I have a string: 0220110000AL0091 and I would like to get back the last 000 for replace by three spaces.
So for: 0220110000AL0091, I want to replace by 0220110 AL0091.
I don't know how to apply the regex between the 7th and 11th characters!
Thanks
You're looking for a negative lookahead.
You will look for a sequence of three zeros, not followed by a zero.
Here is how you could do it in Python (doc here, ctrl+f -> (?!):
>>> import re
>>> s = "0220110000AL0091"
>>> re.sub("000(?!0)", " ", s)
'0220110 AL0091'
>>>
Related
I have a string named
Set-Cookie: BIGipServerApp_Pool_SSL=839518730.47873.0000; path=/
I am trying to extract 839518730.47873.0000 from it. For exact string I am fine with my regex but If I include any digit before 1st = then its all going wrong.
No Digit
>>> m=re.search('[0-9.]+','Set-Cookie: BIGipServerApp_Pool_SSL=839518730.47873.0000; path=/')
>>> m.group()
'839518730.47873.0000'
With Digit
>>> m=re.search('[0-9.]+','Set-Cookie: BIGipServerApp_Pool_SSL2=839518730.47873.0000; path=/')
>>> m.group()
'2'
Is there any way I can extract `839518730.47873.0000' only but doesnt matter what else lies in the string.
I tried
>>> m=re.search('=[0-9.]+','Set-Cookie: BIGipServerApp_Pool_SSL=839518730.47873.0000; path=/')
>>> m.group()
'=839518730.47873.0000'
As well but its starting with '=' in the output and I dont want it.
Any ideas.
Thank you.
If your substring always comes after the first =, you can just use capture group with =([\d.]+) pattern:
import re
result = ""
m = re.search(r'=([0-9.]+)','Set-Cookie: BIGipServerApp_Pool_SSL2=839518730.47873.0000; path=/')
if m:
result = m.group(1) # Get Group 1 value only
print(result)
See the IDEONE demo
The main point is that you match anything you do not need and match and capture (with the unescaped round brackets) the part of pattern you need. The value you need is in Group 1.
You can use word boundaries:
\b[\d.]+
RegEx Demo
Or to make match more targeted use lookahead for next semi-colon after your matched text:
\b[\d.]+(?=\s*;)
RegEx Demo2
Update :
>>> m.group(0)
'839518730.47873.0000'
>>> m=re.search(r'\b[\d.]+','Set-Cookie: BIGipServerApp_Pool_SSL2=839518730.47873.0000; path=/')
>>> m.group(0)
'839518730.47873.0000'
>>>
From Dive into Python3,
re.findall(' s.*? s', "The sixth sick sheikh's sixth sheep's sick.")
It explains that :
The regular expression looks for a space, an s, and then the shortest possible series of any character (.*?), then a space, then another s.
My question is : can .* match the same string as .*? do?
Yes. If the greedy match is identical to the lazy match.
>>> re.findall(' s.*? s', "The sixth sheik") == re.findall(' s.* s', "The sixth sheik")
True
But if greedy match is longer, you will get different results.
>>> re.findall(' s.*? s', "The sixth sick sheik") == re.findall(' s.* s', "The sixth sick sheik")
False
My question is : can .* match the same string as .*? do?
Yes, if there is only one pattern like ' sany s' exists. That is, exactly one match found.
Example:
>>> import re
>>> s = 'foo sgh s'
>>> re.findall(r' s.*? s', s)
[' sgh s']
>>> re.findall(r' s.* s', s)
[' sgh s']
No
check it over here
When remove Question mark:
With python ( regex module ), I am triying to substitute 'x' for each letter 'c' in those strings occurring in a text and:
delimited by 'a', at the left, and 'b' at the right, and
with no more 'a's and 'b's in them.
Example:
cuacducucibcl -> cuaxduxuxibcl
How can I do this?
Thank you.
With the standard re module in Python, you can use a[^ab]+b to match the string which starts and end with a and b and doesn't have any occurence of a or b in between, then supply a replacement function to take care of the replacement of c:
>>> import re
>>> re.sub('a[^ab]+b', lambda m: m.group(0).replace('c', 'x'), 'cuacducucibcl')
'cuaxduxuxibcl'
Document of re.sub for reference.
Use the below regex and then replace the matched c's with x . For this , you need to install external regex module.
>>> import regex
>>> s = 'cuacducucibcl'
>>> regex.sub(r'((?:a|(?<!^)\G)[^abc\n]*)c', r'\1x', s)
'cuaxduxuxibcl'
DEMO
I have tested this Regex
(?<=\))(.+?)(?=\()|(?<=\))(.+?)\b|(.+?)(?=\()
but it doesn't work for strings like this pattern (ef)abc(gh).
I got a result like this "(ef)abc".
But these 3 regexes (?<=\))(.+?)(?=\() , (?<=\))(.+?)\b, (.+?)(?=\()
do work separately for "(ef)abc(gh)", "(ef)abc" ,"abc(ef)" .
can anyone tell me where the problem is or how can I get the expected result?
Assuming you are looking to match the text from between the elements in parenthesis, try this:
^(?:\(\w*\))?([\w]*)(?:\(\w*\))?$
^ - beginning of string
(?:\(\w*\))? - non-capturing group, match 0 or more alphabetic letters within parens, all optional
([\w]*) - capturing group, match 0 or more alphabetic letters
(?:\(\w*\))? - non-capturing group, match 0 or more alphabetic letters within parens, all optional
$ - end of string
You haven't specified what language you might be using, but here is an example in Python:
>>> import re
>>> string = "(ef)abc(gh)"
>>> string2 = "(ef)abc"
>>> string3 = "abc(gh)"
>>> p = re.compile(r'^(?:\(\w*\))?([\w]*)(?:\(\w*\))?$')
>>> m = re.search(p, string)
>>> m2 = re.search(p, string2)
>>> m3 = re.search(p, string3)
>>> print m.groups()[0]
'abc'
>>> print m2.groups()[0]
'abc'
>>> print m3.groups()[0]
'abc'
\([^)]+\)|([^()\n]+)
Try this.Just grab the capture or group.See demo.
https://regex101.com/r/tX2bH4/6
Your problem is that (.+?)(?=\() matches "(ef)abc" in "(ef)abc(gh)".
The easiest solution to this problem is be more explicit about what you are looking for. In this case by exchanging "any character" ., with "any character that is not a parenthesis" [^\(\)].
(?<=\))([^\(\)]+?)(?=\()|(?<=\))([^\(\)]+?)\b|([^\(\)]+?)(?=\()
A cleaner regexp would be
(?:(?<=^)|(?<=\)))([^\(\)]+)(?:(?=\()|(?=$))
Suppose i have a string and i want to match only the part where value is empty and not the part where value is present?
for ex : &lang=&val=1233
I need only &lang and not &val as it has an actual value?
I have this
&(.+)=(?!\s\S)
regex which matches &lang=&val= in the string.
Can anyone help me out
Use following regular expression:
(?:(?<=\?)|&)[^=]+=(?=&|$)
could be explained as:
(?: ....): non-capturing (does not make a group), this may not needed according to your purpose.
\?: escaped ? to match ? literally.
(?<=\?): meaning "preceded by ?": ? is not included to the result.
(?=&|$): meaning "followed by &" or ~at end of the input".
Followings are sample test in Python interactive shell:
>>> pattern = r'(?:(?<=\?)|&)[^=]+=(?=&|$)'
>>> re.findall(pattern, '&lang=&val=')
['&lang=', '&val=']
>>> re.findall(pattern, '&lang=&val=1233')
['&lang=']
>>> re.findall(pattern, '&lang=&val=&val2=123&val3=')
['&lang=', '&val=', '&val3=']
>>> re.findall(pattern, '?lang=&val=&val2=123&val3=')
['lang=', '&val=', '&val3=']
>>> re.findall(pattern, '?lang=blah&val=&val2=123&val3=')
['&val=', '&val3=']
>>> re.findall(pattern, 'www.html.com?user=&lang=eng&code=.in')
do you mean
(&|?)([^&=]+)=(&|$)
(you can use non capturing groups if you need)
but I would just build a hash of all query string parameters and pick the keys without values. it is cheaper.
Try this:
[?&]([^&]+)=(&|$)
The first group will have the name of your parameter.
Note that this regex will also catch an empty first parameter (val1 in foo.php?val1=&val2=ok)
Try this one:
(&([^=]+))=(?=&)