Php SQLSTATE[HY000]:一般错误:1096未使用表
错误是这样的:Php SQLSTATE[HY000]:一般错误:1096未使用表,php,mysql,laravel-5,Php,Mysql,Laravel 5,错误是这样的: SQLSTATE[HY000]: General error: 1096 No tables used (SQL: select `Marcas`.`nombre` as `Marca`, `Modelo`.`nombre` as `Modelo`, `Vehiculos`.`year` as `Anio`, `Vehiculos`.`id` from `Vehiculos` inner join `Reservaciones` on `Vehiculos`.`id` = `Re
SQLSTATE[HY000]: General error: 1096 No tables used (SQL: select `Marcas`.`nombre` as `Marca`, `Modelo`.`nombre` as `Modelo`, `Vehiculos`.`year` as `Anio`, `Vehiculos`.`id` from `Vehiculos` inner join `Reservaciones` on `Vehiculos`.`id` = `Reservaciones`.`id_vehiculo` inner join `Marcas` on `Vehiculos`.`id_marca` = `Marcas`.`id` inner join `Modelo` on `Vehiculos`.`id_modelo` = `Modelo`.`id` where `Vehiculos`.`id_marca` = 1 and `Marcas`.`id_categoria` = 2 and `Vehiculos`.`id` not in (select *))
这是我在laravel提出的问题:
public function checkDisponibilityByDate(Request $req){
$fecha = $req->fecha;
$vehiculos= Vehiculo::join('Reservaciones', 'Vehiculos.id', '=', 'Reservaciones.id_vehiculo')
->join('Marcas', 'Vehiculos.id_marca', '=' , 'Marcas.id')
->join('Modelo', 'Vehiculos.id_modelo', '=' , 'Modelo.id')
->select('Marcas.nombre as Marca', 'Modelo.nombre as Modelo', 'Vehiculos.year as Anio', 'Vehiculos.id')
->where('Vehiculos.id_marca', '=' ,$req->id_marca)
->where('Marcas.id_categoria', '=' ,$req->id_categoria)
->whereNotIn('Vehiculos.id', function($fecha) {
Reservacion::select('Reservaciones.id_vehiculo')
->from('Reservaciones')
->where('Reservaciones.fecha', '=', $fecha);
})
->get();
return $vehiculos->toJson();
}
我希望您能在这方面帮助我,我正在使用Laravel 5我认为如果您想向查询传递参数,您需要使用use$fecha传递它 也可以通过单独构造数组来简化whereNotIn
public function checkDisponibilityByDate(Request $req)
{
$fecha = $req->fecha;
$check = Reservacion::where('fecha', $fecha)->lists('id_vehiculo');
$vehiculos= Vehiculo::join('Reservaciones', 'Vehiculos.id', '=', 'Reservaciones.id_vehiculo')
->join('Marcas', 'Vehiculos.id_marca', '=' , 'Marcas.id')
->join('Modelo', 'Vehiculos.id_modelo', '=' , 'Modelo.id')
->select('Marcas.nombre as Marca', 'Modelo.nombre as Modelo', 'Vehiculos.year as Anio', 'Vehiculos.id')
->where('Vehiculos.id_marca', '=' ,$req->id_marca)
->where('Marcas.id_categoria', '=' ,$req->id_categoria)
->whereNotIn('Vehiculos.id', $check)
})
->get();
return $vehiculos->toJson();
}
希望这有帮助
如果您可以共享您的模型和表结构,那么可能可以建议其他/纯雄辩的方法。请参见查询末尾的“选择”中的“…和Vehiculos.id”。看到问题了吗?如果让我猜的话,那是因为您没有从whereNotIn回调返回任何内容。但我不擅长雄辩,所以可能它有点神奇。尝试更改它以返回Reservation::selectIt在@BarmarTry中不起作用,而在'Vehiculos.id',函数$query use$fecha中{$query->select'id_vehiculo'->from'Reservaciones'->where'Reservaciones.fecha','=',$fecha;。您使用的是哪一个Laravel版本的L5.0或更高版本?您将雄辩的查询生成器与查询生成器混为一谈,这似乎不是最佳实践。如果没有表结构和模型关系代码,就不可能提供替代方法。
public function checkDisponibilityByDate(Request $req)
{
$fecha = $req->fecha;
$check = Reservacion::where('fecha', $fecha)->lists('id_vehiculo');
$vehiculos= Vehiculo::join('Reservaciones', 'Vehiculos.id', '=', 'Reservaciones.id_vehiculo')
->join('Marcas', 'Vehiculos.id_marca', '=' , 'Marcas.id')
->join('Modelo', 'Vehiculos.id_modelo', '=' , 'Modelo.id')
->select('Marcas.nombre as Marca', 'Modelo.nombre as Modelo', 'Vehiculos.year as Anio', 'Vehiculos.id')
->where('Vehiculos.id_marca', '=' ,$req->id_marca)
->where('Marcas.id_categoria', '=' ,$req->id_categoria)
->whereNotIn('Vehiculos.id', $check)
})
->get();
return $vehiculos->toJson();
}