blocks|key|311795|text|最短的路|type|unstyled|depth|inlineStyleRanges|entityRanges|data|311796|re.sub(r'http\S%2B',+'',+stringliteral)|code-block|syntax|javascript|311797|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

the shortest way 

<pre><code>re.sub(r'http\S+', '', stringliteral)
</code></pre>

blocks|key|528612|text|这对我很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|528613|import+re
thestring+=+"text1\ntext2\nhttp://url.com/bla1/blah1/\ntext3\ntext4\nhttp://url.com/bla2/blah2/\ntext5\ntext6"

URLless_string+=+re.sub(r'\w%2B:\/{2}[\d\w-]%2B(\.[\d\w-]%2B)*(?:(?:\/[%5E\s/]*))*',+'',+thestring)
print+URLless_string|code-block|syntax|javascript|528614|结果：|528615|text1
text2

text3
text4

text5
text6|528616|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This worked for me:

<pre><code>import re
thestring = "text1\ntext2\nhttp://url.com/bla1/blah1/\ntext3\ntext4\nhttp://url.com/bla2/blah2/\ntext5\ntext6"

URLless_string = re.sub(r'\w+:\/{2}[\d\w-]+(\.[\d\w-]+)*(?:(?:\/[^\s/]*))*', '', thestring)
print URLless_string
</code></pre>

Result:

<pre><code>text1
text2

text3
text4

text5
text6
</code></pre>

blocks|key|528779|text|删除混在任何文本中的HTTP链接/URL：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|528780|import+re
re.sub(r'''(?i)\b((?:https?://%7Cwww\d{0,3}[.]%7C[a-z0-9.\-]%2B[.][a-z]{2,4}/)(?:[%5E\s()<>]%2B%7C$([%5E\s()<>]%2B%7C(\([%5E\s()<>]%2B$))*\))%2B(?:$([%5E\s()<>]%2B%7C(\([%5E\s()<>]%2B$))*\)%7C[%5E\s`!()\[\]{};:'".,<>?«»“”‘’]))''',+"+",+text)|code-block|syntax|javascript|528781|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<h1>Removal of HTTP links/URLs mixed up in any text:</h1>

<pre><code>import re
re.sub(r'''(?i)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()&lt;&gt;]+|$([^\s()&lt;&gt;]+|(\([^\s()&lt;&gt;]+$))*\))+(?:$([^\s()&lt;&gt;]+|(\([^\s()&lt;&gt;]+$))*\)|[^\s`!()\[\]{};:'".,&lt;&gt;?«»“”‘’]))''', " ", text)
</code></pre>

blocks|key|528657|text|此解决方案适用于http、https和其他普通url类型的特殊字符：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|528658|import+re
def+remove_urls+(vTEXT):
++++vTEXT+=+re.sub(r'(https%7Chttp)?:\/\/(\w%7C\.%7C\/%7C\?%7C\=%7C\&%7C\%25)*\b',+'',+vTEXT,+flags=re.MULTILINE)
++++return(vTEXT)


print(+remove_urls("this+is+a+test+https://sdfs.sdfsdf.com/sdfsdf/sdfsdf/sd/sdfsdfs?bob=%2520tree&jef=man+lets+see+this+too+https://sdfsdf.fdf.com/sdf/f+end"))|code-block|syntax|javascript|528659|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This solution caters for http, https and the other normal url type special characters :

<pre><code>import re
def remove_urls (vTEXT):
 vTEXT = re.sub(r'(https|http)?:\/\/(\w|\.|\/|\?|\=|\&amp;|\%)*\b', '', vTEXT, flags=re.MULTILINE)
 return(vTEXT)


print( remove_urls("this is a test https://sdfs.sdfsdf.com/sdfsdf/sdfsdf/sd/sdfsdfs?bob=%20tree&amp;jef=man lets see this too https://sdfsdf.fdf.com/sdf/f end"))
</code></pre>

blocks|key|311930|text|您真正想要做的是删除所有以http://或https://开头的字符串以及非空格字符的任意组合。下面是我将如何解决这个问题。我的解决方案与@tolgayilmaz非常相似|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|311931|#Define+the+text+from+which+you+want+to+replace+the+url+with+"".
text+='''The+link+to+this+post+is+https://stackoverflow.com/questions/11331982/how-to-remove-any-url-within-a-string-in-python'''

import+re
#Either+use:
re.sub('http://\S%2B%7Chttps://\S%2B',+'',+text)
#OR+
re.sub('http[s]?://\S%2B',+'',+text)|code-block|syntax|javascript|311932|运行上面两个代码的结果是|311933|>>>+'The+link+to+this+post+is+'|311934|我更喜欢第二个，因为它更具可读性。|311935|entityMap^0|D|7|L|8|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

What you really want to do is to remove any string that starts with either <code>http://</code> or <code>https://</code> plus any combination of non white space characters. Here is how I would solve it. My solution is very similar to that of @tolgayilmaz

<pre><code>#Define the text from which you want to replace the url with "".
text ='''The link to this post is https://stackoverflow.com/questions/11331982/how-to-remove-any-url-within-a-string-in-python'''

import re
#Either use:
re.sub('http://\S+|https://\S+', '', text)
#OR 
re.sub('http[s]?://\S+', '', text)
</code></pre>

And the result of running either code above is 

<pre><code>&gt;&gt;&gt; 'The link to this post is '
</code></pre>

I prefer the second one because it is more readable.

blocks|key|528739|text|我知道这个问题已经有人回答了，现在已经很愚蠢了，但我认为这个应该放在这里。这是一个正则表达式，可以匹配任何类型的url。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|528740|[%5E+]%2B\.[%5E+]%2B|code-block|syntax|javascript|528741|它可以像这样使用|528742|re.sub('[%5E+]%2B\.[%5E+]%2B','',sentence)|528743|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I know this has already been answered and its stupid late but I think this should be here. This is a regex that matches any kind of url. 

<pre><code>[^ ]+\.[^ ]+
</code></pre>

It can be used like 

<pre><code>re.sub('[^ ]+\.[^ ]+','',sentence)
</code></pre>

blocks|key|311706|text|你也可以从另一个角度来看待它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|311707|from+urlparse+import+urlparse
[el+for+el+in+['text1',+'FTP://somewhere.com',+'text2',+'http://blah.com:8080/foo/bar#header']+if+not+urlparse(el).scheme]|code-block|syntax|javascript|311708|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could also look at it from the other way around...

<pre><code>from urlparse import urlparse
[el for el in ['text1', 'FTP://somewhere.com', 'text2', 'http://blah.com:8080/foo/bar#header'] if not urlparse(el).scheme]
</code></pre>

blocks|key|311985|text|我认为最通用的URL正则表达式模式是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|311986|URL_PATTERN+=+r'[A-Za-z0-9]%2B://[A-Za-z0-9%25-_]%2B(/[A-Za-z0-9%25-_])*(#%7C\\?)[A-Za-z0-9%25-_&=]*'|code-block|syntax|javascript|311987|有一个小模块可以做你想做的事情：|311988|pip+install+mysmallutils|311989|from+mysutils.text+import+remove_urls

remove_urls(text)|311990|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I think the most general URL regex pattern is this one:
<pre class="lang-py prettyprint-override"><code>URL_PATTERN = r'[A-Za-z0-9]+://[A-Za-z0-9%-_]+(/[A-Za-z0-9%-_])*(#|\\?)[A-Za-z0-9%-_&amp;=]*'
</code></pre>
There is a small module that does what do you want:
<pre class="lang-sh prettyprint-override"><code>pip install mysmallutils
</code></pre>
<pre class="lang-py prettyprint-override"><code>from mysutils.text import remove_urls

remove_urls(text)
</code></pre>

blocks|key|312039|text|一个简单的.*和一个正面的后视应该可以完成这项工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|312040|text="text1\ntext2\nhttp://url.com/bla1/blah1/\ntext3\ntext4\nhttp://url.com/bla2/blah2/\ntext5\ntext6"

req=re.sub(r'http.*?(?=\s)',+"+",+text)
print(req)|code-block|syntax|javascript|312041|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A simple .* with a positive look behind should do the job.
<pre><code>text=&quot;text1\ntext2\nhttp://url.com/bla1/blah1/\ntext3\ntext4\nhttp://url.com/bla2/blah2/\ntext5\ntext6&quot;

req=re.sub(r'http.*?(?=\s)', &quot; &quot;, text)
print(req)
</code></pre>

blocks|key|528833|text|import+re
s+=+'''
text1
text2
http://url.com/bla1/blah1/
text3
text4
http://url.com/bla2/blah2/
text5
text6
http://url.com/bla3/blah3/'''
g+=+re.findall(r'(text\d%2B)',s)
print+('list',g)
for+i+in+g:
++++print+(i)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|528834|输出|unstyled|528835|list+['text1',+'text2',+'text3',+'text4',+'text5',+'text6']
text1
text2
text3
text4
text5
text6++++​|528836|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>import re
s = '''
text1
text2
http://url.com/bla1/blah1/
text3
text4
http://url.com/bla2/blah2/
text5
text6
http://url.com/bla3/blah3/'''
g = re.findall(r'(text\d+)',s)
print ('list',g)
for i in g:
 print (i)
</code></pre>

Out

<pre><code>list ['text1', 'text2', 'text3', 'text4', 'text5', 'text6']
text1
text2
text3
text4
text5
text6 ​
</code></pre>

I want to remove all URLs inside a string (replace them with "")
I searched around but couldn't really find what I want.

Example:

<pre><code>text1
text2
http://url.com/bla1/blah1/
text3
text4
http://url.com/bla2/blah2/
text5
text6
http://url.com/bla3/blah3/
</code></pre>

I want the result to be:

<pre><code>text1
text2
text3
text4
text5
text6
</code></pre>

How to remove any URL within a string in Python

我想删除字符串中的所有URL(用“”替换它们)，我到处搜索，但没有真正找到我想要的。示例：text1text2http://url.com/bla1/blah1/text3text4http://url.com/bla2/blah2/text5text6http://url.com/bla3/blah3/我希望结果是：text1text2text3text4text5text6

问如何在Python中移除字符串中的任何URL
EN

回答 10

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问如何在Python中移除字符串中的任何URLEN

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问如何在Python中移除字符串中的任何URL
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