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))
Related
I need to write a regular expression that has to replace everything except for a single group.
E.g
IN
OUT
OK THT PHP This is it 06222021
This is it
NO MTM PYT Get this content 111111
Get this content
I wrote the following Regular Expression: (\w{0,2}\s\w{0,3}\s\w{0,3}\s)(.*?)(\s\d{6}(\s|))
This RegEx creates 4 groups, using the first entry as an example the groups are:
OK THT PHP
This is it
06222021
Space Charachter
I need a way to:
Replace Group 1,2,4 with String.Empty
OR
Get Group 3, ONLY
You don't need 4 groups, you can use a single group 1 to be in the replacement and match 6-8 digits for the last part instead of only 6.
Note that this \w{0,2} will also match an empty string, you can use \w{1,2} if there has to be at least a single word char.
^\w{0,2}\s\w{0,3}\s\w{0,3}\s(.*?)\s\d{6,8}\s?$
^ Start of string
\w{0,2}\s\w{0,3}\s\w{0,3}\s Match 3 times word characters with a quantifier and a whitespace in between
(.*?) Capture group 1 match any char as least as possible
\s\d{6,8} Match a whitespace char and 6-8 digits
\s? Match an optional whitespace char
$ End of string
Regex demo
Example code
Dim s As String = "OK THT PHP This is it 06222021"
Dim result As String = Regex.Replace(s, "^\w{0,2}\s\w{0,3}\s\w{0,3}\s(.*?)\s\d{6,8}\s?$", "$1")
Console.WriteLine(result)
Output
This is it
My approach does not work with groups and does use a Replace operation. The match itself yields the desired result.
It uses look-around expressions. To find a pattern between two other patterns, you can use the general form
(?<=prefix)find(?=suffix)
This will only return find as match, excluding prefix and suffix.
If we insert your expressions, we get
(?<=\w{0,2}\s\w{0,3}\s\w{0,3}\s).*?(?=\s\d{6}\s?)
where I simplified (\s|) as \s?. We can also drop it completely, since we don't care about trailing spaces.
(?<=\w{0,2}\s\w{0,3}\s\w{0,3}\s).*?(?=\s\d{6})
Note that this works also if we have more than 6 digits because regex stops searching after it has found 6 digits and doesn't care about what follows.
This also gives a match if other things precede our pattern like in 123 OK THT PHP This is it 06222021. We can exclude such results by specifying that the search must start at the beginning of the string with ^.
If the exact length of the words and numbers does not matter, we simply write
(?<=^\w+\s\w+\s\w+\s).*?(?=\s\d+)
If the find part can contain numbers, we must specify that we want to match until the end of the line with $ (and include a possible space again).
(?<=^\w+\s\w+\s\w+\s).*?(?=\s\d+\s?$)
Finally, we use a quantifier for the 3 ocurrences of word-space:
(?<=^(\w+\s){3}).*?(?=\s\d+\s?$)
This is compact and will only return This is it or Get this content.
string result = Regex.Match(#"(?<=^(\w+\s){3}).*?(?=\s\d+\s?$)").Value;
This is my input:
0,0,0,1
1,023,1230,1,0
,1,0,01-09-2018,1,
I want to replace 0s and 1s whose length is 1. Rest of them will be as it is.
I already tried with javascript code i.e. split all the strings with "," as delimiter. Then, checking for strings with length 1 and replacing them as per logic. But that's a tedious method which consumes a lot of time.
I want a Regex that can do the replacements in entire input.
I have already tried with this regex: ((0|1)(?<=,))|((0|1)(?=,)). But the output is wrong
Output will be such:
N,N,N,Y
Y,023,1230,Y,N
,Y,N,01-09-2018,Y,
You can use the following regexps with comma word boundaries:
(?<![^,])1(?![^,])
(?<![^,])0(?![^,])
Replace with the appropriate substring.
They match 1 or 0 only when enclosed with commas or start/end of string positions.
(?<![^,]) - a negative lookbehind that matches a position not immediately preceded with a char other than ,
(?![^,]) - a negative lookahead that matches a position not immediately followed with a char other than ,.
I have not been able to find a proper regex to match any string not starting and ending with some condition.
This matches
AS.E
23.5
3.45
This doesn't match
.263
321.
.ASD
The regex can be alpha-numeric character with optional '.' character and it has to be with in range of 2-4(minimum 2 chars & maximum 4 chars).
I was able to create one ->
^[^\.][A-Z|0-9|\.]{2,4}$
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 match specific strings from beginning to 5th word of article title.
Input string:
The 14 best US colleges in the West are dominated by California — here's who makes the cut.
regex:
/^.*(\bbest\b|\btop\b|\bhot\b).*$/
Currently matched whole article title but want to search till "colleges".
and also need ignore or not matched strings like laptop,hot-spot etc.
You can use this expression
^((?:\w+\s?){1,5}).*
Explanation:
^ assert position at start of the string
\w+ match any word character
\s? match any white space character
{1,5} Quantifier - Between 1 and 5 times, as many times as possible
.* matches any character (except newline)
This matches the first 5 words (and spaces).
^(\w+\s){0,4}\b(best|top|hot)(\s|$)
You want to match string within first five words of input sentence. Then if counted from the start the sentence, there must be 0-4 words before the word you want to match. So you need ^(\w+\s){0,4} before the specific words you want to match. See https://regex101.com/r/nS0dU6/4
regex101 comes to help again.
^(?=(?:\w+\s){0,4}?(?:best|top|hot)\b(?!-))(\w+(?:\s\w+){0,4})
(?=(?:\w+\s){0,4}?(?:best|top|hot)\b(?!-) checks that the keyword is within first 5 (note that (?!-) is added to cater for words such as hot-spot)
(\w+(?:\s\w+){0,4}) then matches the first maximum 5 words
Using this for an example string
+$43073$7
and need the 5 number sequence from it I'm using the Regex expression
#"\$+(?<lot>\d{5})"
which is matching up any +$ in the string. I tried
#"^\$+(?<lot>\d{5})"
as the +$ are always at the beginning of the string. What will work?
If you use anchor ^, you need to include the + symbol at the first and don't forget to escape it because + is a special meta character in regex which repeats the previous token one or more times.
#"^\+\$(?<lot>\d{5})"
And without the anchor, it would be like
#"\$(?<lot>\d{5})"
And get the 5 digit number you want from group index 1.
DEMO
I would match what you want:
\d+
or if you only want digits after "special" characters at the start of input:
^\W+(\d+)
grabbing group 1