如何按祖先检索 Google App Engine 实体

发布于 2024-08-29 08:22:48 字数 847 浏览 6 评论 0原文

我的 Google App Engine 数据存储区中有以下 2 个模型:

class Search(db.Model):
    what = db.StringProperty()

class SearchResult(db.Model):
    search = db.ReferenceProperty(Search)

    title = db.StringProperty()
    content = db.StringProperty()

并且,我尝试在以下函数中检索给定搜索实体的所有 SearchResult 实体:

def get_previous_search_results(what='', where=''):
    search_results = None

    search = db.GqlQuery("SELECT * FROM Search WHERE what = :1", what).fetch(1)
    if search:
        search_results = db.GqlQuery("SELECT * FROM SearchResult WHERE ANCESTOR IS :1", search[0].key()).fetch(10)

    return search_results

但是,它始终返回一个空集。

有什么想法我做错了吗?我已通读 Python Datastore API 文档,这看起来像是执行此操作的正确方法,但它不起作用。

I have the following 2 models in my Google App Engine datastore:

class Search(db.Model):
    what = db.StringProperty()

class SearchResult(db.Model):
    search = db.ReferenceProperty(Search)

    title = db.StringProperty()
    content = db.StringProperty()

And, I am trying to retrieve all SearchResult entities for a given Search entity in the following function:

def get_previous_search_results(what='', where=''):
    search_results = None

    search = db.GqlQuery("SELECT * FROM Search WHERE what = :1", what).fetch(1)
    if search:
        search_results = db.GqlQuery("SELECT * FROM SearchResult WHERE ANCESTOR IS :1", search[0].key()).fetch(10)

    return search_results

However, it always returns an empty set.

Any ideas what I am doing wrong? I've read through the Python Datastore API docs and this seem like the correct way to do this, but it's not working.

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评论(1

纸短情长 2024-09-05 08:22:48

您是否与父级一起创建搜索实体? ReferenceProperty 不会创建祖先关系,并且您可能需要 search.searchresult_set,它将是对具有对搜索对象“search”的引用的 SearchResult 对象的查询。

Are you creating the Search entities with a parent? The ReferenceProperty doesn't create an ancestor relationship, and it seems likely you might want search.searchresult_set, which will be a Query for SearchResult objects that have a reference to the Search object 'search'.

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