这种方法会遍历数据库中的术语列表，检查这些术语是否在传递为参数的文本中，如果有一个，则用搜索页面链接替换它，并将该术语作为参数传递。
术语数量很高（约100000个），因此该过程相当缓慢，但这没关系，因为它是作为cron任务执行的。然而，它会导致脚本内存消耗激增，我找不到原因：
class SearchedTerm(models.Model):
[...]
@classmethod
def add_search_links_to_text(cls, string, count=3, queryset=None):
"""
Take a list of all researched terms and search them in the
text. If they exist, turn them into links to the search
page.
This process is limited to `count` replacements maximum.
WARNING: because the sites got different URLS schemas, we don't
provides direct links, but we inject the {% url %} tag
so it must be rendered before display. You can use the `eval`
tag from `libs` for this. Since they got different namespace as
well, we enter a generic 'namespace' and delegate to the
template to change it with the proper one as well.
If you have a batch process to do, you can pass a query set
that will be used instead of getting all searched term at
each calls.
"""
found = 0
terms = queryset or cls.on_site.all()
# to avoid duplicate searched terms to be replaced twice
# keep a list of already linkified content
# added words we are going to insert with the link so they won't match
# in case of multi passes
processed = set((u'video', u'streaming', u'title',
u'search', u'namespace', u'href', u'title',
u'url'))
for term in terms:
text = term.text.lower()
# no small word and make
# quick check to avoid all the rest of the matching
if len(text) < 3 or text not in string:
continue
if found and cls._is_processed(text, processed):
continue
# match the search word with accent, for any case
# ensure this is not part of a word by including
# two 'non-letter' character on both ends of the word
pattern = re.compile(ur'([^\w]|^)(%s)([^\w]|$)' % text,
re.UNICODE|re.IGNORECASE)
if re.search(pattern, string):
found += 1
# create the link string
# replace the word in the description
# use back references (\1, \2, etc) to preserve the original
# formatin
# use raw unicode strings (ur"string" notation) to avoid
# problems with accents and escaping
query = '-'.join(term.text.split())
url = ur'{%% url namespace:static-search "%s" %%}' % query
replace_with = ur'\1<a title="\2 video streaming" href="%s">\2</a>\3' % url
string = re.sub(pattern, replace_with, string)
processed.add(text)
if found >= 3:
break
return string
您可能也需要这段代码：
class SearchedTerm(models.Model):
[...]
@classmethod
def _is_processed(cls, text, processed):
"""
Check if the text if part of the already processed string
we don't use `in` the set, but `in ` each strings of the set
to avoid subtring matching that will destroy the tags.
This is mainly an utility function so you probably won't use
it directly.
"""
if text in processed:
return True
return any(((text in string) for string in processed))
我只有两个对象的引用可能是嫌疑人：terms和processed。但我看不出它们不被垃圾回收的任何原因。
编辑：
我认为我应该说这个方法是在Django模型方法中调用的。我不知道这是否相关，但这是代码：
class Video(models.Model):
[...]
def update_html_description(self, links=3, queryset=None):
"""
Take a list of all researched terms and search them in the
description. If they exist, turn them into links to the search
engine. Put the reset into `html_description`.
This use `add_search_link_to_text` and has therefor, the same
limitations.
It DOESN'T call save().
"""
queryset = queryset or SearchedTerm.objects.filter(sites__in=self.sites.all())
text = self.description or self.title
self.html_description = SearchedTerm.add_search_links_to_text(text,
links,
queryset)
我可以想象自动Python正则表达式缓存会占用一些内存。但它应该只做一次，而且在每次调用update_html_description时内存消耗都会增加。
问题不仅在于它消耗了大量内存，而且它没有释放：每次调用都会占用大约3%的内存，最终填满并崩溃脚本，显示“无法分配内存”。