如何在Django中使用Bug详细信息序列化Task ID?

我的Django模型

序列化错误模型时,我想序列化“任务”表中的“任务ID”。 当我对Bug进行序列化时,我还希望它与“ taskId”相关联作为JSON响应。

模型->任务模型,错误模型,任务错误模型(一对一)

任务模型

class Task(models.Model):
    TASK_STATUS = (
       ('Not Assigned','Not Assigned'),('Assigned','Assigned'),('Completed','Completed'),('Ready to Check','Ready to Check'),('Bug Found','Bug Found'),)
    taskid = models.IntegerField(primary_key=True)
    taskname = models.CharField(max_length=50)
    task_status = models.CharField(max_length=50,choices=TASK_STATUS,default='Not Assigned')
    description = models.TextField()
    assignee = models.ForeignKey(Employee,on_delete=models.SET_NULL,null=True,blank=True,related_name='tassignee')
    assigner = models.ForeignKey(Employee,related_name='tassigner')
    project_id = models.ForeignKey(Project,on_delete=models.CASCADE,blank=True)

    class Meta:
        ordering = ('taskid',)

    def __str__(self):
        return self.taskid,self.taskname

错误模型

class Bug(models.Model):
    STATUS = (
        ('Assigned',('Reassigned','Reassigned'),('Fixed','Fixed'),)
    SEVERITY = (
        ('Critical','Critical'),('High','High'),('Medium','Medium'),('Low','Low'),)
    PRIORITY = (
        ('High',)
    bugid = models.IntegerField(primary_key=True)
    bugname = models.CharField(max_length=50)
    description = models.TextField()
    status = models.CharField(max_length=50,choices=STATUS,default='assig')
    screenshot= models.URLField(max_length=200)
    priority = models.CharField(max_length=50,choices=PRIORITY,default='M')
    severity = models.CharField(max_length=50,choices=SEVERITY,default='M')
    reporter = models.ForeignKey(Employee,related_name='breporter')
    assignee = models.ForeignKey(Employee,related_name='bassignee')
    bug_date = models.DateField(auto_now_add=True)
    due_date = models.DateField()

TaskBug模型

class TaskBug(models.Model):
    task_id = models.ForeignKey(Task,blank=True)
    bug_id = models.ForeignKey(Bug,blank=True)

    class Meta:
        unique_together = [['task_id','bug_id']]
        ordering = ('task_id',)

    def __str__(self):
        return str(self.task_id)
xiaohou0604 回答:如何在Django中使用Bug详细信息序列化Task ID?

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