Rails 在顶部排序特定值

Question

build_histories（名称、状态、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

我试过了，它返回了

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

但我想将其排序为“已排队”总是在顶部，然后“失败/成功”应该按 updated_at desc 顺序排列

例子

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：我尝试基于this

尝试了以下

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，但在状态栏中我得到的是状态 = 1 或 2 而不是“排队/成功/失败”

请帮助

Answer 1

如果您使用的是MySql，您可以尝试按照field的顺序进行操作

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

对于 postgres

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

Answer 2

试试

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，但我认为它应该可以工作。