Django FileField在调用Python对象时超出了模型最大递归深度 [英] Django FileField in model maximum recursion depth exceeded while calling a Python object

问题描述

我的模型：

class Movie(models.Model):
        title = models.CharField(max_length=200)
        description = models.CharField(max_length=1500)
        publish_date = models.DateField()
        file_1080p = models.FileField(upload_to=generate_filename_movie, validators=[validate_video])
        file_720p = models.FileField(upload_to=generate_filename_movie_720)
        file_480p = models.FileField(upload_to='temp/')
        subtitles = models.FileField(upload_to=generate_filename_subtitle_movie, validators=[validate_subtitles])
        already_converted = bool(False)

我的保存功能：

    def save(self, *args, **kwargs):
            if self.already_converted == False:
                    # todo convert files
                    self.already_converted = True
            if self.subtitles.path.endswith('.srt'): # srt subtitle should process
                    import re
                    m = re.sub(r'([0-9]{2}:[0-9]{2}:[0-9]{2}),([0-9]{3})', r'\1.\2', self.subtitles.read())
                    from django.core.files.base import ContentFile
                    filenameToSave = generate_filename_subtitle_movie(0, self.subtitles.name)
                    self.subtitles.save(filenameToSave, ContentFile("WEBVTT\n\n" + m))
            super(Movie, self).save(*args, **kwargs)

错误发生在这里： self.subtitles.save （filenameToSave，ContentFile（WEBVTT\\\
\\\
+ m））

谢谢。

推荐答案

在 FileField 上调用保存将默认触发它的模型的保存方法。您可以通过向 FileField 的保存方法提供另外的第三个参数来绕过它。

Calling the save method on a FileField will trigger it's model's save method by default. You can bypass it by providing an additional third argument to the FileField's save method.

self.subtitles.save(filenameToSave, ContentFile("WEBVTT\n\n" + m), False)

看看：

https://docs.djangoproject.com/en/ dev / ref / models / fields /＃django.db.models.fields.files.FieldFile.save

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