryBuilder('jobViews')
.addSelect("COUNT(jobViews.user_id) AS jobViews_total_count" )
.leftJoinAndSelect(Jobs, "jobs", "jobs.id = jobViews.job_id")
.where("jobs.user_id != :id", { id: user_id })
.groupBy("jobViews.job_id")**
.orderBy('jobViews_total_count', 'DESC')**
.limit(limit)
.offset(offset)
.getRawMany();
Please if any can help me out in this will be really appreciated
Thanks
Upvotes: 18
Views: 31247
Reputation: 85
Now is the only one way is to use queryBuilder
And .groupBy("user.id")
with .addGroupBy("user.id")
https://orkhan.gitbook.io/typeorm/docs/select-query-builder#adding-group-by-expression
Or write raw
query:
import { getManager } from 'typeorm';
const entityManager = getManager();
const someQuery = await entityManager.query(`
SELECT
fw."X",
fw."Y",
ew.*
FROM "table1" as fw
JOIN "table2" as ew
ON fw."X" = $1 AND ew.id = fw."Y";
`, [param1]);
Upvotes: 0
Reputation: 11
At least in the current version there is no way to do this feature (neither in the documentation nor in the web)
Upvotes: 1