blocks|key|3111122|text|在分隔器上最多拆分一次，然后取第一块：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3111123|sep+=+'...'
stripped+=+text.split(sep,+1)[0]|code-block|syntax|javascript|3111124|你没有说如果分隔符不存在会发生什么。在这种情况下，this和Alex's+solution都将返回整个字符串。|offset|length|3111125|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/904753/4518341^0|0|0|U|F|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Split on your separator at most once, and take the first piece:
<pre><code>sep = '...'
stripped = text.split(sep, 1)[0]
</code></pre>
You didn't say what should happen if the separator isn't present. Both this and <a href="https://stackoverflow.com/a/904753/4518341">Alex's solution</a> will return the entire string in that case.

blocks|key|339305|text|假设您的分隔符是'...'，但它可以是任何字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|339306|text+=+'some+string...+this+part+will+be+removed.'
head,+sep,+tail+=+text.partition('...')

>>>+print+head
some+string|code-block|syntax|javascript|339307|如果找不到分隔符，head将包含所有原始字符串。|offset|length|style|CODE|339308|partition函数是在Python2.5中添加的。|339309|339310|+S.partition(sep)+->+(head,+sep,+tail)|blockquote|339311|339312|在S中搜索分隔符sep，并返回分隔符之前的部分、分隔符本身和分隔符之后的部分。如果找不到分隔符，则返回S和两个空字符串。|339313|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html#str.partition^0|0|0|9|4|0|0|9|0|0|0|1|G|L|H|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|15|8|@$I|16|J|17|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|18|8|@]|9|@$I|19|J|1A|1|1B]]|A|$]]|$1|O|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1D|8|@$I|1E|J|1F|K|L]|$I|1G|J|1H|K|L]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1J|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|W|$X|$5|Y|Z|10|A|$11|12]]]]

Assuming your separator is '...', but it can be any string.
<pre><code>text = 'some string... this part will be removed.'
head, sep, tail = text.partition('...')

&gt;&gt;&gt; print head
some string
</code></pre>
If the separator is not found, <code>head</code> will contain all of the original string.
The <a href="https://docs.python.org/3/library/stdtypes.html#str.partition" rel="noreferrer">partition</a> function was added in Python 2.5.
<blockquote>
<code>S.partition(sep)</code> -&gt; <code>(head, sep, tail)</code>
Searches for the separator sep in S, and returns the part before it,
the separator itself, and the part after it. If the separator is not
found, returns S and two empty strings.
</blockquote>

blocks|key|3111166|text|如果你想删除字符串中最后一次出现分隔符之后的所有内容，我发现这个方法效果很好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3111167|<separator>.join(string_to_split.split(<separator>)[:-1])|offset|length|style|CODE|3111168|例如，如果string_to_split是类似于root/location/child/too_far.exe的路径，而您只需要文件夹路径，则可以按"/".join(string_to_split.split("/")[:-1])拆分，得到的结果是root/location/child|3111169|entityMap^0|0|0|1L|0|5|F|O|V|22|15|3G|J|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@$D|Q|E|R|F|G]|$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$]]

If you want to remove everything after the last occurrence of separator in a string I find this works well:

<code>&lt;separator&gt;.join(string_to_split.split(&lt;separator&gt;)[:-1])</code>

For example, if <code>string_to_split</code> is a path like <code>root/location/child/too_far.exe</code> and you only want the folder path, you can split by <code>"/".join(string_to_split.split("/")[:-1])</code> and you'll get 
<code>root/location/child</code>

blocks|key|339295|text|不使用正则表达式(我假设这就是您想要的)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|339296|def+remafterellipsis(text):
++where_ellipsis+=+text.find('...')
++if+where_ellipsis+==+-1:
++++return+text
++return+text[:where_ellipsis+%2B+3]|code-block|syntax|javascript|339297|或者，使用正则表达式：|339298|import+re

def+remwithre(text,+there=re.compile(re.escape('...')%2B'.*')):
++return+there.sub('',+text)|339299|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|$]]

Without a regular expression (which I assume is what you want):
<pre><code>def remafterellipsis(text):
 where_ellipsis = text.find('...')
 if where_ellipsis == -1:
 return text
 return text[:where_ellipsis + 3]
</code></pre>
or, with a regular expression:
<pre><code>import re

def remwithre(text, there=re.compile(re.escape('...')+'.*')):
 return there.sub('', text)
</code></pre>

blocks|key|341791|text|import+re
test+=+"This+is+a+test...we+should+not+be+able+to+see+this"
res+=+re.sub(r'\.\.\..*',"",test)
print(res)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|341792|输出："This+is+a+test“|unstyled|341793|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre class="lang-py prettyprint-override"><code>import re
test = "This is a test...we should not be able to see this"
res = re.sub(r'\.\.\..*',"",test)
print(res)
</code></pre>

Output: "This is a test"

blocks|key|340705|text|从文件中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|340706|import+re
sep+=+'...'

with+open("requirements.txt")+as+file_in:
++++lines+=+[]
++++for+line+in+file_in:
++++++++res+=+line.split(sep,+1)[0]
++++++++print(res)|code-block|syntax|javascript|340707|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|$]]

From a file:

<pre><code>import re
sep = '...'

with open("requirements.txt") as file_in:
 lines = []
 for line in file_in:
 res = line.split(sep, 1)[0]
 print(res)
</code></pre>

blocks|key|3324200|text|方法find将返回字符串中的字符位置。然后，如果要从角色中移除所有内容，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3324201|mystring+=+"123⋯567"
mystring[+0+:+mystring.index("⋯")]

>>+'123'|code-block|syntax|javascript|3324202|如果要保留字符，请将字符位置加1。|3324203|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The method find will return the character position in a string. Then, if you want remove every thing from the character, do this:

<pre><code>mystring = "123⋯567"
mystring[ 0 : mystring.index("⋯")]

&gt;&gt; '123'
</code></pre>

If you want to keep the character, add 1 to the character position.

blocks|key|344257|text|这是在Python3.7中使用的，在我的例子中，我需要删除字符串变量|type|unstyled|depth|inlineStyleRanges|entityRanges|data|344258|中的点后|blockquote|344259|344260|344261|费用=+45.05|344262|split_string+=+fees.split("."，1)|344263|子字符串=+split_string|344264|打印(子字符串)|344265|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$]]|$1|E|3|-4|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

<blockquote>
This is in python 3.7 working to me
In my case I need to remove after dot in my string variable fees
</blockquote>
fees = 45.05 
split_string = fees.split(&quot;.&quot;, 1)
substring = split_string[0]
print(substring)

blocks|key|341825|text|删除字符串中最后一个字符之后的所有字符的另一种方法(假设您想删除最后一个‘/’之后的所有字符)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|341826|path+=+'I/only/want/the/containing/directory/not/the/file.txt'

while+path[-1]+!=+'/':
++++path+=+path[:-1]|code-block|syntax|javascript|341827|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|$]]

Yet another way to remove all characters after the last occurrence of a character in a string (assume that you want to remove all characters after the final '/').
<pre><code>path = 'I/only/want/the/containing/directory/not/the/file.txt'

while path[-1] != '/':
 path = path[:-1]
</code></pre>

blocks|key|3324122|text|使用re的另一种简单方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3324123|import+re,+clr

text+=+'some+string...+this+part+will+be+removed.'

text=+re.search(r'(\A.*)\.\.\..%2B',url,re.DOTALL%7Cre.IGNORECASE).group(1)

//+text+=+some+string|code-block|syntax|javascript|3324124|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|$]]

another easy way using re will be 

<pre><code>import re, clr

text = 'some string... this part will be removed.'

text= re.search(r'(\A.*)\.\.\..+',url,re.DOTALL|re.IGNORECASE).group(1)

// text = some string
</code></pre>

I have a string. How do I remove all text after a certain character? (In this case <code>...</code>) 
The text after will <code>...</code> change so I that's why I want to remove all characters after a certain one.

How to remove all characters after a specific character in python?

我有一根绳子。如何删除某个字符后面的所有文本？(在本例中为...)后面的文本将会...改变，所以这就是为什么我要删除某个字符之后的所有字符。

问如何删除python中特定字符后面的所有字符？
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问如何删除python中特定字符后面的所有字符？
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