Question

我有一个带有类的grails项目，当从控制器“手动”执行时，我可以删除没有问题。我使用以下代码。

def delete = {
    def projectInstance = Project.get( params.id )
    def employee = projectInstance.employee
    def projectarray = new ArrayList<Project>();
    projectarray += employee.getProjects()
    println("Size of projectarray is " + projectarray.size())
    if(projectInstance) {
        def rolearray = []

        projectarray.remove(projectInstance)
        def temp = new TreeSet<Project>();
        temp += employee.getProjects()
        temp.clear()
        temp.addAll(projectarray)
        employee.projects = temp

        projectInstance.employer = null
        projectInstance.delete(flush:true)
        flash.message = "Project ${params.id} deleted"
        redirect(action:"edit", controller: "employee", id: employee.id)
    }
    else {
        flash.message = "Project not found with id ${params.id}"
        redirect(action:list)
    }
}

这样就可以删除单个实例。

现在我想从不同的控制器中删除员工的所有项目。

这样就存储在员工中：

class Employee implements Comparable
{
static hasMany = [projects:Project]  
static constraints = 
{
}

static mapping = {
    projects cascade:"all-delete-orphan", lazy:false
}

@XmlElementWrapper(name="projectslist")
SortedSet<Project> projects = new TreeSet<Project>();  // make a sortedSet?

}

那么我现在如何从特定员工实例中删除所有项目？

Answer 1

我可能会误解你的问题，因为我无法理解你的一些代码。这似乎没必要。如果你的关系设置正确（即Project belongsTo Employee），这应该足以删除一个项目：

def delete = {
    def projectInstance = Project.get( params.id )
    projectInstance.delete(flush:true)
    flash.message = "Project ${params.id} deleted"
    redirect(action:"edit", controller: "employee", id: employee.id)
}

如果这是一对多，则下次检索员工时项目将会消失。这应该可以删除员工的所有项目：

def delete = {
    def employee = Employee.get( params.id )
    employee.getProjects().clear()
    employee.save(flash:true)
    flash.message = "All projects of employee deleted."
    redirect(action:"edit", controller: "employee", id: employee.id)
}

假设级联：“all-delete-orphan”。如果情况并非如此，那么您可能还需要删除实例，这可能看起来像这样：

def delete = {
    def employee = Employee.get( params.id )

    // Make copy to avoid concurrent modification issues later
    def copy = new TreeSet<Project>(employee.getProjects()); 

    employee.getProjects().clear();
    employee.save(flash:true)

    copy.each{
      $it.delete();
    }     

    flash.message = "All projects of employee deleted."
    redirect(action:"edit", controller: "employee", id: employee.id)
}

我不是一个时髦的专家，所以不确定是否需要副本，或者你是否可以直接迭代集合。似乎总是有一种更加时髦的做事方式。您可能还想查看deleteFrom动态域类方法。根据要删除的关系数量，这可能是一种更有效的grails方法。

Answer 2

您可以使用Grails在声明hasMany关系时生成的removeFrom*方法 - 它等同于addTo*方法：

def employee = Employee.get(params.id)
employee.projects.toList().each { employee.removeFromProjects(it) } // toList() prevents a ConcurrentModifactionException

从grails中的有序集中删除所有项目

2 个答案: