Question

build_histories（name，status，updated_at）

foo1, Queued, 2017-01-08 13:46:59
foo2, Failed, 2017-01-02 13:46:59
foo3, Success,2017-01-03 13:46:59
foo4, Queued, 2017-01-04 13:46:59
foo5, Failed, 2017-01-05 13:46:59
foo6, Success,2017-01-09 13:46:59

我尝试了这个并返回

BuildHistory.joins(app_build: :app).where("build_histories.bundle like ?","%#{params}%").order("CASE WHEN status = 'Queued' THEN 1 ELSE 2 END,  status", "updated_at DESC")

foo1, Queued, 2017-01-08 13:46:59
foo4, Queued, 2017-01-04 13:46:59 
foo2, Failed, 2017-01-02 13:46:59
foo5, Failed, 2017-01-05 13:46:59
foo3, Success,2017-01-03 13:46:59
foo6, Success,2017-01-09 13:46:59

但是我想把它排序为“排队”总是在顶部，之后“失败/成功”应该在updated_at desc订单

实施例

foo1, Queued, 2017-01-08 13:46:59
foo4, Queued, 2017-01-04 13:46:59
foo6, Success,2017-01-09 13:46:59
foo5, Failed, 2017-01-05 13:46:59
foo3, Success,2017-01-03 13:46:59 
foo2, Failed, 2017-01-02 13:46:59

供参考：我试过基于this

尝试以下

BuildHistory.joins(app_build: :app).where("build_histories.bundle like ?","%#{params}%").select("*, CASE WHEN build_histories.status = 'Queued' THEN 1 ELSE 2 END as tmp_order").order("tmp_order, build_histories.updated_at desc")

我能够得到结果但是在状态栏中我得到状态= 1或2而不是“排队/成功/失败”

请帮助

Answer 1

如果您使用MySql，可以按field

尝试按顺序排列

.order("FIELD(status, 'Queued'), updated_at DESC")

对于postgres

.order("status = 'Queued' DESC, updated_at DESC")

Answer 2

尝试

BuildHistory.order("CASE WHEN status = 'Queued' THEN 1 ELSE 2 END", "updated_at DESC")

注意：我自己没有检查过代码，但我认为它应该有用。

Rails在顶部排序特定值

2 个答案: