[英]Regex to extract all urls from string
I have a string like this 我有这样的字符串
http://example.com/path/topage.htmlhttp://twitter.com/p/xyanhshttp://httpget.org/get.zipwww.google.com/privacy.htmlhttps://goodurl.net/
I would like to extract all url / webaddress into a Array. 我想将所有url / webaddress提取到一个数组中。 for example
例如
urls = ['http://example.com/path/topage.html','http://twitter.com/p/xyan',.....]
Here is my approach which didn't work. 这是我的方法不起作用。
import re
strings = "http://example.com/path/topage.htmlhttp://twitter.com/p/xyanhshttp://httpget.org/get.zipwww.google.com/privacy.htmlhttps://goodurl.net/"
links = re.findall('http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+', strings)
print links
// result always same as strings
The problem is that your regex pattern is too inclusive. 问题是您的正则表达式模式太包容了。 It includes all urls.
它包括所有URL。 You can use lookahead by using (?=)
您可以使用(?=)
Try this: 尝试这个:
re.findall("((www\.|http://|https://)(www\.)*.*?(?=(www\.|http://|https://|$)))", strings)
Your problem is that http://
is being accepted as a valid part of a url. 您的问题是
http://
被接受为url的有效部分。 This is because of this token right here: 这是因为这里有此令牌:
[$-_@.&+]
or more specifically: 或更具体地说:
$-_
This matches all characters with the range from $
to _
, which includes a lot more characters than you probably intended to do. 这将匹配
$
到_
范围内的所有字符,其中包含的字符比您可能打算的要多得多。
You can change this to [$\\-_@.&+]
but this causes problems since now, /
characters will not match. 您可以将其更改为
[$\\-_@.&+]
但这会导致问题,因为/
字符不匹配。 So add it by using [$\\-_@.&+/]
. 因此,可以使用
[$\\-_@.&+/]
添加它。 However, this will again cause problems since http://example.com/path/topage.htmlhttp
would be considered a valid match. 但是,这将再次引起问题,因为
http://example.com/path/topage.htmlhttp
将被视为有效匹配。
The final addition is to add a lookahead to ensure that you are not matching http://
or https://
, which just so happens to be the first part of your regex! 最后添加的内容是添加前瞻性以确保您不匹配
http://
或https://
,而这恰好是正则表达式的第一部分!
http[s]?://(?:(?!http[s]?://)[a-zA-Z]|[0-9]|[$\-_@.&+/]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+
A simple answer without getting into much complication: 一个简单的答案而不会引起太多的复杂性:
import re
url_list = []
for x in re.split("http://", l):
url_list.append(re.split("https://",x))
url_list = [item for sublist in url_list for item in sublist]
In case you want to append the string http://
and https://
back to the urls, do appropriate changes to the code. 如果您想将字符串
http://
和https://
附加回URL,请对代码进行适当的更改。 Hope i convey the idea. 希望我传达这个想法。
这是我的
(r’http[s]?://[a-zA-Z]{3}\.[a-zA-Z0-9]+\.[a-zA-Z]+')
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