how can we remove the text before the line that start's with second comma(line 5 in the example),how can i do that using regex?
example :
,
abc,xyz,ggg,nrmr
cde,jjj,kkkk,iiii,tem,posting
234,mm/dd/yy
,
454654,output2,sample
45646,output1,non-sample
16546,225.02
ABC,2.98
expected :
454654,output2,sample
45646,output1,non-sample
16546,225.02
ABC,2.98
It seems you may use
val s = """,
abc,xyz,ggg,nrmr
cde,jjj,kkkk,iiii,tem,posting
234,mm/dd/yy
,
454654,output2,sample
45646,output1,non-sample
16546,225.02
ABC,2.98"""
val res = s.replaceFirst("(?sm)\\A(.*?^,$){2}", "").trim()
println(res)
// =>
// 454654,output2,sample
// 45646,output1,non-sample
// 16546,225.02
// ABC,2.98
See the Scala demo.
Pattern details:
(?sm) - s enables . to match any char in the string including newlines, and m makes ^ and $ match start/end of line respectively
\\A - the start of string
(.*?^,$){2} - 2 occurrences of:
.*? - any 0+ chars as few as possible up to the leftmost
^,$ - line that only contains ,.
Related
I'm trying to come up with with a regex statement that will match and not match the following cases:
CT.test // (1) must match
CT.test (MONT) // (2) must match
CT.test (ABS) // (3) must match
CT.badsf // (4) must not match
CT.test (WOW) // (5) must not match
I've tried CT.test( \(MONT\)| \(ABS\)|^$) but that only matches cases 2 and 3 and not case 1.
What is a regex statement that will match case 1, 2 and 3 and not match cases 4 and 5?
You can use
^CT\.test(?: \((?:MONT|ABS)\))?$
See the regex demo
See the details here:
^ - start of string
CT\.test - (note the escaped dot): a CT.test string
(?: \((?:MONT|ABS)\))? - an optional sequence of
- a space
\( - a ( char
(?:MONT|ABS) - MONT or ABS string
\) - a ) char
$ - end of string.
OTC_omega_20210302.csv
CH_delta_20210302.csv
MD_omega_20210310.csv
CD_delta_20210310.csv
val hdfsPath = "/development/staging/abcd-efgh"
val fs = org.apache.hadoop.fs.FileSystem.get(spark.sparkContext.hadoopConfiguration)
val files = fs.listStatus(new Path(s"${hdfsPath}")).filterNot(_.isDirectory).map(_.getPath)
val regX = "OTC_*[0-9].csv|CH_*[0-9].csv".stripMargin.r
val filteredFiles = files.filter(fName => regX.findFirstMatchIn(fName.getName).isDefined)
What is regex do i need to give if i need any file name that starts with either (OTC_ or CH_ ) and ends with YYYYMMDD.csv ?
As per the above files i need two outputs
OTC_omega_20210302.csv
CH_delta_20210302.csv
Please help
You can use
val regX = "^(?:OTC|CH)_.*[0-9]{8}\\.csv$".r
val regX = """^(?:OTC|CH)_.*[0-9]{8}\.csv$""".r
See the regex demo.
Details:
^ - start of string
(?:OTC|CH) - a non-capturing group matching either OTC or CH char sequences
_ - a _ char
.* - any zero or more chars other than line break chars, as many as possible
[0-9]{8} - eight digits
\. - a literal dot (note . matches any char other than a line break char, you must escape . to make it match a dot)
csv - a csv string
$ - end of string.
I want to get a bloc of lines which contains < or > operator until an empty newline
i try with this regex .*[<>][^,\r\n]+?\(.*\S.*,.*\S.*\).*(?:(\n).*)
You find here my example : https://regex101.com/r/UQYLB5/1/
Expected Result :
MATCH 1 :
BAR18>17M(3,5.2)V
MATCH 2 :
BAR19>1.243037M(3,5.2)V
INFORMATION PROCESS
TAKE B/F: 19V[1]
LIGHT PC CARD:
MATCH 3 :
TEFAL17>1.262259M(4.5,5.5)V
SISS17 : 1789-ID
LIGHT 19/17
MAPPING NICE :
MATCH 4 :
MASCARPONE19>493.818969M(3,5.2)V
BATA17 : CDER78945 -- 1875
LEFT ERREUR - CAME BACK
MATCH 5 :
REPAR_178>748.515487M(4.5,5.5)V
CHAN1 / STEREO MIX
If you don't want to match lines which could consist of spaces only, you could use match either < or > and match at least a non whitespace char \S in the following lines:
^[^<>\r\n]*[<>].*(?:\r?\n[^\r\n\S]*\S.*)*
The pattern will match:
^ Start of string
[^<>\r\n]* Match any char except < `
[<>].* Match either < or > and the rest of the line
(?: Non capture group
\r?\n Match a newline
[^\r\n\S]* Match any char except a newline
\S.* Match a non whitespace char and the rest of the line
)* Close the group and repeat 0+ times
Regex demo
If the first line should also contain a , after matching < or >:
^[^<>\r\n]*[<>][^\r\n,]*,.*(?:\r?\n[^\r\n\S]*\S.*)*
Regex demo
I have been trying to get a regex expression to return me the following in the following situations.
XX -> XX
XXX -> XXX
XX/XX -> XX
XX/XX/XX -> XX/XX
XXX/XXX/XX -> XXX/XXX
I had the following Regex, however they do no work.
^[^/]+ => https://regex101.com/r/xvCbNB/1
=========
([A-Z])\w+ => https://regex101.com/r/xvCbNB/2
They are close but are not there.
Any Help would be appreciated.
You want to get all text from the start till the last occurrence of a specific character or till the end of string if the character is missing.
Use
^(?:.*(?=\/)|.+)
See the regex demo and the regex graph:
Details
^ - start of string
(?:.*(?=\/)|.+) - a non-capturing group that matches either of the two alternatives, and if the first one matches first the second won't be tried:
.*(?=\/) - any 0+ chars other than line break chars, as many as possible upt to but excluding /
| - or
.+ - any 1+ chars other than line break chars, as many as possible.
It will be easier to use a replace here to match / followed by non-slash characters before end of line:
Search regex:
/[^/]*$
Replacement String:
""
Updated RegEx Demo 1
If you're looking for a regex match then use this regex:
^(.*?)(?:/[^/]*)?$
Updated RegEx Demo 2
Any special reason it has to be a regular expression? How about just splitting the string at the slashes, remove the last item and rejoin:
function removeItemAfterLastSlash(string) {
const list = string.split(/\//);
if (list.length == 1) [
return string;
}
list.pop();
return list.join("/");
}
Or look for the last slash an remove it:
function removeItemAfterLastSlash(string) {
const index = string.lastIndexOf("/");
if (index === -1) {
return string;
}
return string.splice(0, index);
}
I have a string:
dkj a * & &*(&(*(
//#HELLO
^%#&UJNWDUK()C*(v 8*J DK*9
//#HE#$^&&(akls#$98akdjl ak##sjdkja
//
%^&*(//#HELLO//#BYE<><>
//#BYE
^%#&UJNWDUK()C*(v 8*J DK*90K )
//#HELLO
&*^J$XUK 8j8 j jk kk8(&*(
//#BYE
and I need to have 2 groups such as each group must start with //HELLO then there should be a next line and any type of text can follow (.*) but it will end with a //BYE preceded by a line:
1)
//#HELLO
^%#&UJNWDUK()C*(v 8*J DK*9
//#HE#$^&&(akls#$98akdjl ak##sjdkja
//
%^&*(//#HELLO//#BYE<><>
//#BYE
2)
//#HELLO
&*^J$XUK 8j8 j jk kk8(&*(
//#BYE
and replaces the original string to this: (basically adding // to each line of each group)
dkj a * & &*(&(*(
////#HELLO
//^%#&UJNWDUK()C*(v 8*J DK*9
////#HE#$^&&(akls#$98akdjl ak##sjdkja
////
//%^&*(//#HELLO//#BYE<><>
////#BYE
^%#&UJNWDUK()C*(v 8*J DK*90K )
////#HELLO
//&*^J$XUK 8j8 j jk kk8(&*(
////#BYE
Here is my current progress:
I have
\/\/#HELLO\n.*?\/\/#BYE[\n$]
However im not sure how to go about the replacement, I'm thinking separating each line per group using \G after the //#HELLO and ending with //#BYE
It's a bit complex, but this will do it:
Search: (?m)(//#HELLO[\r\n]+|\G(?://#BYE|(?=(?:[^#]|#(?!HELLO[\r\n]+))*#BYE)[^\r\n]*[\r\n]*))
Replace: //$1
In Groovy:
String resultString = subjectString.replaceAll(/(?m)(\/\/#HELLO[\r\n]+|\G(?:\/\/#BYE|(?=(?:[^#]|#(?!HELLO[\r\n]+))*#BYE)[^\r\n]*[\r\n]*))/, '//$1');
For grouping into separate lines use the following regex:
//#HELLO\r(.*[\n\r]+)*//#BYE\r?
\r - Newline character
[\n\r] - Enter characters
*? - Non-greedy match
?- Match 1 or 0 times
You can take out the ? at the end if it always ends with a newline.
You can then use the group (The value inside the brackets) to search and replace.