Question

情景是这样的：每个Process都有多个ProcessingSteps

我编写的代码能够在不与ProcessingSteps对应的情况下获取所有进程。

我知道我错过了where子句，我想问一下我们如何在Grails中这样做。

我只想获取相应Process

的每个ProcessingStepUpdate

我有两个域类ProcessingStep和ProcessingStepUpdate

package a.b.c
public class ProcessingStep {
    Process process
}

public class ProcessingStepUpdate{
static belongsTo = [processingStep: ProcessingStep]
ProcessingStep processingStep
}

这是我写的剧本

Process.list(max:1).each {
    //List<ProcessingStep> test2= ProcessingStep.findAllByProcess(it)
    //println it
    def test3 = ProcessingStep.createCriteria().list() {
        eq("process",it)
    }
    println it

    it.list().each {
        //not telling it where to get the list from 
        ProcessingStep.list().each { pstep ->
            def test4 = ProcessingStepUpdate.createCriteria().list() {
                eq("processingStep",pstep)

                // Projections are aggregating, reporting, and
                // filtering functions that can be applied after
                // the query has finished.
                // A common use for projections is to summarize data
                // in a query
                /* projections{
                    groupProperty("processingStep")
                }*/
            }
            println pstep
            //List<ProcessingStepUpdate> test = ProcessingStepUpdate.findAllByProcessingStep(it)
            //List<ProcessingStepUpdate> test = ProcessingStepUpdate.findWhere()
            //println "it"
        }
    }
}

我有一天一直坚持这个问题...... OOPS世界的新手！

Answer 1

我会试着猜测这项任务只是为了孩子们的孩子。然后是这样的：

public class Process {
  static hasMany = [processingSteps: ProcessingStep]
}

public class ProcessingStep {
  static belongsTo = [process: Process]
  static hasMany = [updates: ProcessingStepUpdate]
}

public class ProcessingStepUpdate {
  static belongsTo = [processingStep: ProcessingStep]
}


Process.list().each{ process -> 
  process.processingSteps.each { step ->
    step.updates.each {
      println "Process: $process, Update: $it"
    }
  }
}

甚至

def updates = Process.list()*.processingSteps.flatten()*.updates.flatten()
println updates.join('\n')

看看Groovy Collections，特别是“星点”*。运营商“部分。

Grails- Criteria构建器 - createCriteria

1 个答案: