I have a long file which contains texts and there is a particular type of text which is 24 character long (it can be lowercase alphabet or number),it starts and ends with " and there are parentheses around this.
One of these kind of row looks like this:
"something": ("qwertyuiopasdfghjklz1234"),
and I would like to get:
"something": "qwertyuiopasdfghjklz1234",
I would like to remove the parentheses. I have the following regex: ([a-z0-9"]{26}) which finds this expression but I don't seem to find a way to figure out what to write to replace the row in order to delete the parentheses.
You may use
(:\h*)\(("[a-z0-9]{24}")\)(,)
Replace with $1$2$3, see the regex demo.
Details
(:\h*) - Group 1 ($1): : and 0 or more horizontal whitespaces
\( - a ( char
("[a-z0-9]{24}") - Group 2 ($2): ", 24 lowercase ASCII letters or digits and then a "
\) - a ) char
(,) - Group 3 ($3): a , char.
You may use below code, working for your sample.
Find what: \(|\)
Replace with: nothing
Related
I have tags (only ASCII chars inside brackets) of the following structure: [Root.GetSomething], instead, some contributors ended up submitting contributions with Cyrillic chars that look similar to Latin ones, e.g. [Rооt.GеtSоmеthіng].
I need to locate, and then replace those inconsistencies with the matching ASCII characters inside the brackets.
I tried \[([АаІіВСсЕеРТтОоКкХхМ]+)\]; (\[)([^\x00-\x7F]+)(\]), and some variations of the range but those searches don't see any matches. I seem to be missing something important in the regex execution logic.
You can use a regex matching any "interesting" Cyrillic char in between [ + letters or . + ] and a conditional replacement pattern:
Find What: (?:\G(?!\A)|\[)[a-zA-Z.]*\K(?:(А)|(а)|(І)|(і)|(В)|(С)|(с)|(Е)|(е)|(Р)|(Т)|(т)|(О)|(о)|(К)|(к)|(Х)|(х)|(М))(?=[[:alpha:].]*])
Replace With: (?1A:?2a:?3I:?4i:?5B:?6C:?7c:?8E:?9e:?{10}P:?{11}T:?{12}t:?{13}O:?{14}o:?{15}K:?{16}k:?{17}X:?{18}x:?{19}M)
Make sure Match Case option is ON. See a regex demo with a string:
Details:
(?:\G(?!\A)|\[) - end of the previous successful match or a [ char
[a-zA-Z.]* - zero or more . or ASCII letters
\K - match reset operator that discards the currently matched text from the overall match memory buffer
(?:(А)|(а)|(І)|(і)|(В)|(С)|(с)|(Е)|(е)|(Р)|(Т)|(т)|(О)|(о)|(К)|(к)|(Х)|(х)|(М)) - a non-capturing group containing 19 alternatives each of which is put into a separate capturing group
(?=[[:alpha:].]*]) - a positive lookahead that requires zero or more letters or . and then a ] char immediately to the right of the current location.
The (?1A:?2a:?3I:?4i:?5B:?6C:?7c:?8E:?9e:?{10}P:?{11}T:?{12}t:?{13}O:?{14}o:?{15}K:?{16}k:?{17}X:?{18}x:?{19}M) replacement pattern replaces А with A (\u0410) if Group 1 matched, а (\u0430) with a if Group 2 matched, etc.
I have a one line string that looks like this:
{"took":125,"timed_out":false,"_shards":{"total":10,"successful":10,"skipped":0,"failed":0}}{"took":365,"timed_out":false,"_shards":{"total":10,"successful":10,"skipped":0,"failed":0}}{"took":15,"timed_out":false,"_shards":{"total":10,"successful":10,"skipped":0,"failed":0}}
I would like to extract all the numbers after the "took" part, so in my case the output would look like this:
125
365
15
What I've tried so far is using took":(\d{1,6}),"(.*) as a regex. But since its a one line string, it only extracts the first occurence and ignores the others.
You can use
Find What: took":(\d+)|(?s)(?:(?!took":\d).)*
Replace With: (?{1}$1\n)
Details:
took": - literal text
(\d+) - one or more digits captured into Group 1
| - or
(?s) - set the DOTALL mode on (. matches line break chars now)
(?:(?!took":\d).)* - any single char, zero or more times, as many as possible, that does not start a took": + digit char sequence.
The (?{1}$1\n) conditional replacement pattern replaces this way:
(?{1} - if Group 1 is matched
$1\n - replace the match with Group 1 and a newline
) - else, replace with an empty string.
I am trying to use replace in Sublime using regular expressions but I'm stuck. I tried various combinations but don't seem to be getting there.
This is the input and my desired output:
Input: N_BBP_c_46137_n
Output : BBP
I tried combinations of:
[^BBP]+\b
\*BBP*+\g
But none of the above (and many others) don't seem to work.
To turn N_BBP_c_46137_n into BBP and according to the comment just want that entire long name such as N_BBP_ to be replaced by only BBP* you might also use a capture group to keep BBP.
\bN_(BBP)_\S*
\bN_ Match N preceded by a word boundary
(BBP) Capture group 1, match BBP (or use [A-Z]+ to match 1+ uppercase chars)
_\S* Match _ followed by 0+ times a non whitespace char
In the replacement use the first capturing group $1
Regex demo
You may use
(N_)[^_]*(_c_\d+_n)
Replace with ${1}some new value$2.
Details
(N_) - Group 1 ($1 or ${1} if the next char is a digit): N_
[^_]* - any 0 or more chars other than _
-(_c_\d+_n) - Group 2 ($2): _c_, 1 or more digits and then _n.
See the regex demo.
I have a list of thousands of records within a .txt document.
some of them look like these records
201910031044 "00059" "11.31AG" "Senior Champion"
201910031044 "00060" "GBA146" "Junior Champion"
201910031044 "00999" "10.12G" "ProAM"
201910031044 "00362" "113.1LI" "Abcd"
Whenever a record similar to this occurs I'd like to get rid of the last words/numbers/etc in the last quotation marks (like "Senior Champion", "Junior Champion" etc. There are many possibilities here)
e.g. (before)
201910031044 "00059" "11.31AG" "Senior Champion"
after
201910031044 "00059" "11.31AG"
I tried the following regex but it wouldn't work.
Search: ^([0-9]{17,17} + "[0-9]{8,8}" + "[a-zA-Z0-9]").*$
Replace: \1 (replace string)
OK I forgot the . (dot) sign however even if I do not have a . (dot) sign it would not work. Not sure if it has anything to do when using the + sign used more than once.
I'd like to get rid of the last words/numbers/etc in the last quotation marks
This does the job:
Ctrl+H
Find what: ^.+\K\h+".*?"$
Replace with: LEAVE EMPTY
CHECK Wrap around
CHECK Regular expression
UNCHECK . matches newline*
Replace all
Explanation:
^ # beginning of line
.+ # 1 or more any character but newline
\K # forget all we have seen until this position
\h+ # 1 or more horizontal spaces
".*?" # something inside quotes
$ # end of line
Screen capture (before):
Screen capture (after):
The RegEx looks for the 4th double quote:
^(?:[^"]*\"){4}([^|]*)
You can see this demo: https://regex101.com/r/wJ9yS6/163
You will still need to parse the lines, so probably easier opening in excel or parsing using code as a CSV.
You have a problem with the count of your characters:
you specify that the line should start with exactly 17 digits ([0-9]{17,17}). However, there are only 12 digits in the data 201910031044.
you can specify exactly 12 digits by using {12} or if it could be 12-17, then {12,17}. I'll assume exactly 12 based on the current data.
similarly, for the second column you specify that it's exactly 8 digits surrounded by quotes ("[0-9]{8,8}") but it only has 5 digits surrounded by quotes.
again, you can specify exactly 5 with {5} or 5-8 with {5,8}. I will assume exactly 5.
finally, there is no quantifier for the final field, so the regex tries to match exactly one character that is a letter or a number surrounded by quotes "[a-zA-Z0-9]".
I'm not sure if there is any limit on the number of characters, so I would go with one or more using + as quantifier "[a-zA-Z0-9]+" - if you can have zero or more, then you can use *, or if it's any other count from m to n, then you can use {m,n} as before.
Not a character count problem but the final column can also have dots but the regex doesn't account for. You can just add . inside the square brackets and it will only match dot characters. It's usually used as a wildcard but it loses its special meaning inside a character class ([]), so you get "[a-zA-Z0-9.]+"
Putting it all together, you get
Search: ^([0-9]{12} + "[0-9]{5}" + "[a-zA-Z0-9.]+").*$
Replace: \1
Which will get rid of anything after the third field in Notepad++.
This can be shortened a bit by using \d instead of [0-9] for digits and \s+ for whitespace instead of +. As a benefit, \s will also match other whitespace like tabs, so you don't have to manually account for those. This leads to
Search: ^(\d{12}\s+"\d{5}"\s+"[a-zA-Z0-9.]+").*$
Replace: \1
If you want to get rid of the last words/numbers/etc in the last quotation marks you could capture in a group what is before that and match the last quotation marks and everything between it to remove it using a negated character class.
If what is between the values can be spaces or tabs, you could use [ \t]+ to match those (using \s could also match a newline)
Note that {17,17} and {8,8} may also be written as {17} and {8} which in this case should be {12} and {5}
^([0-9]{12}[ \t]+"[0-9]{5}"[ \t]+"[a-zA-Z0-9.]+")[ \t]{2,}"[^"\r\n]+"
In parts
^ Start of string
( Capture group 1
[0-9]{12}[ \t]+ Match 12 digits and 1+ spaces or tabs
"[0-9]{5}"[ \t]+ Match 5 digits between " and 1+ spaces or tabs
"[a-zA-Z0-9.]+" Match 1+ times any of the listed between "
) Close group
[ \t]{2,} Match 1+ times
"[^"\r\n]+"
In the replacement use group 1 $1
Regex demo
Before
After
I have this two lines of text, that I want to manipulate using Regular Expression and substitute:
Obj.FieldNameA = Reader.GetEnumFromInt32<ClassName>(QueryGenerator,nameof(Obj.));
Obj.FieldNameB=Reader.GetTrimmedStringOrNull(QueryGenerator,nameof(Obj.));
Attached on the first Obj. there is a Field name, so in this case they are FieldNameA,FieldNameB
I want to attach these values to the second Obj. found on the same line, so the text should become:
Obj.FieldNameA = Reader.GetEnumFromInt32<ClassName>(QueryGenerator,nameof(Obj.FieldNameA));
Obj.FieldNameB=Reader.GetTrimmedStringOrNull(QueryGenerator,nameof(Obj.FieldNameB));
I have tested this very simple (and wrong) regex:
Obj\.(\w*).*\n
With substituition as $1
But I don't know how to use substitution...
Sample code here
Some Notes:
After FieldNameA there is always an equal sign that could be preceded or followed by a space.
Before the second Obj. there could be any character, including < ( etc...
Could this be achieved?
You may use
Find: (Obj\.(\w+).*\(Obj\.)\)
Replace: $1$2)
See the regex demo.
You may also add ^ to the start of the regex to match only at the start of a line/string.
Details
^ - start of string
(Obj\.(\w+).*\(Obj\.) - Group 1 ($1 in the replacement):
Obj\. - Obj. text
(\w+) - Group 2 ($2): 1 or more word chars
.* - any 0+ chars other than line break chars as many as possible (you may use .*? to only match the second Obj. on a line, your current input only has two with the second one closer to the end of a line, so .* will work better)
\(Obj\. - (Obj. text
\) - a ) char.