Yii2 how to add clause where > 0 to the relational attribute - php

I have two models: Users, Posts. User has many posts.
public function getPosts()
{
return $this->hasMany(Posts::className(), ['user_id' => 'id']);
}
I need to get only those users who have posts (posts>0).
How do I write query?
Users:find()->where(['>', 'posts', 0])->all()
The code above doesn't work.

Try This query :
Users:find()
->where('id IN (select user_id FROM posts GROUP BY user_id)')
->all();

To get users with at least one post you need to use INNER JOIN:
Users::find()
->innerJoinWith('posts', false)
->groupBy('users.id')
->all();
It should be more efficient that subquery.
If you want to filter by number of posts, you should add HAVING clause with count condition:
Users::find()
->innerJoinWith('posts', false)
->groupBy('users.id')
->having(new \yii\db\Expression('COUNT(*) > :posts_nr', ['posts_nr' => 2]))
->all();
But such queries may be really heavy on large databases - you should consider adding additional column with posts count to users table.

$orders = Users::find()->with('posts')->all();
Inner join should remove users where posts are null

Related

Laravel multiple select filter for categories, only get one select as output

In laravel 8, I am filtering blog articles by category. However when I select multiple categories in my select menu. I do get the proper request . for example: articles/category/?category_ids=3,4
But it will only output one selected filter. If I select 2 filters it just selects that next filter as if I only selected that one. (I also use Axios but the request is done proper, so its in my Controller)
Here is my code I tried:
$data['articles'] = Article::whereHas('categories', function ($query) use($category_ids){
$query->whereHas('category_id', '=', $category_ids)->where('premium',0);
;})->get();
I also tried:
$data['articles'] = Article::whereHas('categories', function ($query) use($category_ids){
$query->whereIn('category_id', [$category_ids])->where('premium',0);
;})->get();
So how do I get to query both or more category id's ?
I am using a pivot table:
Articles can have many Categories
Categories can have many Articles
I use article_category as a pivot table
When checking for relationship existence in many-to-many relations, the check is still to be done against the id in the categories table.
Try this
$category_ids = collect(explode(',', $request->category_ids))
->map(fn($i) => trim($i))
->all();
$data['articles'] => Article::whereHas('category', fn($query) =>
$query->whereIn('categories.id', $category_ids)
->where('categories.premium', 0)
)->get();
You can explode the categories and then make the query like this.
$categories = explode(',',$request->categories);
$data['articles'] = Article::whereHas('categories', function ($query) use($categories){
$query->whereIn('category_id', $categories)->where('premium',0);
})->get();

Yii2 model to exclude related records

I have two related tables, posts and hidden_posts, where posts.id corresponds to hidden_posts.post_id.
In my posts model I have this relation to return a record if the post should be hidden:
public function getHiddenPosts()
{
return $this->hasOne(HiddenPost::className(), ['post_id' => 'id']);
}
Now I need to return all posts that are NOT hidden. So I am looking for the equivalent of this pseudo code:
return $this->hasNone(HiddenPost::className(), ['post_id' => 'id'])->all();
Which is saying, "show me all posts that are not in the hidden_posts table".
So does this use an outer join query or is there a statement that I can't find do do this in one line?
You can do it this way. Get all posts that are not listed in Hidden table:
$posts = Post::find()
->andFilterWhere(['not in',
'post.id',
HiddenPost::find()
->select(['hidden_post.post_id'])
->all();
In any case, it is best to proceed from the raw SQL statement. Write a statement that satisfies your results and transfer it to ActiveRecord query.
Post items could be retrieved using an inner join
$res = Post::find()
->select('post.*')
->innerJoin('hdn_post', '`post`.`id` = `hdn_post`.`post_id`')
->all();
It could be good practice using yii2 owned function instead of adding queries inside the model such as using select queries in your model.
Instead you can use ORM functions in yii2 has already done by gii inner functions created for to make u=your work easy.
Add * #property YourModel $hidden_post
and inside this model add you post_id such as ( * #property integer $post_id ) to create relation.
public function getHiddenPosts($hidden_post) {
return $this->find()->joinWith('hidden_post')
->where(['hidden_post' => $hidden_post])
->all();
}
You could retrive the Post items using an inner join
$posts = Post::find()
->select('post.*')
->innerJoin('hidden_post', '`post`.`id` = `hidden_post`.`post_id`')
->all();
for not hidden then use left join and check for null result of related table
$posts = Post::find()
->select('post.*')
->leftJoin('hidden_post', '`post`.`id` = `hidden_post`.`post_id`')
->where('ISNULL(`hidden_post`.console_id)')
->all();

get user with most articles and its amount with active record or query

I want to get the user that wrote the most articles. I do so fine in two ways with ActiveRecord like the following:
$table = Articles::find()
->select('articles.*, COUNT(*) AS cnt')
->with('user','userDetails')
->groupBy('articles.user_id')
->orderBy(('cnt DESC'))
->limit(10)
->offset($offset)
->all();
and with a query like the following:
$query = (new Query())
->select('articles.user_id, COUNT(*) AS num_articles')
->from('articles')
->join('LEFT JOIN', 'user_details', 'user_details.user_id = articles.user_id')
->groupBy('articles.user_id')
->orderBy('num_articles DESC')
->limit(10)
->offset($offset)
->all();
The problem is that the ActiveRecord gives me further needed informations userDetails that I need. But I do not get the amount of articles of user that should be on cnt
With the Query I get the user_id and the amount of articles. But I do not get it working by joining with userDetails. All of these does not work: LEFT JOIN, RIGHT JOIN, INNER JOIN.
I am interested in resolving both for learning, but for concrete I need help with the ActiveRecord problem.
Okay well I solved it for the ActiveRecord. The ActiveRecords needs a public $var; in the Model. So to get the amount you have to add the mentioned public... to your Model so that in my case:
public $cnt; extends the ActiveRecord of Articles
now I can access it with the given Request in my Question. But this just solves the first point. I am still interested in the second way for more complex Queries.
I dont have much idea about active record but I think the below is something what you are looking for
select * from user_details where user_id in
(select A.user from
(select user_id as user, COUNT(*) AS num_articles
from articles group by user_id order by num_articles desc LIMIT 10)A
);
for second point you should include required column from joined table to select statement:
$query = (new Query())
->select('articles.user_id, COUNT(*) AS num_articles, user_details.username as username')
->from('articles')
->join('LEFT JOIN', 'user_details', 'user_details.user_id = articles.user_id')
->groupBy('articles.user_id')
->orderBy('num_articles DESC')
->limit(10)
->offset($offset)
->all();

Laravel many to many loading related models with count

I am trying to link 4 tables and also add a custom field calculated by counting the ids of some related tables using laravel.
I have this in SQL which does what I want, but I think it can be made more efficient:
DB::select('SELECT
posts.*,
users.id AS users_id, users.email,users.username,
GROUP_CONCAT(tags.tag ORDER BY posts_tags.id) AS tags,
COUNT(DISTINCT comments.id) AS NumComments,
COUNT(DISTINCT vote.id) AS NumVotes
FROM
posts
LEFT JOIN comments ON comments.posts_id = posts.id
LEFT JOIN users ON users.id = posts.author_id
LEFT JOIN vote ON vote.posts_id = posts.id
LEFT JOIN posts_tags ON posts_tags.posts_id = posts.id
LEFT JOIN tags ON tags.id = posts_tags.tags_id
GROUP BY
posts.id,
posts.post_title');
I tried to implement it using eloquent by doing this:
$trending=Posts::with(array('comments' => function($query)
{
$query->select(DB::raw('COUNT(DISTINCT comments.id) AS NumComments'));
},'user','vote','tags'))->get();
However the NumComments value is not showing up in the query results.
Any clue how else to go about it?
You can't do that using with, because it executes separate query.
What you need is simple join. Just translate the query you have to something like:
Posts::join('comments as c', 'posts.id', '=', 'c.id')
->selectRaw('posts.*, count(distinct c.id) as numComments')
->groupBy('posts.id', 'posts.post_title')
->with('user', 'vote', 'tags')
->get();
then each post in the collection will have count attribute:
$post->numComments;
However you can make it easier with relations like below:
Though first solution is better in terms of performance (might not be noticeable unless you have big data)
// helper relation
public function commentsCount()
{
return $this->hasOne('Comment')->selectRaw('posts_id, count(*) as aggregate')->groupBy('posts_id');
}
// accessor for convenience
public function getCommentsCountAttribute()
{
// if relation not loaded already, let's load it now
if ( ! array_key_exists('commentsCount', $this->relations)) $this->load('commentsCount');
return $this->getRelation('commentsCount')->aggregate;
}
This will allow you to do this:
$posts = Posts::with('commentsCount', 'tags', ....)->get();
// then each post:
$post->commentsCount;
And for many to many relations:
public function tagsCount()
{
return $this->belongsToMany('Tag')->selectRaw('count(tags.id) as aggregate')->groupBy('pivot_posts_id');
}
public function getTagsCountAttribute()
{
if ( ! array_key_exists('tagsCount', $this->relations)) $this->load('tagsCount');
$related = $this->getRelation('tagsCount')->first();
return ($related) ? $related->aggregate : 0;
}
More examples like this can be found here http://softonsofa.com/tweaking-eloquent-relations-how-to-get-hasmany-relation-count-efficiently/
as of laravel 5.3 you can do this
withCount('comments','tags');
and call it like this
$post->comments_count;
laravel 5.3 added withCount

Laravel 4 select column from another table in subquery

I am attempting to do the equivalent of this:
select p.id, p.title, b.brand,
(select big from images where images.product_id = p.id order by id asc limit 1) as image
from products p
inner join brands b on b.id = p.brand_id
Here is where I am at now, but it of course doesn't work:
public function getProducts($brand)
{
// the fields we want back
$fields = array('p.id', 'p.title', 'p.msrp', 'b.brand', 'p.image');
// if logged in add more fields
if(Auth::check())
{
array_push($fields, 'p.price_dealer');
}
$products = DB::table('products as p')
->join('brands as b', 'b.id', '=', 'p.brand_id')
->select(DB::raw('(select big from images i order by id asc limit 1) AS image'), 'i.id', '=', 'p.id')
->where('b.active', '=', 1)
->where('p.display', '=', 1)
->where('b.brand', '=', $brand)
->select($fields)
->get();
return Response::json(array('products' => $products));
}
I don't really see anything in the docs on how to do this, and I can't seem to piece it together from other posts.
In "regular" SQL, the subquery is treated AS a column, but I am not sure how to string that together here. Thanks for any help on this.
I strongly recommend you to use Eloquent, instead of pure SQL. It's one of the most beautful things in Laravel. Two models and relations and it's done! If you need to use pure SQL like that, put it all in DB::raw. It's easier, simpler and (ironically) less messy!
With the models, you could use relations between the two tables (represented by the models itself) and say (so far I understood) that Brands belongs to Products, and Images belongs to Product. Take a look at Eloquent's documentation on Laravel. Probably will be more clearly.
Once the relations are done, you can only say that you wanna get
$product = Product::where(function ($query) use ($brand){
$brand_id = Brand::where('brand', '=', $brand)->first()->id;
$query->where('brand_id', '=', $brand_id);
})
->image()
->get();
That and a better look at Eloquent's documentation should help you to do the job.
P.S.: I didn't test the code before send it and wrote it by head, but i think it works.

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