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I have a String pattern: "hh:mm:ss dd:MM:yyyy to hh:mm:ss dd:MM:yyyy" and I want to extract date String from it.
Example:
S = "00:00:00 19/08/2022 to 23:59:59 19/08/2022"
Split into S1 = "00:00:00 19/08/2022" and S2 = "23:59:59 19/08/2022".
I'm trying to use String.split function but can't figure out the regex yet. Can somebody help?
I'm using Java 8.
Just split on \s+to\s+:
String pattern = "00:00:00 19/08/2022 to 23:59:59 19/08/2022";
String[] parts = pattern.split("\\s+to\\s+");
System.out.println(Arrays.toString(parts));
This prints:
[00:00:00 19/08/2022, 23:59:59 19/08/2022]
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I have a google sheet where i have long text string in each cell.
following is one of the text string
VACANCY...Test Hotels is hiring for the below position.#Job_Title : Director of a Revenue/ Revenue Manager#Hotel_Name : Signature Hotel#Job_Location : Dubai#Nationality : Selective#Experience : Mandatory hotel experience#Salary_Range : Unspecified#Benefits : Unspecified- Candidate should be currently in UAE and has relevant UAE Hotel experience.Please specify “Applying Position” in the subject line.Email CV: test#test.com#jobseekers #vacancy #Dubai #jobs #recruiters #hotels #manager #revenue
I want to add white spaces before certain words or character so it look neat . for example i want to add white space before "#", "Salary", " job location" etc.
Ho can i do that
Please be reminded that you have to escape any metacharacter
=REGEXREPLACE(A1,"(#|Salary|job location)"," $1")
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Data:
BUY 2 FOR 5(STORES)
BUY 2 FOR 10(STORES)
What I tried:
regexp_extract(DATA, '.*? (\\d+) .*$', 2)
Desired result:
5
10
Like this:
regexp_extract(DATA, '^[^0-9]+?\\d+[^0-9]+?(\\d+)', 1);
or
regexp_extract(DATA, '^\\D+?\\d+\\D+?(\\d+)', 1);
Regex means: one or more Non-digits at the beginning, one of more digits, one or more non-digits, and finally the capturing group of digits, you need to extract the group number one.
One more solution is to split string by non-didits and take 2nd element:
select split(DATA, '[^0-9]+')[2];
Or even simpler:
select split(DATA, '\\D+')[2]; --\\D+ means one or more non-digits
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So I have a CSV file with some anomaly for eg
2019-07-25 00:00:00,1014488,2019-07-25 12:24:12,112629,Amy,Flutmus,84004,GM,0001,2.99,312,FFO & CS PLATE ||22,10999,90027,90062||Sand w/ Options,1,0,0.2,18.85,0,1
i want to replace , between these characters || ||.
So I'm expecting
2019-07-25 00:00:00,1014488,2019-07-25 12:24:12,112629,Amy,Flutmus,84004,GM,0001,2.99,312,FFO & CS PLATE ,22,*10999*90027*90062,Sand w/ Options,1,0,0.2,18.85,0,1
You could use re.sub to capture all your strings between || and then replacing the ,s with *s:
import re
value = "2019-07-25 00:00:00,1014488,2019-07-25 12:24:12,112629,Amy,Flutmus,84004,GM,0001,2.99,312,FFO & CS PLATE ||22,10999,90027,90062||Sand w/ Options,1,0,0.2,18.85,0,1"
pattern = re.compile(r'\|\|(.+)\|\|')
cleaned_value = pattern.sub(lambda match: match.group().replace(",", "*"), value)
print(cleaned_value.replace(r'||', ','))
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Example: DATA500 replaced with DATA[500]
This should work. It replaces groups of one or more digits, the "\d+" part, with the captured string surrounded by [], the "[$1]" part.
$a = "bob123bob123";
$a =~ s/(\d+)/[$1]/g;
print "$a";
# bob[123]bob[123]