Is it normal for regex to add ; (semicolon) in extract? For example, in SharePoint workflow I have this [^/]*$ RegEx (with Extract selected) get the file name and put it in a collection. The RegEx puts a ; at end of the extract string.
For example:
input: http://www.abc.com/sites/sales/document library/2013 june sale 05450 12 (draft).pdf
Regex: [^/]*$
output: 2013 june sale 05450 12 (draft).pdf;
I think since it puts in a Collection Variable that's why it puts the trailing ;
What are you all opinions?
Since you are already using a negative character class you can as well expand it to:
[^/;]*$
It looks a little weird since you are selecting multiple characters using the star (*) but if you claim it works for you then this one should too.
Related
I have a small issue, I am trying to get specific characters from a long string using regex but I am having trouble.
Workflow
Prometheus --> Grafana --> Variable (using regex)
I can't use anything other than Regex expressions to achieve this result
I am currently using this expression to grab the long string from some json output:
.*channel_id="(.*?)".*
FROM THIS
{account_id="XXXXXXX-xxxx-xxxx-xxxx-xxxxxxxxxx",account_name="testalpha",channel_id="s0022110430col0901241usa",channel_abbr="s0022109430col}
This returns a string that's ALWAYS 24 characters long:
s0022110430col0901241usa
PROBLEM:
I need to grab the 3 letters 'col' and 'usa' as they are the two teams that are playing, ideally I would be able to pipe the results from the first regex to get these values (the position is key, since the first value will ALWAYS be the 12-14th characters and the second value is the last 3 characters) if I could output these values in uppercase with the string "vs" in between to create a string such as:
COL vs USA
or
ARG vs BRA
I am open to any and every suggestion anyone may have
Thank you!
PS - The uppercase thing is 'nice to have' BUT not needed
I'm still learning RegEx, so this is all I could come up with:
For the col (first team):
(?<=(channel_id=".{11}))\w{3}
For the usa (second team):
(?<=(channel_id=".{21}))\w{3}
Can you define the channel_id?
It begins with 's' and then there are many numbers. If they are always numbers, you can use this regex:
channel_id=".[0-9]+([a-z]+)[0-9]+([a-z]+)
You will get 2 groups, one with "col" and the other with "usa".
Edit:
Or if you just know, that you have always the same size, you can use something like:
channel_id=".{11}([a-z]+).{7}([a-z]+)
I am new to regex and having difficulty obtaining values that are caught in between spaces.
I am trying to get the values "field 1" "abc/def try" from the sameple data below just using regex
Currently im using (^.{18}\s+) to skip the first 18 characters, but am at at loss of how to do grab values with spaces between.
A1234567890 field 1 abc/def try
02021051812 12 test test 12 pass
3333G132021 no test test cancel
any help/pointers will be appreciated.
If this text has fixed-width columns, you can match and trim the column values knowing the amount of chars between start of string and the column text.
For example, this regex will work for the text you posted:
^(.*?)\s*(?<=.{19})(.*?)\s*(?<=^.{34})(.*?)\s*(?<=^.{46})
See the regex demo.
So, Column 2 starts at Position 19, Column 3 starts at Position 34 and Column 4 (end of string here) is at Position 46.
However, this regex is not that efficient, and it would be really great if the data format is fixed on the provider's side.
Given the not knowing if the data is always the same length I created the following, which will provide you with a group per column you might want to use:
^((\s{0,1}\S{1,})*)(\s{2,})((\s{0,1}\S{1,})*)(\s{2,})((\s{0,1}\S{1,})*)
Regex demo
I have a filename like this:
0296005_PH3843C5_SEQ_6210_QTY_BILLING_D_DEV_0000000000000183.PS.
I needed to break down the name into groups which are separated by a underscore. Which I did like this:
(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)(\d{16})(.*)
So far so go.
Now I need to extract characters from one of the group for example in group 2 I need the first 3 and 8 decimal ( keep mind they could be characters too ).
So I had try something like this :
(.*?)_([38]{2})(.*?) _(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)(\d{16})(.*)
It didn’t work but if I do this:
(.*?)_([PH]{2})(.*?) _(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)(\d{16})(.*)
It will pull the PH into a group but not the 38 ? So I’m lost at this point.
Any help would be great
Try the below Regex to match any first 3 char/decimal and one decimal
(.?)_([A-Z0-9]{3}[0-9]{1})(.?)(.*?)(.?)_(.?)(.*?)(.?)_(.?)
Try the below Regex to match any first 3 char/decimal and one decimal/char
(.?)_([A-Z0-9]{3}[A-Z0-9]{1})(.?)(.*?)(.?)_(.?)(.*?)(.?)_(.?)
It will match any 3 letters/digits followed by 1 letter/digit.
If your first two letter is a constant like "PH" then try the below
(.?)_([PH]+[0-9A-Z]{2})(.?)(.*?)(.?)_(.?)(.*?)(.?)_(.?)
I am assuming that you are trying to match group2 starting with numbers. If that is the case then you have change the source string such as
0296005_383843C5_SEQ_6210_QTY_BILLING_D_DEV_0000000000000183.PS.
It works, check it out at https://regex101.com/r/zem3vt/1
Using [^_]* performs much better in your case than .*? since it doesn't backtrack. So changing your original regex from:
(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)(\d{16})(.*)
to:
([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_(.*?)(\d{16})(.*)
reduces the number of steps from 114 to 42 for your given string.
The best method might be to actually split your string on _ and then test the second element to see if it contains 38. Since you haven't specified a language, I can't help to show how in your language, but most languages employ a contains or indexOf method that can be used to determine whether or not a substring exists in a string.
Using regex alone, however, this can be accomplished using the following regular expression.
See regex in use here
Ensuring 38 exists in the second part:
([^_]*)_([^_]*38[^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_(.*?)(\d{16})(.*)
Capturing the 38 in the second part:
([^_]*)_([^_]*)(38)([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_(.*?)(\d{16})(.*)
I am trying to use Regex to parse a series of strings to extract one or more text dates that may be in multiple formats. The strings will look something like the following:
24 Aug 2016: nno-emvirt010a/b; 16 Aug 2016 nnt-emvirt010a/b nnd-emvirt010a/b COSI-1.6.5
24.16 nno-emvirt010a/b nnt-emvirt010a/b nnd-emvirt010a/b EI.01.02.03\
9/23/16: COSI-1.6.5 Logs updated at /vobs/COTS/1.6.5/files/Status_2016-07-27.log, Status_2016-07-28.log, Status_2016-08-05.log, Status_2016-08-08.log
I am not concerned about validating the individual date fields; just extracting the date string. The part I am unable to figure out is how to not match on number sequences that match the pattern but aren’t dates (‘1.6.5’ in ex. (1) and 01.02.03 in ex. (2)) and dates that are part of a file name (2016-07-27 in ex. (3)). In each of these exception cases in my input data, the initial numbers are preceded by either a period(.), underscore (_) or dash (-), but I cannot determine how to use this to edit the pattern syntax to not match these strings.
The pattern I have that partially works is below. It will only ignore the non date matches if it starts with 1 digit as in example 1.
/[^_\.\(\/]\d{1,4}[/\-\.\s*]([1-9]|0[1-9]|[12][0-9]|3[01]|[a-z]{3})[/\-\.\s*]\d{1,4}/ig`
I am not sure about vba check if this works . seems they have given so much options : https://www.safaribooksonline.com/library/view/regular-expressions-cookbook/9781449327453/ch04s04.html
^(?:(1[0-2]|0?[1-9])/(3[01]|[12][0-9]|0?[1-9])|↵
(3[01]|[12][0-9]|0?[1-9])/(1[0-2]|0?[1-9]))/(?:[0-9]{2})?[0-9]{2}$
^(?:
# m/d or mm/dd
(1[0-2]|0?[1-9])/(3[01]|[12][0-9]|0?[1-9])
|
# d/m or dd/mm
(3[01]|[12][0-9]|0?[1-9])/(1[0-2]|0?[1-9])
)
# /yy or /yyyy
/(?:[0-9]{2})?[0-9]{2}$
According to the test strings you've presented, you can use the following regex
See this regex in use here
(?<=[^a-zA-Z\d.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:-\d{2}){2})|\d{2}\.\d{2})(?=[^a-zA-Z\d.])
This regex ensures that specific date formats are met and are preceded by nothing (beginning of the string) or by a non-word character (specifically a-z, A-Z, 0-9) or dot .. The date formats that will be matched are:
24 Aug 2016
24.16
9/23/16
The regex could be further manipulated to ensure numbers are in the proper range according to days/month, etc., however, I don't feel that is really necessary.
Edits
Edit 1
Since VBA doesn't support lookbehinds, you can use the following. The date is in capture group 1.
(?:[^a-zA-Z\d.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:-\d{2}){2})|\d{2}\.\d{2})(?=[^a-zA-Z\d.])
Edit 2
As per bulbus's comment below
(?:[^\w.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d{2,4})|(?:(?:\d{1,2}\/){2}\d{2,4})|(?:\d{2,4}(?:-\d{2}){2})|\d{2}\.\d{2})
Took liberty to edit that a bit.
replaced [^a-zA-Z\d.] with [^\w.], comes with added advantage of excluding dates with _2016-07-28.log
Due to 1 removed trailing condition (?=[^a-zA-Z\d.]).
Forced year digits from \d+ to \d{2,4}
Edit 3
Due to added conditions of the regex, I've made the following edits (to improve upon both previous edits). As per the OP:
The edited pattern above works in all but 2 cases:
it does not find dates with the year first (ex. 2016/07/11)
if the date is contained within parenthesis in the string, it returns the left parenthesis as part of the date (ex. match = (8/20/2016)
Can you provide the edit to fix these?
In the below regexes, I've changed years to \d+ in order for it to work on any year greater than or equal to 0.
See the code in use here
(?:[^\w.]|^)((?:\d{1,2}\s+[A-Z][a-z]{2}\s+\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:\/\d{1,2}){2})|(?:\d+(?:-\d{2}){2})|\d{2}\.\d+)
This regex adds the possibility of dates in the XXXX/XX/XX format where the date may appear first.
The reason you are getting ( as a match before the regex is the nature of the Full Match. You need to, instead, grab the value of the first capture group and not the whole regex result. See this answer on how to grab submatches from a regex pattern in VBA.
Also, note that any additional date formats you need to catch need to be explicitly set in the regex. Currently, the regex supports the following date formats:
\d{1,2}\s+[A-Z][a-z]{2}\s+\d+
12 Apr 17
12 Apr 2017
(?:\d{1,2}\/){2}\d+
1/4/17
01/04/17
1/4/2017
01/04/2017
\d+(?:\/\d{1,2}){2}
17/04/01
2017/4/1
2017/04/01
17/4/1
\d+(?:-\d{2}){2}
17-04-01
2017-04-01
\d{2}\.\d+ - Although I'm not sure what this date format is even used for and how it could be considered efficient if it's missing month
24.16
I have a file that I need to reformat and remove "extra" blank lines.
I am using the Perl syntax regular expression search and replace functionality of UltraEdit and need the regular expression to put in the "Find What:" field.
Here is a sample of the file I need to re-format.
All current text
REPLACE with all the following:
Winter 2011 Class Schedule
Winter 2011 Class Registration Dates: Dec. 6, 2010 – Jan. 1, 2011
Winter 2011 Class Session Dates: Jan. 5 – Feb. 12, 2011
DANCE
Adventures in Ballet & Tap
3 – 6 years Instructor: Ann Newby
Tots ages 3 – 6 years old develop a greater sense of rhythm, flexibility and coordination as they explore the basic elements of movement.
Saturdays 9 - 10 a.m. Jan. 8 – Feb. 12 Six-week fees: $30
African Storytelling
3 – 6 years Instructor: Ann Newby
Tots ages 3 – 6 years old explore storytelling and fables through spoken word, music, movement and visual arts experiences.
Saturdays 10 – 11 a.m. Jan. 8 – Feb. 12 Six-week fee: $30
African Dance / Children
You'll notice that some of the double blank lines have spaces or tabs or both in them.
After the search and replace has been run I should have a file that looks like this.
All current text
REPLACE with all the following:
Winter 2011 Class Schedule
Winter 2011 Class Registration Dates: Dec. 6, 2010 – Jan. 1, 2011
Winter 2011 Class Session Dates: Jan. 5 – Feb. 12, 2011
DANCE
Adventures in Ballet & Tap
3 – 6 years Instructor: Ann Newby
Tots ages 3 – 6 years old develop a greater sense of rhythm, flexibility and coordination as they explore the basic elements of movement.
Saturdays 9 - 10 a.m. Jan. 8 – Feb. 12 Six-week fees: $30
African Storytelling
3 – 6 years Instructor: Ann Newby
Tots ages 3 – 6 years old explore storytelling and fables through spoken word, music, movement and visual arts experiences.
Saturdays 10 – 11 a.m. Jan. 8 – Feb. 12 Six-week fee: $30
African Dance / Children
Replacing
^(\s*\r\n){2,}
With
\r\n
Is what I ended up with.
This only selects blank lines in multiples of two or more and replaces them with one.
It depends what the line endings are. Assuming \n, replace this:
([ \t]*\n){3,}
with \n\n.
Try this perl oneliner perl -00pe0, if you want in place editing, just add -i option
Replacing
\n\s*\n\s*
with
\n\n
should do the trick
For completeness I want to reference here the large post Remove / delete blank and empty lines in the user forums of UltraEdit which contains at bottom after all the explanations for newbies the solution for reducing two or more lines with nothing (empty lines) or just whitespaces (blank lines) to one empty line independent on line terminator type.
And some words on what Alan Moore wrote in his answer:
UltraEdit's Perl regular expression support is not crippled by its line-based architecture. Perl regular expression engines have a flag which determine if a dot matches all characters except newline characters like carriage return (CR) and line feed (LF) or really all characters including CR and LF. This makes the difference if a text file is interpreted as large byte stream or as a sequence of lines for Perl regular expression finds/replaces. In UltraEdit the flag is set by default to not include \r (CR) and \n (LF) by a dot in the regular expression search string. But this behavior can be easily changed in UltraEdit by starting the regular expression string with (?s) which changes the value of the flag match_not_dot_newline as posted in UltraEdit user forums at topic "." in Perl regular expressions doesn't include CRLFs?
A Perl regular expression replace working for files with
carriage return + line feed (DOS/Windows) or
only line feed (Unix, Mac OS 10.0 and later versions) or
only carriage return (Mac OS 9 and previous versions)
as line ending with optionally trailing spaces and tabs at end of a paragraph (one or more lines) and with two or more lines without (empty line) or with whitespaces (blank line) below the paragraph could be done with search string \h*(\r?\n|\r)(?:\h*\1){2,} and \1\1 as replace string.
Explanation:
\h* matches any horizontal whitespace character according to Unicode 0 or more times. This first part of the search expression matches horizontal whitespace characters at end of a line like horizontal tabs, normal spaces, no-break-spaces and some other not often used spaces.
The usage of \s is not good as this character class matches any whitespace character including the vertical whitespace characters carriage return and line feed.
(\r?\n|\r) ... is an OR expression with two arguments in a marking group. The first argument matches a line feed optionally with a preceding carriage return while the second argument matches just a carriage return. So this expression matches all three common types of line terminations completely correct. It is important for the rest of the search and the replace to match always either CR+LF (both together) or just LF or just CR.
(?:\h*\1) ... is a non marking group which matches 0 or more horizontal whitespaces and the newline as found before back-referenced with \1, i.e. CR+LF or just LF or just CR. So this part of the expression finds an empty or blank line.
{2,} ... is a multiplier for the previous expression in the non marking group which means at least two times. So after end of a paragraph there must be two or more empty or blank lines. Only one empty or blank line below a paragraph is not enough for a positive match of search expression.
The replace string \1\1 references twice the first found line break.
The advantage of this regular expression in comparison to the others posted here is that the line ending type must not be known. The search expression finds that out and found line ending is referenced in the replace string. And probably existing trailing whitespaces at end of a paragraph and whitespaces on next line are removed also by this regular expression replace if there are two or more empty or blank lines below a paragraph.
{2,} can be replaced by + in search string if trimming whitespaces at end of a paragraph and on next empty or blank line should be also done on running this Perl regular expression replace. But please note that in this case the replace makes replaces which do not change anything at all if there are not trailing whitespaces at end of a paragraph and next line is an empty line.
In Vim, Using
:%!cat -s
I find this is the easiest way to delete extra empty line so far.
I'm not sure what UltraEdit lets you get away with in the "replace" area, but if you cannot use a newline (I've had this problem before) but can use capture references, this might work:
Find : \s*(\r\n)\s*(\r\n)\s*\r\n
Replace : $1$2
Not tested extensively, but seems to work on the sample you provided.
See this thread for what's causing the problem. As I understand it, UltraEdit regexes are greedy at the character level (i.e., within a line), but non-greedy at the line level (roughly speaking). I don't have access to UE, but I would try writing the regex so it has to match something concrete after the last blank line. For example:
search: (\r\n[ \t]*){2,}(\S)
replace: $1$2
This matches and captures two or more instances of a line separator and any horizontal whitespace that follows it, but it only retains the last one. The \S should force it to keep matching until it finds a line with at least one non-whitespace character.
I admit that I don't have a whole lot of confidence in this solution; UltraEdit's regex support is crippled by its line-based architecture. If you want an editor that does regexes right, and you don't want to learn a whole new regex syntax (like vim's), get EditPadPro.
Should also work with spaces on blank lines
Search - /\n^\s*\n/
Replace - \n\n
On my Intellij IDE what was search for \n\n and Replace it by \n