how can I search in TWO columns if key1 in column1 is equal to value1 and key2 is equal to value2 in column2...
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one column works
whereJsonContains('VehicleApplications'
,['ModelName' => $model, 'YearID' => $year] )->paginate(10);
mysql json laravel mariadb
edited Nov 22 at 20:49
pirho
3,586101830
asked Nov 22 at 17:33
AlbearBirra
1
add a comment |
up vote
0
down vote
favorite
one column works
whereJsonContains('VehicleApplications'
,['ModelName' => $model, 'YearID' => $year] )->paginate(10);
mysql json laravel mariadb
edited Nov 22 at 20:49
pirho
3,586101830
asked Nov 22 at 17:33
AlbearBirra
1
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
one column works
whereJsonContains('VehicleApplications'
,['ModelName' => $model, 'YearID' => $year] )->paginate(10);
mysql json laravel mariadb
edited Nov 22 at 20:49
pirho
3,586101830
asked Nov 22 at 17:33
AlbearBirra
1
one column works
whereJsonContains('VehicleApplications'
,['ModelName' => $model, 'YearID' => $year] )->paginate(10);
mysql json laravel mariadb
mysql json laravel mariadb
edited Nov 22 at 20:49
pirho
3,586101830
asked Nov 22 at 17:33
AlbearBirra
1
edited Nov 22 at 20:49
pirho
3,586101830
asked Nov 22 at 17:33
AlbearBirra
1
edited Nov 22 at 20:49
pirho
3,586101830
edited Nov 22 at 20:49
pirho
3,586101830
edited Nov 22 at 20:49
pirho
3,586101830
3,586101830
asked Nov 22 at 17:33
AlbearBirra
1
asked Nov 22 at 17:33
AlbearBirra
1
asked Nov 22 at 17:33
AlbearBirra
1
1
add a comment |
add a comment |
1 Answer
1
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0
down vote
Make sure you define your variable types if someone knows of a cleaner query post it
$make = (string) $make;
$model = (string) $model;
$year = (int) $year;
$category = (int) $category;
whereJsonContains('VehicleApplications',
['ModelName' => $model, 'YearID' => $year]
)->
whereJsonContains('pcItems',
['pcCategoryID' => $category]
)->paginate(10);
answered Nov 23 at 13:37
AlbearBirra
1
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
up vote
0
down vote
Make sure you define your variable types if someone knows of a cleaner query post it
$make = (string) $make;
$model = (string) $model;
$year = (int) $year;
$category = (int) $category;
whereJsonContains('VehicleApplications',
['ModelName' => $model, 'YearID' => $year]
)->
whereJsonContains('pcItems',
['pcCategoryID' => $category]
)->paginate(10);
answered Nov 23 at 13:37
AlbearBirra
1
add a comment |
up vote
0
down vote
Make sure you define your variable types if someone knows of a cleaner query post it
$make = (string) $make;
$model = (string) $model;
$year = (int) $year;
$category = (int) $category;
whereJsonContains('VehicleApplications',
['ModelName' => $model, 'YearID' => $year]
)->
whereJsonContains('pcItems',
['pcCategoryID' => $category]
)->paginate(10);
answered Nov 23 at 13:37
AlbearBirra
1
add a comment |
up vote
0
down vote
up vote
0
down vote
Make sure you define your variable types if someone knows of a cleaner query post it
$make = (string) $make;
$model = (string) $model;
$year = (int) $year;
$category = (int) $category;
whereJsonContains('VehicleApplications',
['ModelName' => $model, 'YearID' => $year]
)->
whereJsonContains('pcItems',
['pcCategoryID' => $category]
)->paginate(10);
answered Nov 23 at 13:37
AlbearBirra
1
Make sure you define your variable types if someone knows of a cleaner query post it
$make = (string) $make;
$model = (string) $model;
$year = (int) $year;
$category = (int) $category;
whereJsonContains('VehicleApplications',
['ModelName' => $model, 'YearID' => $year]
)->
whereJsonContains('pcItems',
['pcCategoryID' => $category]
)->paginate(10);
answered Nov 23 at 13:37
AlbearBirra
1
answered Nov 23 at 13:37
AlbearBirra
1
answered Nov 23 at 13:37
AlbearBirra
1
answered Nov 23 at 13:37
AlbearBirra
1
1
add a comment |
add a comment |
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