我在laravel中試過這個(gè)，但它沒有用。$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 中編寫這個(gè) MySQL 查詢？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
2 回答
千萬里不及你
TA貢獻(xiàn)1784條經(jīng)驗(yàn) 獲得超9個(gè)贊
你的問題不清楚orderBy。根據(jù)您的問題：
如何在 Laravel 5.7 中編寫這個(gè) MySQL 查詢？
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
回答 ：
$test = ReportDoctorCompanyDailyTestModal::select(
'report_doctor_company_daily_test.test_id',
'testing_types.testing_name',
'doctor.first_name',
'doctor.last_name',
DB::raw('SUM(report_doctor_company_daily_test.test_price)')
)
->join('report_doctor_company_daily_test', 'testing_types.id', '=', 'report_doctor_company_daily_test.test_id')
->join('doctor', 'report_doctor_company_daily_test.doctor_id', '=', 'doctor.id')
->groupBy('testing_types.id');
dd($test->toSql());
結(jié)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`
婷婷同學(xué)_
TA貢獻(xiàn)1844條經(jīng)驗(yàn) 獲得超8個(gè)贊
use 可以使用->orderBy();語(yǔ)句。
例如；
$doctor_daily=ReportDoctorCompanyDailyTestModal()->orderBy()->get();
然后你可以按升序獲取數(shù)據(jù)。如果你想要降序，可以使用
->orderBy('id','DESC');- 2 回答
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