Hi,
Am getting stuck in string manipulation
I have three sentence such as 1.“Sam opened Trust October 31, 2018” ,2.“it is when November 30, 2018”, 3.“thony December 31,2018” in above the three sentences i have i like to remove the month, date and year from that difference sentence is there any possible way pls give some idea to crack this.
Regards,
Spark
Use can use regex to remove the required portion from the string.
Try the below.
In Assign activity
result_string = System.Text.RegularExpressions.Regex.Replace(INPUT_STRING, “(Jan(uary)?|Feb(ruary)?|Mar(ch)?|Apr(il)?|May|Jun(e)?|Jul(y)?|Aug(ust)?|Sep(tember)?|Oct(ober)?|Nov(ember)?|Dec(ember)?)\s*(\d{1,2}\s*,\s*\d{4})”, “”)
Replace INPUT_STRING with your string.
Regards,
Ranjith.
About to type the same.
Thanks Dude working fine i need to know the explanation for this pls…
Regards,
Spark
I have used regular expression to remove the required pattern.
I will explain about the pattern used for your case.
Pattern : (Jan(uary)?|Feb(ruary)?|Mar(ch)?|Apr(il)?|May|Jun(e)?|Jul(y)?|Aug(ust)?|Sep(tember)?|Oct(ober)?|Nov(ember)?|Dec(ember)?)\s*(\d{1,2}\s*,\s*\d{4})
Example : it is when November 30, 2018
(Jan(uary)?|Feb(ruary)?|Mar(ch)?|Apr(il)?|May|Jun(e)?|Jul(y)?|Aug(ust)?|Sep(tember)?|Oct(ober)?|Nov(ember)?|Dec(ember)?) - Will check for the month name which can be first three characters(Jan) or full name (January)
Ex : (November will be matched for the above example)
\d{1,2} - Will match number with 1 or 2 digits (30 - will be matched)
\d{4} - match for 4 digit number (2018 will be matched)
\s - matches empty space
* - represent zero or more occurrence.
For detailed understanding about regular expression please go the below document.
Regular-Expressions.pdf (920.0 KB)
Thank you so much @ranjith i clearly understood what you done.
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