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Hi I want to know how can i do this query in Laravel 8 , I tried adding the join clause but not work as expected, i need join clause? Or maybe there is another form to do it. I search others examples but i don´t see anythat help me. The query is the next:
DB::table('escandallo_p as esc')
->select("esc.material", "esc.referencia", "esc.ancho", "esc.proveedor", "esc.punto",
"esc.precio", "esc.consumo", "esc.veces", "esc.001", "esc.002", "esc.003", "esc.004",
"esc.005", "esc.006", "esc.007", "esc.008", "esc.009", "esc.010", "esc.011", "esc.um", "esc.merma", "esc.importe", "esc.tipo", "esc.xtalla", "esc.fase",
DB::raw("(select anulado from prototipos_p as p where p.prototipo = '".$scandal[0]->prototipo."' and p.tipo = 'c' and p.referencia = esc.referencia )"),
// ignore
//original query "(select anulado from prototipos_p as p where p.prototipo = ",$request->prototipo," and p.tipo = 'c' and p.referencia = esc.referencia ) as 'anulado'",
// "(select clase from prototipos_p as p where p.prototipo = ",$request->prototipo," and p.tipo = 'c' and p.referencia = esc.referencia ) as 'clase'")
//Converted query ->select('pro.anulado')->where('pro.prototipo', $request->prototipo)
// ->where("p.prototipo", "=", $request->prototipo)
->where("esc.id_escandallo", "=", $request->id_escandallo)
->where("esc.id_version", "=", $request->version)
->orderBy("id","asc")
->get();
!!!! I need to pass the esc.referencia to the sub select query
The second select is the conversion of the select inside "" ( i know this is wrong is only for explain it).
Thank you in advance for any suggestion.
Best regards
EDIT: I can solve my problem with DB::raw, but if anyone know others methos are welcome!
You need to pass callback to the join query to add the extra query to the laravel's join method,
Example from Laravel Doc:
DB::table('users')
->join('contacts', function ($join) {
$join->on('users.id', '=', 'contacts.user_id')
->where('contacts.user_id', '>', 5);
})
->get();
It is explained in Laravel's doc, Advanced Join Clauses
There is Subquery support too Subquery Joins,
Eg:
$latestPosts = DB::table('posts')
->select('user_id', DB::raw('MAX(created_at) as last_post_created_at'))
->where('is_published', true)
->groupBy('user_id');
$users = DB::table('users')
->joinSub($latestPosts, 'latest_posts', function ($join) {
$join->on('users.id', '=', 'latest_posts.user_id');
})
->get();
These two might help you to achieve what you are trying
After test joins, joinSub, whereIn and other forms of doing this, I solved my problem using the DB::raw():
DB::table('escandallo_p as esc')
->select('parameters',....,
DB::raw("(SELECT column //(ONLY ONE)
FROM table
WHERE column = '".$parameter."' ...) AS nombre"),
)
->where('column', "=", $parameter)
->orderBy("id","asc")
->get();
I am attempting to build a query on the fly depending upon what is entered in a form. When I do so, I check what the sql statement is in the Console. I have OR and AND contitions, so I use a call back to build the query. the SQL statement that results is:
select `id`, `activity_name`, `activity_name_web`, `year`, `season`, `sex`,
`active_on_web`, `begin_active_on_web`, `end_active_on_web`, `begin_age`,
`end_age`, `begin_date`, `end_date` from `activities` where (season='fall'
OR season='summer' is null) order by `begin_date` desc, `activity_name` asc
limit 25 offset 0 []
notice the words "is null". I can't determine how this phrase is being added.
The Code:
$qry_where = "1";
$arry_Season = array();
if($this->a_fall)
$arry_Season[] = "LOWER(`season`)='fall'";
if($this->a_spring)
$arry_Season[] = "LOWER(`season`)='spring'";
if($this->a_summer)
$arry_Season[] = "LOWER(`season`)='summer'";
if($this->a_winter)
$arry_Season[] = "LOWER(`season`)='winter'";
if(count($arry_Season)>0) {
$qry_where = $arry_Season[0];
if(count($arry_Season)>1){
array_shift($arry_Season);
foreach($arry_Season as $season){
$qry_where .= " OR " . $season;
}
}
}
$activities = Activity::select('id','activity_name','activity_name_web','year','season', 'sex', 'active_on_web',
'begin_active_on_web', 'end_active_on_web', 'begin_age', 'end_age', 'begin_date','end_date')
->where(function($query) use ($qry_where) {
if($qry_where == "1")
$query->where('id','>',0);
else {
$query->where(DB::raw($qry_where));
}
})
->orderBy('begin_date','desc')->orderBy('activity_name','asc')
->paginate(25);
So, if the users checks the box labelled "fall", I add it to the query builder.
so - Where is the phrase "is null" coming from.
select `id`, `activity_name`, `activity_name_web`, `year`, `season`, `sex`,
`active_on_web`,`begin_active_on_web`, `end_active_on_web`, `begin_age`,
`end_age`, `begin_date`, `end_date` from `activities`
where (season='fall' OR season='summer' is null)
order by `begin_date` desc, `activity_name` asc limit 25 offset 0 []
btw, say I check fall and summer, then I ddd($qry_where) just before I build the query, I get this:
"LOWER(`season`)='fall' OR LOWER(`season`)='summer'"
I also need to mention that I am using a livewire component.
thanks for the help in advance.
#apokryfos thank you for your quick answer. It was driving me crazy. It seems when I passed a variable into the ->where(DB::raw()), it evaluated the variable too late for the ->when() to see that the variable contained a comparison?
In any case, I tweeked the solution that #apokryfos a bit and came up with the following:
$activities = Activity::select('id','activity_name','activity_name_web','year','season', 'sex', 'active_on_web',
'begin_active_on_web', 'end_active_on_web', 'begin_age', 'end_age', 'begin_date','end_date',
'cost_member','cost_non_member')
->when(count($seasons) > 0, function($query) use ($seasons) {
$query->whereIn('season', $seasons);
})
->orderBy('begin_date','desc')->orderBy('activity_name','asc')
->paginate($this->items_on_page);
the ->whereIn() could not process the "DB::raw('LOWER(season)', $seasons)" as suggested. If i want to make sure that the data in the [activities].[season] column, I will either have to make sure the seasons are input into the table in lower case, or do a ->select(DB:raw('LOWER(season)') in the SQL statement. I like the first option.
thank you again for your input apokryfos
You did a where(<raw query>) which defaults to do a WHERE <raw query> is not NULL since WHERE assumes a comparison. Here's what I think is a more concise version of your query:
$seasons = array_filter([
$this->a_spring ? 'spring' : null,
$this->a_summer ? 'summer' : null,
$this->a_fall ? 'fall' : null
$this->a_winter ? 'winter' : null
]);
$activities = Activity::select('id','activity_name','activity_name_web','year','season', 'sex', 'active_on_web',
'begin_active_on_web', 'end_active_on_web', 'begin_age', 'end_age', 'begin_date','end_date')
->when(count($seasons) > 0, function($query) use ($seasons) {
$query->whereIn(DB::raw('LOWER(season)'), $seasons);
})->when(count($seasons) == 0, function ($query) {
$query->where('id', > 0); // Not sure you need this part
})
->orderBy('begin_date','desc')->orderBy('activity_name','asc')
->paginate(25);
I have 2 queries and I would like to know the elements (date_debut) of the second query which exists in the first query (dateincal). The elements of the second query can appear one or more times in the first query.
once the dates (date debut) (dateincal) have been found i want to be able to then retrieve also the other information for the element found
$feries = Jourferie::Select('dateincal', 'description')
->where('dateincal', '>=', Carbon::now()->startOfMonth())
->where('dateincal', '<=', Carbon::now()->endOfMonth())
->get();
$plannings = Planning::Select('date_debut', 'id', 'agent_id', 'site_id')
->where('id', '!=', 0)
->where('statut', 'provisoire')
->get();
I don't know if that way can help you:
foreach($feries as $ferie){
$myresult = $plannings->contains('date_debut',$ferie->dateincal)
/* do things */
}
I have a table that contains, amongst other columns, a column of browser versions. And I simply want to know from the record-set, how many of each type of browser there are. So, I need to end up with something like this: Total Records: 10; Internet Explorer 8: 2; Chrome 25: 4; Firefox 20: 4. (All adding up to 10)
Here's my two pence:
$user_info = Usermeta::groupBy('browser')->get();
Of course that just contains the 3 browsers and not the number of each. How can I do this?
This is working for me:
$user_info = DB::table('usermetas')
->select('browser', DB::raw('count(*) as total'))
->groupBy('browser')
->get();
This works for me (Laravel 5.1):
$user_info = Usermeta::groupBy('browser')->select('browser', DB::raw('count(*) as total'))->get();
Thanks Antonio,
I've just added the lists command at the end so it will only return one array with key and count:
Laravel 4
$user_info = DB::table('usermetas')
->select('browser', DB::raw('count(*) as total'))
->groupBy('browser')
->lists('total','browser');
Laravel 5.1
$user_info = DB::table('usermetas')
->select('browser', DB::raw('count(*) as total'))
->groupBy('browser')
->lists('total','browser')->all();
Laravel 5.2+
$user_info = DB::table('usermetas')
->select('browser', DB::raw('count(*) as total'))
->groupBy('browser')
->pluck('total','browser');
If you want to get collection, groupBy and count:
$collection = ModelName::groupBy('group_id')
->selectRaw('count(*) as total, group_id')
->get();
Cheers!
Open config/database.php
Find strict key inside mysql connection settings
Set the value to false
Works that way as well, a bit more tidy.
getQuery() just returns the underlying builder, which already contains the table reference.
$browser_total_raw = DB::raw('count(*) as total');
$user_info = Usermeta::getQuery()
->select('browser', $browser_total_raw)
->groupBy('browser')
->pluck('total','browser');
Try with this
->groupBy('state_id','locality')
->havingRaw('count > 1 ')
->having('items.name','LIKE',"%$keyword%")
->orHavingRaw('brand LIKE ?',array("%$keyword%"))
$post = Post::select(DB::raw('count(*) as user_count, category_id'))
->groupBy('category_id')
->get();
This is an example which results count of post by category.
Laravel Version 8
Removed the dependency of DB
$counts = Model::whereIn('agent_id', $agents)
->orderBy('total', 'asc')
->selectRaw('agent_id, count(*) as total')
->groupBy('agent_id')
->pluck('total','agent_id')->all();
Another way would be this:
$data = Usermeta::orderBy('browser')->selectRaw('browser, count(*) as total')->get()
Since this is the top result when i search for eloquent count with groupby returns only first
using "illuminate/database": "^9.38" in composer. so "should" be the latest at the time of this post
I honestly have no idea why they think that returning the first record is the right option.. ie
IMHO the current implementation doesn't make sense for queries including groupBy statements.
Why should be N_1 be the "right" result?
Since #taylorotwell rightly pointed out some performance issues with counting the subquery results, why don't we fix that on the php side, by checking if there are any group statements, and if so, performing a N_1 + N_2 + .... + N_M ?
https://github.com/laravel/ideas/issues/1693#issuecomment-621167890
Wrapping the query and doing a count seems to work for me
$records = ...
$record_count = DB::table( "fake" );
$record_count->fromSub($records->select(DB::raw(1)),"query");
$record_count->count();
create a "fake" query builder
add a sub "from" from (...) query set the "sub query" to select 1 instead of returning huge column data. not sure if this is needed but in my mind it seems like a good idea
do the normal ->count()
returns the "expected" result since it executes:
select count(*) as aggregate from (select 1 from ... group by ...) as `query`
The sub is the "query" that $records would normaly execute
My use-case is for pagination (not using laravel). so i get the record count then pass it to the paginator then call ->forPage()->get()
https://github.com/laravel/framework/issues/44081#issuecomment-1301816710
Laravel Eloquent query that uses the GROUP BY clause with advanced aggregation functions and conditional statements.
GroupBy on one column.
$data = Employee::select(
'department',
DB::raw('SUM(salary) as total_salary'),
DB::raw('COUNT(*) as total_employees'),
DB::raw('SUM(IF(bonus > 13000, 1, 0)) as employees_with_bonus')
)
->groupBy('department')
->havingRaw('total_employees > 5 AND total_salary > 10000')
->orHavingRaw('department_rank LIKE ?', array("%$keyword%"))
->get();
In the above query, if the bonus is greater than 13000, the IF function returns 1 otherwise it returns 0.
GroupBy on two columns: The groupBy method takes multiple arguments
$data = Employee::select(
'department',
'location',
DB::raw('SUM(salary) as total_salary'),
DB::raw('COUNT(*) as total_employees'),
DB::raw('SUM(IF(bonus > 1000, 1, 0)) as employees_with_bonus')
)
->groupBy('department', 'location')
->havingRaw('total_employees > 5 AND total_salary > 10000')
->get();
GroupBy and JOIN: Laravel Eloquent query that joins two tables and uses grouping and aggregate functions. (inner join)
$data = Employee::select(
'employees.department',
'employees.location',
DB::raw('SUM(employees.salary) as total_salary'),
DB::raw('COUNT(*) as total_employees'),
DB::raw('SUM(IF(employees.bonus > 1000, 1, 0)) as employees_with_bonus')
)
->join('departments', 'employees.department', '=', 'departments.name')
->groupBy('employees.department', 'employees.location')
->havingRaw('total_employees > 5 AND total_salary > 10000')
->get();
Raw MySQL Query:
SELECT
department,
SUM(salary) AS total_salary,
COUNT(*) AS total_employees,
SUM(IF(bonus > 1000, 1, 0)) AS employees_with_bonus
FROM
employees
GROUP BY
department
HAVING
total_employees > 5 AND total_salary > 13000;
If you want to get sorted data use this also
$category_id = Post::orderBy('count', 'desc')
->select(DB::raw('category_id,count(*) as count'))
->groupBy('category_id')
->get();
In Laravel 8 you can use countBy() to get the total count of a group.
Check the documentation on the same.
https://laravel.com/docs/8.x/collections#method-countBy
Simple solution(tested with Laravel 9 and Spatie/Permissions).
Controller:
//Get permissions group by guard name(3 in my case: web, admin and api)
$permissions = Permission::get()->groupBy('guard_name');
View:
#foreach($permissions as $guard => $perm)
<div class="form-group">
<label for="permission">Permissions ({{ ucfirst($guard) }}) {{ count($perm) }}</label>
<select name="permission[]" id="permission" class="form-control #error('permission') is-invalid #enderror" multiple>
#foreach($perm as $value)
<option value="{{ $value->id }}">{{ $value->name }}</option>
#endforeach
</select>
#error('permission')
<div class="invalid-feedback">
{{ $message }}
</div>
#enderror
</div>
#endforeach
Here is a more Laravel way to handle group by without the need to use raw statements.
$sources = $sources->where('age','>', 31)->groupBy('age');
$output = null;
foreach($sources as $key => $source) {
foreach($source as $item) {
//get each item in the group
}
$output[$key] = $source->count();
}
I'm trying to group my entity by a field (year) and do a count of it.
Code:
public function countYear()
{
$qb = $this->getEntityManager()->createQueryBuilder();
$qb->select('b.year, COUNT(b.id)')
->from('\My\Entity\Album', 'b')
->where('b.year IS NOT NULL')
->addOrderBy('sclr1', 'DESC')
->addGroupBy('b.year');
$query = $qb->getQuery();
die($query->getSQL());
$result = $query->execute();
//die(print_r($result));
return $result;
}
I can't seem to say COUNT(b.id) AS count as it gives an error, and
I do not know what to use as the addOrderby(???, 'DESC') value?
There are many bugs and workarounds required to achieve order by expressions as of v2.3.0 or below:
The order by clause does not support expressions, but you can add a field with the expression to the select and order by it. So it's worth repeating that Tjorriemorrie's own solution actually works:
$qb->select('b.year, COUNT(b.id) AS mycount')
->from('\My\Entity\Album', 'b')
->where('b.year IS NOT NULL')
->orderBy('mycount', 'DESC')
->groupBy('b.year');
Doctrine chokes on equality (e.g. =, LIKE, IS NULL) in the select expression. For those cases the only solution I have found is to use a subselect or self-join:
$qb->select('b, (SELECT count(t.id) FROM \My\Entity\Album AS t '.
'WHERE t.id=b.id AND b.title LIKE :search) AS isTitleMatch')
->from('\My\Entity\Album', 'b')
->where('b.title LIKE :search')
->andWhere('b.description LIKE :search')
->orderBy('isTitleMatch', 'DESC');
To suppress the additional field from the result, you can declare it AS HIDDEN. This way you can use it in the order by without having it in the result.
$qb->select('b.year, COUNT(b.id) AS HIDDEN mycount')
->from('\My\Entity\Album', 'b')
->where('b.year IS NOT NULL')
->orderBy('mycount', 'DESC')
->groupBy('b.year');
what is the error you get when using COUNT(b.id) AS count? it might be because count is a reserved word. try COUNT(b.id) AS idCount, or similar.
alternatively, try $qb->addOrderby('COUNT(b.id)', 'DESC');.
what is your database system (mysql, postgresql, ...)?
If you want your Repository method to return an Entity you cannot use ->select(), but you can use ->addSelect() with a hidden select.
$qb = $this->createQueryBuilder('q')
->addSelect('COUNT(q.id) AS HIDDEN counter')
->orderBy('counter');
$result = $qb->getQuery()->getResult();
$result will be an entity class object.
Please try this code for ci 2 + doctrine 2
$where = " ";
$order_by = " ";
$row = $this->doctrine->em->createQuery("select a from company_group\models\Post a "
.$where." ".$order_by."")
->setMaxResults($data['limit'])
->setFirstResult($data['offset'])
->getResult();`