[英]Django Generic Relations and ORM Queries
說我有以下型號:
class Image(models.Model):
image = models.ImageField(max_length=200, upload_to=file_home)
content_type = models.ForeignKey(ContentType)
object_id = models.PositiveIntegerField()
content_object = generic.GenericForeignKey()
class Article(models.Model):
text = models.TextField()
images = generic.GenericRelation(Image)
class BlogPost(models.Model):
text = models.TextField()
images = generic.GenericRelation(Image)
找到所有至少附有一個Image的文章的處理器和內存最有效的方法是什么?
我這樣做了:
Article.objects.filter(pk__in=Image.objects.filter(content_type=ContentType.objects.get_for_model(Article)).values_list('object_id', flat=True))
哪個有效,但除了丑陋之外需要永遠。
我懷疑使用原始SQL有一個更好的解決方案,但這超出了我的范圍。 對於它的價值,上面生成的SQL如下:
SELECT `issues_article`.`id`, `issues_article`.`text` FROM `issues_article` WHERE `issues_article`.`id` IN (SELECT U0.`object_id` FROM `uploads_image` U0 WHERE U0.`content_type_id` = 26 ) LIMIT 21
編輯: czarchaic的建議有更好的語法,但更糟(更慢)的性能。 他的查詢生成的SQL如下所示:
SELECT DISTINCT `issues_article`.`id`, `issues_article`.`text`, COUNT(`uploads_image`.`id`) AS `num_images` FROM `issues_article` LEFT OUTER JOIN `uploads_image` ON (`issues_article`.`id` = `uploads_image`.`object_id`) GROUP BY `issues_article`.`id` HAVING COUNT(`uploads_image`.`id`) > 0 ORDER BY NULL LIMIT 21
編輯: Jarret Hardie的萬歲! 這是他應該明顯的解決方案生成的SQL:
SELECT DISTINCT `issues_article`.`id`, `issues_article`.`text` FROM `issues_article` INNER JOIN `uploads_image` ON (`issues_article`.`id` = `uploads_image`.`object_id`) WHERE (`uploads_image`.`id` IS NOT NULL AND `uploads_image`.`content_type_id` = 26 ) LIMIT 21
由於泛型關系,您應該能夠使用傳統的查詢集語義來查詢此結構以獲得反向關系:
Article.objects.filter(images__isnull=False)
這將為與多個Image
相關的任何Article
生成重復Article
,但您可以使用distinct()
QuerySet方法消除它:
Article.objects.distinct().filter(images__isnull=False)
我認為你最好的選擇是使用聚合
from django.db.models import Count
Article.objects.annotate(num_images=Count('images')).filter(num_images__gt=0)
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