I'm wondering if it is possible to use "groups" in ReGeX used in Open Refine GREL syntax. I mean, I'd like to replace all the dots followed and preceded by a character WITH the same character and dot but followed by a space and then the character.
Something like:
s.replace(/(.{1})\..({1})/,/(1).\s(2)/)
It should, but your last argument needs to be a string, not a regular expression. Internally Refine uses Java's Matcher#replaceAll method which accepts a string argument.
I think I found out how to deal with this. You need to put $X in your string value to address a Xth capture group.
It should be like this:
s.replace(/.?(#capcure group 1).?(#capcure group 2).*?/), " some text $1 some text $2 some text")
Related
I am a complete Reg-exp noob, so please bear with me. Tried to google this, but haven't found it yet.
What would be an appropriate way of writing a Regular expression matching files starting with a dot, such as .buildpath or .htaccess?
Thanks a lot!
In most regex languages, ^\. or ^[.] will match a leading dot.
The ^ matches the beginning of a string in most languages. This will match a leading .. You need to add your filename expression to it.
^\.
Likewise, $ will match the end of a string.
You may need to substitute the \ for the respective language escape character. However, under Powershell the Regex I use is: ^(\.)+\/
Test case:
"../NameOfFile.txt" -match '^(\\.)+\\\/'
works, while
"_./NameOfFile.txt" -match '^(\\.)+\\\/'
does not.
Naturally, you may ask, well what is happening here?
The (\\.) searches for the literal . followed by a +, which matches the previous character at least once or more times.
Finally, the \\\/ ensures that it conforms to a Window file path.
It depends a bit on the regular expression library you use, but you can do something like this:
^\.\w+
The ^ anchors the match to the beginning of the string, the \. matches a literal period (since an unescaped . in a regular expression typically matches any character), and \w+ matches 1 or more "word" characters (alphanumeric plus _).
See the perlre documentation for more info on Perl-style regular expressions and their syntax.
It depends on what characters are legal in a filename, which depends on the OS and filesystem.
For example, in Windows that would be:
^\.[^<>:"/\\\|\?\*\x00-\x1f]+$
The above expression means:
Match a string starting with the literal character .
Followed by at least one character which is not one of (whole class of invalid chars follows)
I used this as reference regarding which chars are disallowed in filenames.
To match the string starting with dot in java you will have to write a simple expression
^\\..*
^ means regular expression is to be matched from start of string
\. means it will start with string literal "."
.* means dot will be followed by 0 or more characters
I have a string like: Title Name_2021-04-13_A+B+C_Division.txt. I need to extract the A+B+C. The A+B+C may be other letters. I believe that using Regex would be the simplest way to do this. In other words I need to get the substring between underscore 2 and underscore 3 of string. All of my code is written in vb.net. I have tried:
boatClass = Regex.Match(myFile, "(?<=_)(.*)(?=_)").ToString
I know this is not right but I think it is close. What do I need to add or change?
The regex code that will extract a substring between the second and third underscore of a string is:
(?:[^_]+_){2}([^_]+)
However, I chose to use the split function:
myString.Split("_"c)(2)
I have two types of URL's which I would need to clean, they look like this:
["//xxx.com/se/something?SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"]
["//www.xxx.com/se/car?p_color_car=White?SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"]
The outcome I want is;
SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"
I want to remove the brackets and everything up to SE, the URLS differ so I want to remove:
First URL
["//xxx.com/se/something?
Second URL:
["//www.xxx.com/se/car?p_color_car=White?
I can't get my head around it,I've tried this .*\/ . But it will still keep strings I don't want such as:
(1 url) =
something?
(2 url) car?p_color_car=White?
You can use
regexp_replace(FinalUrls, r'.*\?|"\]$', '')
See the regex demo
Details
.*\? - any zero or more chars other than line breakchars, as many as possible and then ? char
| - or
"\]$ - a "] substring at the end of the string.
Mind the regexp_replace syntax, you can't omit the replacement argument, see reference:
REGEXP_REPLACE(value, regexp, replacement)
Returns a STRING where all substrings of value that match regular
expression regexp are replaced with replacement.
You can use backslashed-escaped digits (\1 to \9) within the
replacement argument to insert text matching the corresponding
parenthesized group in the regexp pattern. Use \0 to refer to the
entire matching text.
I have a column in Openrefine, which I would like to add a character string in each of its rows, based on the position in the string.
For example:
I have an 8th character number string: 85285296 and would like to add "-" at the fourth place: "8528-5296".
Anyone can help me find the specific function in OpenRefine?
Thanks
Tzipy
The simplest approach is to just use the expression language's built-in string indexing and concatenation:
value[0,4]+'-'+value[4,8]
or more generally, if you don't know that your value is exactly 8 characters long:
value[0,4]+'-'+value[4,999]
Possible solution (not sure if it's the most straightforward):
value.replace(/(\d{4})(.+)/, "$1-$2")
This means : if $1 represents the content of the first parenthesis/group in the regular expression before and $2 the content of the second one, replaces each value in the column with $1-$2.
Some other options:
value.splitByLengths(4,4).join("-")
value.match(/(\d{4})(\d{4})/).join("-")
value.substring(0,4)+"-"+value.substring(4,8)
I think 'splitByLengths' is the neatest, but I might use 'match' instead because it fails with an error if your starting string isn't 8 digits - which means you don't accidentally process data that doesn't conform to your assumption of what data is in the column - but you could use a facet/filter to check this with any of the others
I'm trying to figure out the base regex to capture the middle of a google url out of a sql database.
For example, a few links:
https://www.google.com/cars/?year=2016&model=dodge+durango&id=1234
https://www.google.com/cars/?year=2014&model=jeep+cherokee+crossover&id=6789
What would be the regex to capture the text to get dodge+durango , or jeep+cherokee+crossover ? (It's alright that the + still be in there.)
My Attempts:
1)
\b[=.]\W\b\w{5}\b[+.]?\w{7}
, but this clearly does not work as this is a hard coded scenario that would only work like something for the dodge durango example. (would extract "dodge+durango)
2) Using positive lookback ,
[^+]( ?=&id )
but I am not fully sure how to use this, as this only grabs one character behind the & symbol.
How can I extract a string of (potentially) any length with any amount of + delimeters between the "model=" and "&id" boundaries?
seems like you could use regexp_replace and access match groups:
regexp_replace(input, 'model=(.*?)([&\\s]|$)', E'\\1')
from here:
The regexp_replace function provides substitution of new text for
substrings that match POSIX regular expression patterns. It has the
syntax regexp_replace(source, pattern, replacement [, flags ]). The
source string is returned unchanged if there is no match to the
pattern. If there is a match, the source string is returned with the
replacement string substituted for the matching substring. The
replacement string can contain \n, where n is 1 through 9, to indicate
that the source substring matching the n'th parenthesized
subexpression of the pattern should be inserted, and it can contain \&
to indicate that the substring matching the entire pattern should be
inserted. Write \ if you need to put a literal backslash in the
replacement text. The flags parameter is an optional text string
containing zero or more single-letter flags that change the function's
behavior. Flag i specifies case-insensitive matching, while flag g
specifies replacement of each matching substring rather than only the
first one
I may be misunderstanding, but if you want to get the model, just select everything between model= and the ampersand (&).
regexp_matches(input, 'model=([^&]*)')
model=: Match literally
([^&]*): Capture
[^&]*: Anything that isn't an ampersand
*: Unlimited times