如何编写Laravel订单查询？

Posted 2021-05-07

我在laravel中尝试过，但是没有用。

$doctor_daily=ReportDoctorCompanyDailyTestModal::select('test_id',DB::raw('SUM(test_price) as num'))
                                                       ->join('testing_types','report_doctor_company_daily_test.test_id','=','testing_types.id')
                                                       ->join('doctor','doctor.id','=','report_doctor_company_daily_test.doctor_id')
                                                       ->groupBy('test_id')
                                                       ->get();

如何在Laravel 5.7中编写此mysql查询？

SELECT
    report_doctor_company_daily_test.test_id,
    testing_types.testing_name,
    doctor.first_name,
    doctor.last_name,
    SUM( report_doctor_company_daily_test.test_price ) 
FROM
    testing_types
    INNER JOIN report_doctor_company_daily_test ON testing_types.id = report_doctor_company_daily_test.test_id
    INNER JOIN doctor ON report_doctor_company_daily_test.doctor_id = doctor.id 
GROUP BY
    testing_types.id

答案

使用可以使用->orderBy();语句。

例如；

$doctor_daily=ReportDoctorCompanyDailyTestModal()->orderBy()->get();

然后您可以通过升序获取数据。如果要降序排列，可以使用

->orderBy('id','DESC');

另一答案

您对orderBy的问题不清楚。根据您的问题：

如何在Laravel 5.7中编写此MySQL查询？

SELECT
    report_doctor_company_daily_test.test_id,
    testing_types.testing_name,
    doctor.first_name,
    doctor.last_name,
    SUM( report_doctor_company_daily_test.test_price )  FROM
    testing_types
    INNER JOIN report_doctor_company_daily_test ON testing_types.id = report_doctor_company_daily_test.test_id
    INNER JOIN doctor ON report_doctor_company_daily_test.doctor_id = doctor.id  GROUP BY
    testing_t
ypes.id

---

答案：

$test = ReportDoctorCompanyDailyTestModal::select(
        'report_doctor_company_daily_test.test_id',
        'testing_types.testing_name',
        'doctor.first_name',
        'doctor.last_name',
        DB::raw('SUM(report_doctor_company_daily_test.test_price)')
    )
    ->join('report_doctor_company_daily_test', 'testing_types.id', '=', 'report_doctor_company_daily_test.test_id')
    ->join('doctor', 'report_doctor_company_daily_test.doctor_id', '=', 'doctor.id')
    ->groupBy('testing_types.id');

dd($test->toSql());

结果：

select 
     `report_doctor_company_daily_test`.`test_id`, 
     `testing_types`.`testing_name`, 
     `doctor`.`first_name`, 
     `doctor`.`last_name`, 
     SUM(report_doctor_company_daily_test.test_price) 

from 
    `testing_types` 

inner join `report_doctor_company_daily_test` 
    on `testing_types`.`id` = `report_doctor_company_daily_test`.`test_id` 

inner join `doctor` 
    on `report_doctor_company_daily_test`.`doctor_id` = `doctor`.`id` 

group by `testing_types`.`id`

另一答案

简单地说，您可以像使用它

->orderBy('id','DESC')->get();