This is my input:
0,0,0,1
1,023,1230,1,0
,1,0,01-09-2018,1,
I want to replace 0s and 1s whose length is 1. Rest of them will be as it is.
I already tried with javascript code i.e. split all the strings with "," as delimiter. Then, checking for strings with length 1 and replacing them as per logic. But that's a tedious method which consumes a lot of time.
I want a Regex that can do the replacements in entire input.
I have already tried with this regex: ((0|1)(?<=,))|((0|1)(?=,)). But the output is wrong
Output will be such:
N,N,N,Y
Y,023,1230,Y,N
,Y,N,01-09-2018,Y,
You can use the following regexps with comma word boundaries:
(?<![^,])1(?![^,])
(?<![^,])0(?![^,])
Replace with the appropriate substring.
They match 1 or 0 only when enclosed with commas or start/end of string positions.
(?<![^,]) - a negative lookbehind that matches a position not immediately preceded with a char other than ,
(?![^,]) - a negative lookahead that matches a position not immediately followed with a char other than ,.
Related
Here's the list of my random numbers
r7l5ndecvz9
0qb6rtxsd6ui
dj5iuzpq5vn
rysquf0jkek
435vw5h2qag
And what I want to do is to match or trim the rest after the number of characters (lets say 4), and I want to output as this:
r7l5
0qb6
dj5i
rysq
435v
See, I only want to match the rest of the characters after the 4th character. I tried this regex expression .{4}\b but it only matches the last 4 characters. I also tried this expression ^(\S\S\S\S) but it only matches the first 4 characters. I want to match the rest AFTER the 4th character. Any idea what I'm doing wrong?
Given the sample inputs and outputs, you want to capture the first four characters. The regex for this is ^(.{4})
The caret matches the start of the line, the dot matches any character, the 4 in braces indicates 4 occurrences, and the parentheses capture the result for later access.
Here's working Python code:
import re
inputs = ['r7l5ndecvz9','0qb6rtxsd6ui','dj5iuzpq5vn','rysquf0jkek','435vw5h2qag']
for str in inputs:
match = re.search('^(.{4})',str)
print(match.group(0))
I have trouble understanding why my regex query takes one extra character besides the symbols I have told regex to include into the query, so this is my regex:
([\-:, ]{1,})[^0-9]
This is my test text:
Test- Product-: 1 --- 3 hour ,--kayak:--rental
It always includes the first character of each starting word, like P on Product or h on hour, how can I prevent regex from including those first characters?
I am trying to get all dashes, double points, comma and spaces excluding numbers or any characters.
The [^0-9] part of your regex matches any char but a digit, so you should remove it from your pattern.
There is no need to wrap the character class with a capturing group, and {0,1} is equal to +, so the whole regex can be shortened to
[-:, ]+
Note that - in the initial and end positions inside a character class does not have to be escaped.
I have the following regular expression:
/^[a-f0-9]{8}$/ --- This expression extracts an 8 character string as a md5 hash, for example: if I have the following string "hello world .305eef9f x1xxx 304ccf9f test1232" it will return "304ccf9f"
I also have the following regular expression:
/.[^.]*$/ --- This expression extracts a string after the last period (included), for example, if I have "hello world.this.is.atest.case9.23919sd3xxxs" it will return ".23919sd3xxxs"
Thing is, I've readen a bit about regex but I can't join both expressions in order to find the md5 string after the last period (included), for example:
topLeftLogo.93f02a9d.controller.99f06a7s ----> must return ".99f06a7s"
Thanks in advance for your time and help!
/^[a-f0-9]{8}$/ --- This expression extracts an 8 character string as a md5 hash
Yes but it doesn't return "304ccf9f" from "hello world .305eef9f x1xxx 304ccf9f test1232" because ^ in regex means start of string. How is it possible for it to match in middle of a string?
/.[^.]*$/ --- This expression extracts a string after the last period
No. It will do if you escape first dot only \.
To combine these two you have to replace ^ with \.:
\.[a-f0-9]{8}$
To match your characters 8 times after the last dot in this range [a-f0-9] you might use (if supported) a positive lookahead (?!.*\.) to match your values and assert that what follows does not contain a dot:
\.[a-f0-9]{8}(?!.*\.)
Regex demo
If you want to match characters from a-z instead of a-f like 99f06a7s you could use [a-z0-9]
About the first example
This regex ^[a-f0-9]{8}$ will match one of the ranges in the character class 8 times from the start until the end of the string due to the anchors ^ and $. It would not find a match in hello world .305eef9f x1xxx 304ccf9f test1232 on the same line.
About the second example
.[^.]*$ will match any character zero or more times followed by matching not a dot. That would for example also match a single a and is not bound to first matching a dot because you have to escape the dot to match it literally.
I'm adding this just in case people needs to solve a similar casuistic:
Case 1: for example, we want to get the hexadecimal ([a-f0-9]) 8 char string from our filename string
between the last period and the file extension, in order, for example, to remove that "hashed" part:
Example:
file.name2222.controller.2567d667.js ------> returns .2567d667
We will need to use the following regex:
\.[a-f0-9]{8}(?=\.\w+$)
Case 2: for example, we want the same as above but ignoring the first period:
Example:
file.name2222.controller.2567d667.js ------> returns 2567d667
We will need to use the following regex
[a-f0-9]{8}(?=\.\w+$)
I have not been able to find a proper regex to match any string not starting and ending with some condition.
This matches
AS.E
23.5
3.45
This doesn't match
.263
321.
.ASD
The regex can be alpha-numeric character with optional '.' character and it has to be with in range of 2-4(minimum 2 chars & maximum 4 chars).
I was able to create one ->
^[^\.][A-Z|0-9|\.]{2,4}$
but with this I couldn't achieve mask '.' character at the end of regex.
Thanks.
Maybe not the most optimized but a working one. Created step by step:
The first character should be alphanumeric
^[a-zA-Z0-9]
0, 1 or 2 character alphanumeric or . but not matching end of string
[a-zA-Z0-9\.]{0,2}
an alphanumeric character matching end of string
[a-zA-Z0-9]$
Concatenate all of this to obtain your regex
^[a-zA-Z0-9][a-zA-Z0-9\.]{0,2}[a-zA-Z0-9]$
Edit: This regex allows multiple dots (up to 2)
If I guessed correctly, you want to match all words that are
Between 2 and 4 characters long ...
... and start and end with a character from [A-Z0-9] ...
... and have characters from [A-Z0-9.] in the middle ...
... and are not preceded or followed by a ..
Try this regex to match all these substrings in a text:
(?<=^|[^.])[A-Z0-9][A-Z0-9.]{0,2}[A-Z0-9](?=$|[^.])
However, note that this will match the AA in .AAAA.. If you don't want this match, then please give more details on your requirements.
When you are only interested in the number of matches, but not the matched strings, then you could use
(^|[^.])[A-Z0-9][A-Z0-9.]{0,2}[A-Z0-9]($|[^.])
If you have one string, and want to know whether that string completely matches or not, then use
^[A-Z0-9][A-Z0-9.]{0,2}[A-Z0-9]$
If there may be at most one . inside the match, replace the part [A-Z0-9.]{0,2} with ([A-Z0-9]?[A-Z0-9.]?|[A-Z0-9.]?[A-Z0-9]?).
You can use this pattern to match what you say,
^[^\.][a-zA-Z0-9\.]{2,4}[^\.]$
Check the result here..
https://regex101.com/r/8BNdDg/3
I want to extract from the following regex (?<=^\d+\s*).*?\t trying to extract from the following text just the resources\blahblah:
10 _Resources\index.test FAIL
11 _Resources\index.test FAIL
12 Resources\index.test FAIL
13set\Relicensing Statement.test FAIL
but it captures the following text:
0 _Resources\index.test
1 _Resources\index.test
2 Resources\index.test
3set\Relicensing Statement.test
I just want the lines like Resources\index.test and not the starting numbers, no spaces, why is failing? If I just execute ^\d+\s*and matches with the any number of digits and space, but do not works with prefix.
Since you commented you were using Notepad++, how about matching ^\d+\s*([^\t]*).*$ and replacing by \1 ?
From NSRegularExpression (I saw it was tagged):
Look-behind assertion. True if the parenthesized pattern matches text
preceding the current input position, with the last character of the
match being the input character just before the current position. Does
not alter the input position. The length of possible strings matched
by the look-behind pattern must not be unbounded (no * or +
operators.)
The same problem holds in most of the languages.
Can't you extract $1 from (?:^\d+\s*)(.*?\t)?