我想要创建类似于SQL子查询的别名,这样我就可以像普通SQL字段一样使用它。例如,我有三个表:user、user_comments和user_comment_answers。如果我想得到当前用户评论的所有答案,我会这样做: COUNT(*) as comment_answers_count WHEREuser_comment_id IN (SELECT id FROM user_comments WHERE u
如何使这个查询来查询构建器Laravel5.4?select * from gb_employee where employee_id not in (select gb_emp_client_empid from gb_emp_client_lineswhere gb_emp_client_clientid =1) ;
SELECT `id` FROM `jobs` WHERE `job_type` IN ( 'topemployercheckbox1','topemployercheckbox2') OR `job_category` IN ( 'Contract','Intern') ;
我需要将这个MySQL查询转换为Laravel5.4查询生成器查询。
我试图用Eloquent在Laravel5.4中执行这个查询,但我没有使子查询正常工作。这是原始的SQL查询: inner join projects_categories pc on p.id = pc.project_idwhere pc.name in (select pc.name from projects p
i
我们有一份用Laravel5.2写的旧申请书。我们目前正试图升级到5.4,而阻碍我们的最后一件事是whereHas错误。下面是原始sql查询:select * from `jobs` where exists (select * from `channels` where `jobs`deleted_at` is null limit 24 offset 0
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select users.id, users.nickname, users.comment from users (SELECT friends.friend_id from friends WHERE friends.user_id = '.我遵循了Laravel Query Builder WHERE NOT IN,并像这样转换了我的查询