How do I make a part of my Regex non-greedy? - regex

FINAL SOLUTION
/(\d{1,5}\s[^\d].{5,20}(dr|drive)(\.|\s|\,))/i
ORIGINAL QUESTION
Regex
/([0-9]{1,5}.{5,20}(dr|drive)(\.|\s|\,))/i
Pattern
PO Box 66 23 Britton Drive Bloomfield CT 06002
This regex is returning '66 23 Britton Drive'. I want to return '23 Britton Drive'. I have tried the following variations of the Regex:
/(([0-9]{1,5}.{5,20})?(dr|drive)(\.|\s|\,))/i - adding a new capturing group and making it uncreedy
/([0-9]{1,5}.{5,20}?(dr|drive)(\.|\s|\,))/i - making the length of in between characters ungreedy
/([0-9]{1,5}.{5,20}(dr|drive)(\.|\s|\,))/Ui - adding ungreedy modifier
More Patterns That Don't Work
PO Box 156 430 S Wheeling Dr. Wheeling, IL 60090
Patterns That Do Work
1195 Columbia Dr PO Box 1256 Longview, WA 98632
3400 SW Washington drive PO Box 1349 Peoria, IL 61654

^[^\d]+\d{2} (\d{2} [^ ]+ [^ ]+)
Debuggex Demo
Obviously, i'd need more patterns to make this work for all situations
Small fix for your current RegEx
(\d{1,5} [^\d]{5,20}(dr|drive)(\.|\s|\,))

Related

Find a String from a varying number block to the end

I have nearly 8000 lines of the following text:
DIL 2 M 006 SC SCHÜTZ 083 1 Stck
25215-1 BIN-SORT 2152310251724-1 BIN-SORT getestet 048 133 Stck
RBBE60-T3dsg 21S003 SEALING 6X8.9X2.4 MM 082 3 Stck
I am only interested in the 3 digit block at the end and the number behind.
So this should be the output:
083 1
048 133
082 3
It could be, that the same number e.g. 048 appears at the beginning of the line. this shouldn't be a hit.
Unfortunatelly i have no idea how to extract this strings with the help of notpad++.
This expression,
.*(\d{3}\s+\d+).*
with a replacement of $1 is likely to work here.
The expression is explained on the top right panel of this demo if you wish to explore/simplify/modify it.
You may try the following find and replace, in regex mode:
Find: ^.*?(\d+ \d+) \S*$
Replace: $1
The logic here is to use .* to consume everything up until the last two consecutive digits in the line. Then, we replace with only the captured two digits.
Demo

Advanced regex: What would be the regex for this pattern?

Want to identify names of all authors in the following text:
#misc{diaz2006automatic,
title={AUTOMATIC ROCKING DEVICE},
author={Diaz, Navarro David and Gines, Rodriguez Noe},
year={2006},
month=jul # "~12",
note={EP Patent 1,678,025}
}
#article{standefer1984sitting,
title={The sitting position in neurosurgery: a retrospective analysis of 488 cases},
author={Standefer, Michael and Bay, Janet W and Trusso, Russell},
journal={Neurosurgery},
volume={14},
number={6},
pages={649--658},
year={1984},
publisher={LWW}
}
#article{gentsch1992identification,
title={Identification of group A rotavirus gene 4 types by polymerase chain reaction.},
author={GenTSCH, JoN R and Glass, RI and Woods, P and Gouvea, V and Gorziglia, M and Flores, J and Das, BK and Bhan, MK},
journal={Journal of Clinical Microbiology},
volume={30},
number={6},
pages={1365--1373},
year={1992},
publisher={Am Soc Microbiol}
}
For the above text, regex should match:
match1 - Diaz, Navarro David
match2 - Gines, Rodriguez Noe
match3 - Standefer, Michael
match4 - Janet W
match5 - Trusso, Russell
...and so on
Although what you want should be easily achievable by capturing the contents between { and } for all lines starting with author= and then just splitting it using \s*(?:,|\band\b)\s* regex which will give you all the author names.
But just in case, your regex engine is PCRE based, you can use this regex, whose group1 content will give you the author names like you want.
^\s*author={|(?!^)\G((?:(?! and|, )[^}\n])+)(?: *and *)?(?:[^\w\n]*)
This regex exploits \G operator to match lines starting with author= and then starts matching the names which shouldn't contain and or , within it using (?!^)\G((?:(?! and|, )[^}\n])+)(?: *and *)?(?:[^\w\n]*) regex part
Regex Demo

Regex to match blocks of text with key phrases in the middle

VB2010: I have text that consists of blocks of text that start with day and time DD HHMM and end only at the next day/time.
Here is my sample text:
18 2131 Z50000 ZZ-AAA
PR
PR
AGM TPS P773QQ 1500 DCA 22FEB
21,77,23,M10,F,26,3100,2
OK
18 2134 Z50000 ZZ-AAA
PR
QU HMKKDBB
.DDVZAZC 182134
ARR
FI US1500/AN P773QQ/DA KDCA/AD KMIA/IN 2026/FB 152/LA /LR
DT DDL DCAV 182134 M33A
- OS KMIA /GNO6541/R200RR
18 2134 Z50000 ZZ-AAA
PR
PR
ARR OPN P773QQ 1500 DCA 22FEB
0757
OK
18 2135 Z50000 ZZ-AAA
PR
PR
ARR M58 P773QQ 1500 DCA 22FEB
212
UNKNOWN POL/SPOL
QU HMKKDBB
.DDVZAZC 182134
ARR
FI US1500/AN P773QQ/DA KDCA/AD KMIA/IN 2026/FB 152/LA /LR
DT DDL DCAV 182134 M33A
- OS KMIA /GNO6541/R200RR
18 2136 Z50000 ZZ-AAA
PRF 1500/18 MIA IN 0152 333
18 2137 Z50000 ZZ-AAA
PR
PRZ 1500/18 MIA IN 2026 N/A 333
My goal is to get only the blocks of text that have key phrases ^FI and ^DT in the middle. The matching groups should contain only two blocks. The one from 18 2134 and end at M33A and then from 18 2135 to M33A.
I have tried:
This works for the most part except it starts the match at the prior block.
RegexOptions.Singleline Or RegexOptions.Multiline Or RegexOptions.IgnoreCase
^\d\d \d{4}(.*?)^FI US(.*?)^DT DDL(.*?)\r
This one I took from another post but cant seem to wrap my head around. It matches only the first part of every block.
RegexOptions.Multiline Or RegexOptions.IgnoreCase
^\d\d \d{4}.*\r[\s\S]*?(?=(?:^\d\d \d{4}|$))
Haven't used regex in a while so any help appreciated.
You may use
(?ms)^\d\d +\d{4}\b(?:(?!^(?:\d\d +\d{4}\b|FI|DT)).)*?^(?:FI|DT).*?(?=^\d\d +\d{4}\b|\Z)
See the regex demo (Though it is a PCRE regex test, it will work the same in .NET).
Pattern details
(?ms) - multiline and singleline options
^ - start of a line
\d\d +\d{4}\b - 2 digits, 1 or more spaces and 4 digits as a whole word
(?:(?!^(?:\d\d +\d{4}\b|FI|DT)).)*? - any char, 0+ repetitions, as few as possible, that does not start the sequence: start of a line, 2 digits, 1 or more spaces and 4 digits as a whole word, or FI or DT
^(?:FI|DT) - FI or DT at the start of a line
.*? - any 0+ chars, as few as possible
(?=^\d\d +\d{4}\b|\Z) - a positive lookahead that requires ^\d\d +\d{4}\b (start of a line, 2 digits, 1 or more spaces and 4 digits as a whole word) or \Z (end of string) to match immediately to the right of the current location.
This regex should find what you need, if single line enabled
[0-3]\d\s+[0-2]\d[0-5]\d.*?(FI.*?)\n(DT.*?)\n
Explanation:
[0-3]\d\s+[0-2]\d[0-5]\d day hour and minute check
.*? ungreedy capturing, . includes newline
(FI.*?)\n first group, FI line, until line break
(DT.*?)\n second group, same deal

Regex match line starting with whitespace and first character is non-digit

I am trying to create a regex that will only match lines that start with whitespace, then have 1-4 non-digits as the first characters, and then at least one or more spaces after the digits. The purpose of this regex is to use it in the "Find and Replace" option of Notepad++ to remove any lines that do not start with space(s) and then have a number as the first character in the line.
What I have now is allowing me to match the lines that start with whitespace and are followed with a group of digits and another space. However, these are the lines I want to keep. How can I modify the following regex so that it will match everything else other than these lines?
/^([\s]+\d[\s]|[\s]+\d\d[\s]|[\s]+\d\d\d[\s])/gm
Here's an example of the data we're using the regex on. The regex should only match the lines that DO NOT start with 1, 2, 49, 50, 99 and 100. Note that the lines that start with "40th" and "5/23/2017" should match.
Page 1
40th Marathon and 25th Marathon Relay
5/23/2017 USATF Certified Marathon (#RE98723UB) Downtown/City, ST
Timing: Race Services See our Calendar of Events at www.website.com
Results questions: http://www.website.com/fixresults
=====================================================================================
**** FINAL RESULTS IN NETTIME ORDER ****
Place Div/Tot Div Halfway 22miles Guntime Nettime Pace Name
===== ======== ===== ======= ======= ======= ======= ===== =======
1 1/153 M0139 1:15:08 2:05:50 2:29:20 2:29:20 5:42 Eric
2 2/153 M0139 1:15:07 2:06:29 2:29:56* 2:29:56 5:44 Bryan
Record 2:17:35 by Randy in 1986
49 8/77 M4049 1:36:48 2:54:03 3:37:02 3:36:59 8:17 Joshua
50 28/153 M0139 1:49:45 3:03:56 3:37:38# 3:37:22 8:18 Brian
# Under USATF OPEN guideline
99 1/16 M6069 1:56:30 3:15:24 3:51:06 3:50:46 8:49 Paul
100 3/35 F5059 1:50:06 3:11:37 3:51:03 3:50:47 8:49 Ashley
101 4/35 F5059 1:55:26 3:16:37 3:56:03 3:55:57 9:14 Joan
* Under USATF Age-Group guideline
% For an Explanation of AgeGraded Percentages, See Here: http://www.website.com/agegrading
So if we used the regex in Notepad++ to find the matching strings/lines and replace (delete) them, the desired end result would be as follows (in other words, the following lines would NOT match the regex):
1 1/153 M0139 1:15:08 2:05:50 2:29:20 2:29:20 5:42 Eric
2 2/153 M0139 1:15:07 2:06:29 2:29:56* 2:29:56 5:44 Bryan
49 8/77 M4049 1:36:48 2:54:03 3:37:02 3:36:59 8:17 Joshua
50 28/153 M0139 1:49:45 3:03:56 3:37:38# 3:37:22 8:18 Brian
99 1/16 M6069 1:56:30 3:15:24 3:51:06 3:50:46 8:49 Paul
100 3/35 F5059 1:50:06 3:11:37 3:51:03 3:50:47 8:49 Ashley
101 4/35 F5059 1:55:26 3:16:37 3:56:03 3:55:57 9:14 Joan
Any assistance would be greatly appreciated.
See regex in use here
^(?! +\d+ ).*\n*
^ Assert position at the start of the line
(?! +\d+ ) Negative lookahead ensuring what follows is not one or more spaces, then one or more digits, then a space
.* Match any character (except \n) any number of times
\n* Matches any number of newline characters
Result:
1 1/153 M0139 1:15:08 2:05:50 2:29:20 2:29:20 5:42 Eric
2 2/153 M0139 1:15:07 2:06:29 2:29:56* 2:29:56 5:44 Bryan
49 8/77 M4049 1:36:48 2:54:03 3:37:02 3:36:59 8:17 Joshua
50 28/153 M0139 1:49:45 3:03:56 3:37:38# 3:37:22 8:18 Brian
99 1/16 M6069 1:56:30 3:15:24 3:51:06 3:50:46 8:49 Paul
100 3/35 F5059 1:50:06 3:11:37 3:51:03 3:50:47 8:49 Ashley
101 4/35 F5059 1:55:26 3:16:37 3:56:03 3:55:57 9:14 Joan
If this is to use in the Find/Replace dialog then you can use a cunning trick...
^(pattern_I_want_to_keep)$|^.*$
And replace it with
\1
Anything that doesn't match what you want to keep will be removed, although it will leave a blank line. They can be removed with a plugin or another regex.
This is simpler to read than concocting a match for what you don't want to keep, or using a negative lookahead.

regex return everything up to the first space after nth character

I have a list of product names and I want to shorten them (Short Name). I need a regex that will return the first word if it is more than 5 characters and the first two words if it is 5 characters or less.
Product Name Short Name
BABY WIPES MIS /ALOE BABY WIPES
PKU GEL PAK PKU GEL
CA ASCORBATE TAB 500MG CA ASCORBATE
SOD SUL/SULF CRE 10-2% SOD SUL/SULF
ASPIRIN TAB 81MG EC ASPIRIN
IRON TAB 325MG IRON TAB
PEDA PEDA
I initially used:
^([^ \t]+).*
but it only returns the first word so BABY WIPES MIS /ALOE would be BABY. I then tried:
.....([^ \t]+)
But this appears to not work for names less than 5 characters. Any help would be greatly appreciated.
Brief
Your try is close, however, since you negated spaces and tabs, you were unable to move past the first word.
Code
See code in use here
^(\S{1,5}[ \t]*?\S+).*$
Note: The link uses the following shortened regex. \h may not work in your flavour of regex, which is why the code above is posted as well.
^(\S{1,5}\h*?\S+).*$
Super-simplified it becomes ^\S{1,5}\h*?\S+ (without capture groups and .*$ as the OP initially used.)
Results
Input
BABY WIPES MIS /ALOE
PKU GEL PAK
CA ASCORBATE TAB 500MG
SOD SUL/SULF CRE 10-2%
ASPIRIN TAB 81MG EC
IRON TAB
PEDA
Output
BABY WIPES
PKU GEL
CA ASCORBATE
SOD SUL/SULF
ASPIRIN
IRON TAB
PEDA
Explanation
^ Assert position at the start of a line
(\S{1,5}[ \t]*?\S+) Capture group doing the following
\S{1,5} Match any non-whitespace character between 1 and 5 times
[ \t]*? Match space or tab characters any number of times, but as few as possible (note in PCRE regex, this can be replaced with \h*? to make it shorter)
\S+ Match any non-whitespace character between one and unlimited times
.* Match any character (except newline character assuming s modifier is off - it should be for this problem)
$ Assert position at the end of a line
You can use a regex like this:
^\S{1,5} \S+|^\S+
or
^\S{1,5} ?\S*
Working demo
By the way, if you want to replace a full line with the shortened version, then you can use this regex instead:
(^\S{1,5} \S+|^\S+).*
or
(^\S{1,5} ?\S*).*
With the replacement string $1 or \1 depending on your regex engine.
Working demo