I have three tables
Job Model
+---------------------------+
| id | name |
+---------------------------+
| 1 | web design |
+---------------------------+
| 2 | desktop development |
+---------------------------+
Applicant Model
+----------------------------------------------------------+
| id | job_id | user_id | desc |
+----------------------------------------------------------+
| 1 | 2 | 1 | I am an expert developer |
+----------------------------------------------------------+
| 2 | 2 | 2 | I am good |
+----------------------------------------------------------+
User Model
+----------------+
| id | name |
+----------------+
| 1 | john |
+----------------+
| 2 | steve |
+----------------+
Using eloquent laravel,
How do I get applicants with job_id "2" and their user name together.
Tried a lot but unfortunately to no avail.
You would use a query something like this:
$jobs = Job::where('id', 2)
->join('applicants', 'jobs.id', '=', 'applicants.job_id')
->join('users', 'applicants.user_id', '=', 'users.id')
->select(['user.name', 'job.name', 'applicant.desc'])
->get();
You will have to change it to your requirements but should about what you need.
You can get more info here:
https://laravel.com/docs/5.7/queries
Hope that helps
Related
I am trying to figure out how to make this query work. These two tables do not have a direct relation (i.e. hasOne, hasMany, etc). I am looking to only get back records from client_vendor_relationship that do NOT have a collection_opt_in. Since I don't have a way of doing ->whereDoesntHave(), I am not sure how to get this data back.
client_vendor_relationship
| id | client_id | vendor_id | active |
|----|-----------|-----------|--------|
| 1 | 23484 | 1872 | 1 |
| 2 | 5643 | 345 | 1 |
| 3 | 431 | 4443 | 1 |
collection_opt_in
| id | client_id | vendor_id | year |
|----|-----------|-----------|------|
| 1 | 23484 | 23484 | 2020 |
| 2 | 23484 | 23484 | 2019 |
| 3 | 431 | 4443 | 2019 |
Current Query
$relationships = ClietVendorRelationship::where('active', 1)
->leftJoin('collection_opt_in', function($join){
$join->on('client_vendor_relationship.client_id', '=', 'collection_opt_in.client_id')
->on('client_vendor_relationship.vendor_id', '=', 'collection_opt_in.vendor_id');
})
->where() // Not sure what to put here to ONLY get rows back that the left join didnt find entries for
->groupBy('client_vendor_relationship.id')
->get();
The end goal of this above query would to be to only get the row back from client_vendor_relationship with the id of 2. I need to do this in Eloquent. I know I could easily just do a collection filter but the front end table I am using requires an eloquent query to be returned.
why using left join? while there is another straightforward ways?
you can use 'whereNotExists':
$relationships = ClietVendorRelationship::where('active', 1)
->whereNotExists(function ( $query){
$query->select('collection_opt_in.id')->from('collection_opt_in')->
whereColumn('collection_opt_in.client_id', '=', 'client_vendor_relationship.client_id')->
whereColumn('collection_opt_in.vendor_id', '=', 'client_vendor_relationship.vendor_id');
})
->groupBy('client_vendor_relationship.id')
->get();
also i must say i don't recommend using group by without aggregation
I have this following 2 tables:
Table: Products
+----+---------+-------------+
| id | fn_id | created_at |
+----+---------+-------------+
| 1 | 4 | SOME TIME |
| 2 | 5 | SOME TIME |
| 3 | 6 | SOME TIME |
| 4 | 10 | SOME TIME |
| 5 | 10 | SOME TIME |
+----+---------+-------------+
Table Fn
+----+---------+-------------+
| id | fn_id | created_at |
+----+---------+-------------+
| 1 | 10 | SOME TIME |
| 2 | 11 | SOME TIME |
| 3 | 12 | SOME TIME |
| 4 | 14 | SOME TIME |
+----+---------+-------------+
And a User Input which is giving me a timestamp ($user_timestamp).
Now I need to get all produtcs, where
products.fn_id is 10
fn.fn_id is 10
fn.created == $user_timestamp
The products model has this relation:
public function fn() {
return $this->hasMany('App\FN', 'fn_id', 'fn_id');
}
Now I've tried multiple things like a where query where I want to check if the fn_id on both are "10" and the created_at value of the fn table is equal to $user_timestamp.
However, I wasn't able to do it.
$products = Products::with('fn')->where('fn.fn_id', function ($query) use ($user_timestamp) {
$query->where([['fn_id', 10],['created_at', $user_timestamp]]);
})->get();
You would have to use whereHas to constrain Products based on fn.
$products = Products::with('fn')->whereHas('fn', function($q) use($user_timestamp){
$q->where('fn_id', 10)->where('created_at', $user_timestamp);
})->get();
try this
$id_to_search=10;
$products = Products::with('fn')->where('fbn_id',$id_to_search)->whereHas('fn', function($q) use($user_timestamp,$id_to_search){
$q->where('fn_id', $id_to_search)->where('created_at', $user_timestamp);
})->get();
I'm doing a Laravel query to query a list of elements that his relationship is grater than 0.
The table concerts:
| ID | NAME |
|----|-----------|
| 1 | Concert A |
| 2 | Concert B |
| 3 | Concert C |
And the Position table.
| ID | concert_id | user_id | Content |
|----|----------------|------------|----------|
| 1 | 1 | 1 | xxx |
| 2 | 1 | 2 | yyy |
| 3 | 3 | 1 | zzz |
| 4 | 3 | 2 | www |
| 5 | 1 | 3 | xyx |
| 6 | 3 | 3 | rer |
The query that I need to do is, get the concerts where their position has in the content a value like $some_query$.
The code I've is the following:
$result = $this->model->whereHas('positions')
->with(['positions' => function ($query) {
$query->where('content', 'like', "%{$content}%");
}])->get();
But as far as I can tell, this will bring all the concerts, and also this is going to bring only the positions that has the desired content.
So I've two problems, the first one is I need to get the concerts that his queried positions are greather than 0.
And the second one is that also I need to bring all the positions, not only the queried ones.
Basically the query is just a way to know which concerts I need to bring.
Is there possible to achieve this on a single query?
You'll have to move your query to the whereHas.
If I understood well, this is what you want:
$result = $this->model
// Take the Concerts that has positions with a similar content
->whereHas('positions', function ($query) use ($content) {
$query->where('content', 'like', "%{$content}%");
})
// Take (all) the Positions of these concerts
->with('positions')
->get();
Now I'm using the query below in hive to split a row into multiple rows, but I also want to group a "Product" column based on "Category" column each group will match by the order of the group and have ";" to sperate each group and have "," separate item in the group.
SELECT id, customer, prodcut_split
FROM orders lateral view explode(split(product,';')) products AS prodcut_split
Here is my data look like now
| id | Customer| Category | Product |
+----+----------+---------------------------+-----------------------------------+
| 1 | John | Furniture; Technology | Bookcases, Chairs; Phones, Laptop |
| 2 | Bob | Office supplies; Furniture| Paper, Blinders; Tables |
| 3 | Dylan | Furniture | Tables, Chairs, Bookcases |
my desired result will look like:
| id | Customer| Category | Product |
+----+----------+----------------+-----------+
| 1 | John | Furniture | Bookcases |
| 1 | John | Furniture | Chairs |
| 1 | John | Technology | Phones |
| 1 | John | Technology | Laptop |
| 2 | Bob | Office supplies| Paper |
| 2 | Bob | Office supplies| Blinders |
| 2 | Bob | Furniture | Tables |
| 3 | Dylan | Furniture | Tables |
| 3 | Dylan | Furniture | Chairs |
| 3 | Dylan | Furniture | Bookcases |
I have tried this one and it's work well, all credit goes to this question: Hive - Split delimited columns over multiple rows, select based on position
select id,customer ,category, products
from
(
SELECT id, category, product
FROM tale_name
lateral VIEW posexplode(split(category,';')) category AS pos_category, category_split
lateral VIEW posexplode(split(product,';')) product AS pos_product, product_split
WHERE pos_category = pos_product) a
lateral view explode(split(product_split,',')) product_split AS products
i need help with a query, can't get it to work correctly. What i'm trying to achieve is to have a select box displaying the number of records associated with a particular theme, for some theme it works well for some it displays (0) when infact there are 2 records, I'm wondering if someone could help me on this, your help would be greatly appreciated, please see below my actual query + table structure :
SELECT theme.id_theme, theme.theme, calender.start_date,
calender.id_theme1,calender.id_theme2, calender.id_theme3, COUNT(*) AS total
FROM theme, calender
WHERE (YEAR(calender.start_date) = YEAR(CURDATE())
AND MONTH(calender.start_date) > MONTH(CURDATE()) )
AND (theme.id_theme=calender.id_theme1)
OR (theme.id_theme=calender.id_theme2)
OR (theme.id_theme=calender.id_theme3)
GROUP BY theme.id_theme
ORDER BY theme.theme ASC
THEME table
|---------------------|
| id_theme | theme |
|----------|----------|
| 1 | Yoga |
| 2 | Music |
| 3 | Taichi |
| 4 | Dance |
| 5 | Coaching |
|---------------------|
CALENDAR table
|---------------------------------------------------------------------------|
| id_calender | id_theme1 | id_theme2 | id_theme3 | start_date | end_date |
|-------------|-----------|-----------|-----------|------------|------------|
| 1 | 2 | 4 | | 2015-07-24 | 2015-08-02 |
| 2 | 4 | 1 | 5 | 2015-08-06 | 2015-08-22 |
| 3 | 1 | 3 | 2 | 2014-10-11 | 2015-10-28 |
|---------------------------------------------------------------------------|
LISTBOX
|----------------|
| |
| Yoga (1) |
| Music (1) |
| Taichi (0) |
| Dance (2) |
| Coaching (1) |
|----------------|
Thanking you in advance
I think that themes conditions should be into brackets
((theme.id_theme=calender.id_theme1)
OR (theme.id_theme=calender.id_theme2)
OR (theme.id_theme=calender.id_theme3))
Hope this help