Function or Variable in Where Statement?
1
Is there a way to put a function or variable in the first value of a where statement?
For example,
$equipments=DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
*->where($this->countOrders('equipments.equipmentID'),'<','1')*
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
The count orders function is
public static function countOrders($id)
{
//
$members=DB::table('orders')
->leftjoin('users', 'users.id', '=', 'orders.orderUser_id')
->leftjoin('equipments', 'equipments.equipmentID', '=', 'orders.equipment_id')
->where('orders.equipment_id','=', $id)
->where('orders.status','=','ACCEPTED')
->count();
return $members;
}
php mysql laravel
asked Nov 14 '18 at 9:17
John Beltran
264
add a comment |
1
Is there a way to put a function or variable in the first value of a where statement?
For example,
$equipments=DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
*->where($this->countOrders('equipments.equipmentID'),'<','1')*
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
The count orders function is
public static function countOrders($id)
{
//
$members=DB::table('orders')
->leftjoin('users', 'users.id', '=', 'orders.orderUser_id')
->leftjoin('equipments', 'equipments.equipmentID', '=', 'orders.equipment_id')
->where('orders.equipment_id','=', $id)
->where('orders.status','=','ACCEPTED')
->count();
return $members;
}
php mysql laravel
asked Nov 14 '18 at 9:17
John Beltran
264
add a comment |
1
1
1
Is there a way to put a function or variable in the first value of a where statement?
For example,
$equipments=DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
*->where($this->countOrders('equipments.equipmentID'),'<','1')*
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
The count orders function is
public static function countOrders($id)
{
//
$members=DB::table('orders')
->leftjoin('users', 'users.id', '=', 'orders.orderUser_id')
->leftjoin('equipments', 'equipments.equipmentID', '=', 'orders.equipment_id')
->where('orders.equipment_id','=', $id)
->where('orders.status','=','ACCEPTED')
->count();
return $members;
}
php mysql laravel
asked Nov 14 '18 at 9:17
John Beltran
264
Is there a way to put a function or variable in the first value of a where statement?
For example,
$equipments=DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
*->where($this->countOrders('equipments.equipmentID'),'<','1')*
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
The count orders function is
public static function countOrders($id)
{
//
$members=DB::table('orders')
->leftjoin('users', 'users.id', '=', 'orders.orderUser_id')
->leftjoin('equipments', 'equipments.equipmentID', '=', 'orders.equipment_id')
->where('orders.equipment_id','=', $id)
->where('orders.status','=','ACCEPTED')
->count();
return $members;
}
php mysql laravel
php mysql laravel
asked Nov 14 '18 at 9:17
John Beltran
264
asked Nov 14 '18 at 9:17
John Beltran
264
asked Nov 14 '18 at 9:17
John Beltran
264
asked Nov 14 '18 at 9:17
John Beltran
264
asked Nov 14 '18 at 9:17
John Beltran
264
264
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
You can use "whereRaw" method to do that.
$equipments = DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
->whereRaw('? < ?', [$this->countOrders('equipments.equipmentID'), 1])
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
answered Nov 14 '18 at 9:26
Kenny
19025
It's still outputting the equipment, despite having a value of 1 when I print out the result of the function in the view. Maybe the function in the whereRaw isn't actually printing a number, or something else, :/
– John Beltran
Nov 14 '18 at 10:01
I tested by inputting just random ids into the function parameter, and it seems to work, but when run with equipments.equipmentID, it outputs everything, even the ids that I worked on my test
– John Beltran
Nov 14 '18 at 10:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You can use "whereRaw" method to do that.
$equipments = DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
->whereRaw('? < ?', [$this->countOrders('equipments.equipmentID'), 1])
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
answered Nov 14 '18 at 9:26
Kenny
19025
It's still outputting the equipment, despite having a value of 1 when I print out the result of the function in the view. Maybe the function in the whereRaw isn't actually printing a number, or something else, :/
– John Beltran
Nov 14 '18 at 10:01
I tested by inputting just random ids into the function parameter, and it seems to work, but when run with equipments.equipmentID, it outputs everything, even the ids that I worked on my test
– John Beltran
Nov 14 '18 at 10:10
add a comment |
1
You can use "whereRaw" method to do that.
$equipments = DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
->whereRaw('? < ?', [$this->countOrders('equipments.equipmentID'), 1])
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
answered Nov 14 '18 at 9:26
Kenny
19025
It's still outputting the equipment, despite having a value of 1 when I print out the result of the function in the view. Maybe the function in the whereRaw isn't actually printing a number, or something else, :/
– John Beltran
Nov 14 '18 at 10:01
I tested by inputting just random ids into the function parameter, and it seems to work, but when run with equipments.equipmentID, it outputs everything, even the ids that I worked on my test
– John Beltran
Nov 14 '18 at 10:10
add a comment |
1
1
1
You can use "whereRaw" method to do that.
$equipments = DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
->whereRaw('? < ?', [$this->countOrders('equipments.equipmentID'), 1])
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
answered Nov 14 '18 at 9:26
Kenny
19025
You can use "whereRaw" method to do that.
$equipments = DB::table('equipments')
->leftjoin('orders', function ($join) {
$join->on('orders.equipment_id', '=', 'equipments.equipmentID')
->where('orders.orderUser_id','=', Auth::user()->id);
})
->leftjoin('business_users', 'business_users.id', '=' , 'equipments.user_id')
->where('equipments.equipmentArchived','=','0')
->whereRaw('? < ?', [$this->countOrders('equipments.equipmentID'), 1])
->orderBy('equipments.equipmentListedDate','DESC')
->distinct()
->paginate(10);
answered Nov 14 '18 at 9:26
Kenny
19025
answered Nov 14 '18 at 9:26
Kenny
19025
answered Nov 14 '18 at 9:26
Kenny
19025
answered Nov 14 '18 at 9:26
Kenny
19025
19025
It's still outputting the equipment, despite having a value of 1 when I print out the result of the function in the view. Maybe the function in the whereRaw isn't actually printing a number, or something else, :/
– John Beltran
Nov 14 '18 at 10:01
I tested by inputting just random ids into the function parameter, and it seems to work, but when run with equipments.equipmentID, it outputs everything, even the ids that I worked on my test
– John Beltran
Nov 14 '18 at 10:10
add a comment |
It's still outputting the equipment, despite having a value of 1 when I print out the result of the function in the view. Maybe the function in the whereRaw isn't actually printing a number, or something else, :/
– John Beltran
Nov 14 '18 at 10:01
I tested by inputting just random ids into the function parameter, and it seems to work, but when run with equipments.equipmentID, it outputs everything, even the ids that I worked on my test
– John Beltran
Nov 14 '18 at 10:10
It's still outputting the equipment, despite having a value of 1 when I print out the result of the function in the view. Maybe the function in the whereRaw isn't actually printing a number, or something else, :/
– John Beltran
Nov 14 '18 at 10:01
It's still outputting the equipment, despite having a value of 1 when I print out the result of the function in the view. Maybe the function in the whereRaw isn't actually printing a number, or something else, :/
– John Beltran
Nov 14 '18 at 10:01
I tested by inputting just random ids into the function parameter, and it seems to work, but when run with equipments.equipmentID, it outputs everything, even the ids that I worked on my test
– John Beltran
Nov 14 '18 at 10:10
I tested by inputting just random ids into the function parameter, and it seems to work, but when run with equipments.equipmentID, it outputs everything, even the ids that I worked on my test
– John Beltran
Nov 14 '18 at 10:10
add a comment |
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