,Ray Balwierczak,4/11/2017,,895 Forest Hill Rd,Apalachin,NY,13732,y,,
i want to select only 13732 from the line. I came up with this regex
(\d)(\s*\d+)*(\,y,,)
But its also selecting the ,y,, .if i remove it that part from regex, the regex also gets valid for the date. please help me on this.
Generally, if you want to match something without capturing it, use zero-length lookaround (lookahead or lookbehind). In your case, you can use lookahead:
(\d)(\s*\d+)*(?=\,y,,)
The syntax (?=<stuff>) means "followed by <stuff>, without matching it".
More information on lookarounds can be found in this tutorial.
Regex: \D*(\d{5})\D*
Explanation: match 5 digits surrounded by zero or more non-digits on both sides. Then you can extract group containing the match.
Here's code in python:
import re
string = ",Ray Balwierczak,4/11/2017,,895 Forest Hill Rd,Apalachin,NY,13732,y,,"
search = re.search("\D*(\d{5})\D*", string)
print search.group(1)
Output:
13732
Related
For example we have a string:
asd/asd/asd/asd/1#s_
I need to match this part: /asd/1#s_ or asd/1#s_
How is it possible to do with plain regex?
I've tried negative lookahead like this
But it didn't work
\/(?:.(?!\/))?(asd)(\/(([\W\d\w]){1,})|)$
it matches this '/asd/asd/asd/asd/asd/asd/1#s_'
from this 'prefix/asd/asd/asd/asd/asd/asd/1#s_'
and I need to match '/asd/1#s_' without all preceding /asd/'s
Match should work with plain regex
Without any helper functions of any programming language
https://regexr.com/
I use this site to check if regex matches or not
here's the possible strings:
prefix/asd/asd/asd/1#s
prefix/asd/asd/asd/1s#
prefix/asd/asd/asd/s1#
prefix/asd/asd/asd/s#1
prefix/asd/asd/asd/#1s
prefix/asd/asd/asd/#s1
and asd part could be replaced with any word like
prefix/a1sd/a1sd/a1sd/1#s
prefix/a1sd/a1sd/a1sd/1s#
...
So I need to match last repeating part with everything to the right
And everything to the right could be character, not character, digit, in any order
A more complicated string example:
prefix/a1sd/a1sd/a1sd/1s#/ds/dsse/a1sd/22$$#!/123/321/asd
this should match that part:
/a1sd/22$$#!/123/321/asd
Try this one. This works in python.
import re
reg = re.compile(r"\/[a-z]{1,}\/\d+[#a-z_]{1,}")
s = "asd/asd/asd/asd/1#s_"
print(reg.findall(s))
# ['/asd/1#s_']
Update:
Since the question lacks clarity, this only works with the given order and hence, I suppose any other combination simply fails.
Edits:
New Regex
reg = r"\/\w+(\/\w*\d+\W*)*(\/\d+\w*\W*)*(\/\d+\W*\w*)*(\/\w*\W*\d+)*(\/\W*\d+\w*)*(\/\W*\w*\d+)*$"
I'm having an issue with Regex.
I'm trying to match T0000001 (2, 3 and so on).
However, some of the lines it searches has what I can describe as positioners. These are shown as a question mark, followed by 2 digits, such as ?21.
These positioners describe a new position if the document were to be printed off the website.
Example:
T123?214567
T?211234567
I need to disregard ?21 and match T1234567.
From what I can see, this is not possible.
I have looked everywhere and tried numerous attempts.
All we have to work off is the linked image. The creators cant even confirm the flavour of Regex it is - they believe its Python but I'm unsure.
Regex Image
Update
Unfortunately none of the codes below have worked so far. I thought to test each code in live (Rather than via regex thinking may work different but unfortunately still didn't work)
There is no replace feature, and as mentioned before I'm not sure if it is Python. Appreciate your help.
Do two regex operations
First do the regex replace to replace the positioners with an empty string.
(\?[0-9]{2})
Then do the regex match
T[0-9]{7}
If there's only one occurrence of the 'positioners' in each match, something like this should work: (T.*?)\?\d{2}(.*)
This can be tested here: https://regex101.com/r/XhQXkh/2
Basically, match two capture groups before and after the '?21' sequence. You'll need to concatenate these two matches.
At first, match the ?21 and repace it with a distinctive character, #, etc
\?21
Demo
and you may try this regex to find what you want
(T(?:\d{7}|[\#\d]{8}))\s
Demo,,, in which target string is captured to group 1 (or \1).
Finally, replace # with ?21 or something you like.
Python script may be like this
ss="""T123?214567
T?211234567
T1234567
T1234434?21
T5435433"""
rexpre= re.compile(r'\?21')
regx= re.compile(r'(T(?:\d{7}|[\#\d]{8}))\s')
for m in regx.findall(rexpre.sub('#',ss)):
print(m)
print()
for m in regx.findall(rexpre.sub('#',ss)):
print(re.sub('#',r'?21', m))
Output is
T123#4567
T#1234567
T1234567
T1234434#
T123?214567
T?211234567
T1234567
T1234434?21
If using a replace functionality is an option for you then this might be an approach to match T0000001 or T123?214567:
Capture a T followed by zero or more digits before the optional part in group 1 (T\d*)
Make the question mark followed by 2 digits part optional (?:\?\d{2})?
Capture one or more digits after in group 2 (\d+).
Then in the replacement you could use group1group2 \1\2.
Using word boundaries \b (Or use assertions for the start and the end of the line ^ $) this could look like:
\b(T\d*)(?:\?\d{2})?(\d+)\b
Example Python
Is the below what you want?
Use RegExReplace with multiline tag (m) and enable replace all occurrences!
Pattern = (T\d*)\?\d{2}(\d*)
replace = $1$2
Usage Example:
In Geany, I want to match the titles of books. One example:
Michael Lewis, Liar's Poker, Hodder & Stoughton Ltd, London, 1989
I try to do so with this regex code:
,\s.*?,
This regex matches too much. it matches: [, Liar's Poker,] and [,London,].
I want to have a regex that only matches the title.
I think you need this regex with no global modifier. If you set global modifier i.e. g then it will return further matches like you have experienced.
,\s*([^,]+)
Demo
As you want to ignore further matches thus you may try this too:
^.*?,\s*([^,]+).*$
You will get Liar's Poker in group 1
Demo 2
/(, \w+[']?\w? \w+,)/g
this regex will get you this
[", Liar's Poker,"]
you will have to do additional processing to remove those leading and trailing commas. Try it out and see if this works for you.
I am trying to replace street with st only if street isn't followed by any alphabet. Replacement is allowed if after street EITHER there is non-alphabet character OR end of string.
I am trying to achieve this in Postgresql 9.5 regex_replace function. Sample query i wrote:
select regexp_replace('super streetcom','street(?!=[a-z])','st');
street here shouldn't have been replaced by st since street is followed by 'c'. So the expected output is 'super streetcom' but the output i am getting is 'super stcom'.
Any help for why i am getting the unexpected output and what can be the right way to achieve the intended result.
A lookahead construct looks like (?!...), all what follows ?! is a lookahead pattern that the engine will try to match, and once found, the match will be failed.
It seems you need to match a whole word street. Use \y, a word boundary:
select regexp_replace('super streetcom street','\ystreet\y','st');
See the online demo
From the docs:
\y matches only at the beginning or end of a word
This looks like a syntax issue. Try: ?! instead of ?!= .
e.g.
select regexp_replace('super street','street(?![a-z])','st');
will return
super st
Hi, my fellow RegEx'ers ;)
I'm trying to match multiple Texts between every two quotes
Here's my text:
...random code
someArray[] = ["Come and",
"get me,",
"or fail",
"trying!",
"Yours truly"]
random code...
So far, I managed to get the correct matches with two patterns, executed after each other:
(?s)someArray\[\].*?=.*?\[(.*?)\]
this extracts the text between the two brackets and on the result, I use this one:
"(.*?)"
This is working just fine, but I'd love to get the Texts in one regex.
Any help is highly appreciated!
Consider using \G. With its help, you may match "(.*?)" preceded by either someArray[] = [ or previous match of "(.*?)" (well, strictly speaking previous match of entire regex). Then just grab first capture groups from all matches:
(?:(?s).*someArray\[\].*?=.*?\[|\G[^"\]]+)"(.*?)"
Demo: https://regex101.com/r/eBQWdU/3
How you grab the first capture groups from depends on the language you're using regex in. For example in PHP you may do something like this:
preg_match_all('/(?:(?s).*someArray\[\].*?=.*?\[|\G[^"\]]+)"(.*?)"/', $input, $matches);
$array_items = $matches[1];
Demo: https://ideone.com/mZgU1x