在Django DAG和Dependencies中创建项目管理应用程序

Question

我在一个项目管理应用程序中工作。应用程序需要的一点是创建具有依赖关系的任务的可能性，即任务2依赖于任务1，任务3依赖于任务2和任务B，等等。这将生成DAG（有向无环图）

问题是如何使用Django默认ORM存储这个“图形”

我提供了一个解决方案，它使用多对多关系（任务与自身）和触发器来避免创建循环，但我仍然不知道如何实现一些操作，例如从单个节点获取整个图形

有人对如何很好地实现它有想法吗？例如，这将允许执行关键路径方法的实现

编辑：

使用@Daniel提出的解决方案，我得到了下面的函数。我们可以看到，为了获得整个图形，对数据库进行了多次调用。该函数在循环期间也进入无限循环

def get_task_graph(task, visited_nodes = [], graph = {}):
    # Visited nodes are nodes that all parents were visited
    # print(task)

    
    
    previous_tasks = task.previous_tasks.all()

    
    still_have_unmarked_parents = set(previous_tasks).difference(set(visited_nodes))
    if still_have_unmarked_parents:
        print(f"Task {task} still have unmarked parents")
        
        for previous_task in still_have_unmarked_parents:
            

            visited_nodes, graph =  get_task_graph(previous_task, visited_nodes=visited_nodes, graph=graph)
    
    next_tasks = task.next_tasks.all()
 
    visited_nodes.append(task)
    print(f"Marking {task} as visited")
    graph[task] = []
    for next_task in next_tasks:
        if next_task not in graph[task]:
            graph[task].append(next_task)
    
        
        visited_nodes, graph = get_task_graph(next_task, visited_nodes=visited_nodes, graph=graph)
        print(f"\t Adding {next_task} to {task}")

    return visited_nodes, graph

谢谢

Answer 1

原始答案：许多关系

让我们创建一个简单的Task模型：

class Task(models.Model):
   name = models.CharField(...)
   previous_tasks = models.ManyToMany('Task', ..., related_name='next_tasks')

现在让我们创建taks_a，并且task_b-task_b将取决于task_a：

# create task_a, task_b:
task_a = Task.objects.create(name='First Task')
task_b = Task.objects.create(name='Second task')

# add task_a as a dependency for task_b:
task_b.previous_tasks.add([task_a])
task_b.save()

现在，我们可以像这样访问这些任务：

# get the tasks to do after task_a is complete:
task_a = Task.objects.get(id='<task-a-id>')
tasks_to_do_after_task_a = task_a.next_tasks.all() # returns a queryset with <task_b>

# get the tasks we need to complete before task_b:
task_b = Task.objects.get(id='<task-b-id>')
tasks_that_task_b_depends_on = task_b.previous_tasks.all() # returns a queryset with <task a>

最后，让我们假设task_a实际上有一些不同的任务依赖于它：

# iterate over all the tasks that depend on task a:
for tasks in task_a.next_tasks.all():

     # do some logic, i.e. find longest task:
     ...

# filter for a specific next_task on task_a:
some_task = task_a.next_tasks.filter(some_keyword='some_value')

# assume we have a duration time-field on tasks:      
next_longest_task = task_a.next_tasks.order_by('-duration')[0] # gets the next task with the longest duration

从那里，您可以根据需要实现路径查找逻辑/其他算法

编辑：递归访问字段

好的-使用上面的结构，我们可以定义如下函数：

def get_task_list(task):

    # get the next_tasks queryset, it will either contain tasks or be an empty queryset:
    next_tasks = task.next_tasks.all()

    # check if the current task has any children:
    if next_tasks:

        # do some logic, check for longest tasks / add next tasks to a list
        ...

        # pass each next task back into the function:
        for next_task in next_tasks:
            return get_task_list(next_task)

    # exit if there are no more next tasks:
    else:

        # do some logic on exiting, calculate the time to complete all tasks:
        ...

        return <optional-content>

您可以对递归函数做更多的研究，但这应该可以让您开始。您可以在函数中巧妙地查找以前的任务或查找下一个任务。您还可以将其作为方法添加到Task模型中，以便在任务实例上调用它