原始答案：多对多关系

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')

现在创建taks_a并task_b-task_b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，假设task_a实际上有一些与其相关的任务：

# 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，您可以根据需要实施寻路逻辑/其他算法。

编辑：递归访问字段

好的-使用上述结构，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>

您可以对递归函数进行更多研究，但这应该可以帮助您入门。您可以聪明地查看函数中的上一个任务或下一个任务。您也可以将此作为方法添加到Task模型中，以便可以在任务实例上调用它。