根据关联（has_many通过）列对记录进行排序

Question

我遇到了一个问题，我必须根据job_trade_contractors的min time_out字段对工作进行排序。对于job_trade_contractors，time_out字段可能为零。我有类似的关联：

1：job.rb

has_many :job_trades, dependent: :destroy
has_many :job_trade_contractors, through: :job_trades

2：job_trade.rb

has_many :job_trade_contractors, dependent: :destroy

3：job_trade_contractors.rb

belongs_to :job_trade, touch: true

当前，我正在通过以下代码对作业进行排序：

Job.left_joins(:job_trade_contractors).order("job_trade_contractors.time_out asc")

它可以根据time_out正确地对作业进行排序，但我需要的是对job_trade_contractors属于特定作业的最小time_out值进行排序。

假设我们有2个工作，每个工作有5个job_trade_contractors。我们将需要根据关联的job_trade_contractors中的最小time_out值对作业进行排序。

示例 职位

j1 -> 1: job_trade_contractor -> id: 1 time_out: 21-01-2020 2: job_trade_contractor -> id: 2 time_out: 20-01-2020 3: job_trade_contractor -> id: 3 time_out: 11-01-2020 4: job_trade_contractor -> id: 4 time_out: 19-01-2020 5: job_trade_contractor -> id: 5 time_out: 23-01-2020 j2 -> 1: job_trade_contractor -> id: 1 time_out: 21-01-2020 2: job_trade_contractor -> id: 2 time_out: 20-01-2020 3: job_trade_contractor -> id: 3 time_out: 11-01-2020 4: job_trade_contractor -> id: 4 time_out: 10-01-2020 5: job_trade_contractor -> id: 5 time_out: 25-01-2020

结果应为

j2 j1

因为我们的job_trade_contractor的time_out值是：j2中的min_time为10-01-2020 和11年1月11日在j1

Answer 1

我们可以通过以下方式实现排序，这可能对您有所帮助

has_many :job_trade_contractors, through: :job_trades, order: 'job_trade_contractors.time_out'

或

has_many :job_trade_contractors, ->  { order(time_out: :asc) }, through: :job_trades

Answer 2

您可以尝试

Job.joins(job_trades: :job_trade_contractors).order('job_trade_contractors.time_out').uniq