Split string by word contain space in Java [closed] - regex

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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]

Related

extract string have dot using regex substr oracle [closed]

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Input string:
(select FRL_ATTRIBUTE_DESCRIPTION from FRL_ATTRIBUTES WHERE FRL_ATTRIBUTE_LEVEL =9 AND FRL_Lines_PL.FRL_ATTRIBUTE_09 = FRL_ATTRIBUTE)
Ouput :
FRL_Lines_PL.FRL_ATTRIBUTE_09
i need to extract FRL_Lines_PL.FRL_ATTRIBUTE_09 alias name string using regex_substr

You said you want to extract a string that contains a dot and provided example. OK, if that's exactly what you have & need, then
SQL> with test (col) as
2 (select '(select FRL_ATTRIBUTE_DESCRIPTION from FRL_ATTRIBUTES WHERE FRL_ATTRIBUTE_LEVEL =9 AND FRL_Lines_PL.FRL_ATTRIBUTE_09 = FRL_ATTRIBUTE)' from dual)
3 select regexp_substr(col, '\w+\.\w+') result
4 from test;
RESULT
-----------------------------
FRL_Lines_PL.FRL_ATTRIBUTE_09
SQL>

How to use regular expression in Hive to extract the second integer? [closed]

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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

Find and Replace , between 2 characters [closed]

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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'||', ','))

Regex to use in HTML source [closed]

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I need a regex to extract the number 330 on ...163&angleKreludor=330&viewID=.... This number vary from day to day...it could be an Integer with up to 3 digits, but it can be a double number like 127.57 with 2 decimal places ... so I would need to capture anything between angleKreludor= and &viewID.... Here is the complete HTML:
var swf = new SWFObject('http://images.lulaser.com/shenkuu/lunar/shenkuu_calendar_v1.swf?angleNeopia=163&angleKreludor=330&viewID=2&lang=pt', 'flash_36175654223', '550', '500', '6', '#FFFFFF');
swf.addParam('quality', 'high');
swf.addParam('scale', 'exactfit');
swf.addParam('menu', 'false');
swf.addParam('allowScriptAccess', 'always');
swf.addParam('swLiveConnect', 'true');
swf.addParam('bgcolor', 'white');
swf.write();
P.S: This is needed to use in Javascript code in Selenium IDE ... I tried in the past and Selenium IDE does not accept look-aheads nor look-behinds

You can search for digits with positive look-behind for the angleKreludor= key
(?<=angleKreludor=)(\d+)
DEMO
For JavaScript use non-captuinrg group
(?:angleKreludor=)(\d+)
var s = 'http://images.lulaser.com/shenkuu/lunar/shenkuu_calendar_v1.swf?angleNeopia=163&angleKreludor=330&viewID=2&lang=pt';
var nr = s.match(/(?:angleKreludor=)(\d+)/);
console.log(nr[1]);
DEMO

I have a string in perl, how do I replace it with the same string but with [ ] around only the numbers? [closed]

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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]