how to print substring between quotes in a string in Java [closed] - substring

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such string:
interface Loopback0 description "Loopback Interface for network management" ip address 10.20.30.40 255.255.255.255 no ip proxy-arp
how to print what is between " "?
Loopback Interface for network management

Assuming that the " character occurs only 2 times or if you want the first pair:
# Get the indices of the " characters
ind1 = exampleStr.find('"')
ind2 = exampleStr.find('"', ind1+1)
# Get the substring between the two indices
result = exampleStr[ind1+1:ind2]

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How to set a text content of a specific column as a variable using Batch or PowerShell? [closed]

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For example, I have a file/file output with the following content:
2022-10-16 14:33 1,860,477 Mausi-~1.JPG Mausi-wife.JPG
There are spaces between these five blocks in between, i.e. between date and time and number and file name and another file name.
I would now like to set only the fourth column of this as a variable. Is this possible?
Split the string on spaces, update the desired value in the resulting array, and then stitch back together with -join:
$row = '2022-10-16 14:33 1,860,477 Mausi-~1.JPG Mausi-wife.JPG'
# split into individual field values
$fields = $row.Split(' ')
# update/overwrite the 4th item in the resulting string array
$fields[3] = "New value"
# stitch row back together with `-join`
$row = $fields -join ' '

How to extract only version number from a string [closed]

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I am extracting oracle version from windows using powershell command and i get result as 10.2.0.3.0Patch2, however i need to extract only numeric value i.e. 10.2.0.3.0 (only version number). Any way we can do it ?
Version info extracted is =
10.2.0.3.0 Production, 10.2.0.3.0Patch2 Production, 10.2.0.5.0 Production, 11.2.0.4.0 Production
You can use a regular expression to extract a substring. Example:
"10.2.0.3.0Patch2" | Select-String '((?:\d{1,3}\.){4}\d{1,3})' | ForEach-Object {
$_.Matches[0].Groups[1].Value
}
# Outputs the string '10.2.0.3.0'
You can read more about regular expressions by reading the about_Regular_Expressions help topic:
PS C:\> help about_Regular_Expressions

val a = month(start_date),year(to-date) [closed]

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I have a requirement, that I have a string like below input and I want string like below output. can anyone please help me ?
example 1
val input = "month(start_date),year(to_date),month(to_date)"
output = "start_date,to-date"
example 2
input = "abc(start),xyz(end)"
output = "start,end"
You need a regex to get the value inside parenthesis
val input = "month(start_date),year(to_date),month(to_date)"
val regex = "(?<=\\()[^)]+(?=\\))".r
val output = regex.findAllIn(input).toSet.mkString(",")
for regex explanation you can find it here How do I match the contents of parenthesis in a scala regular expression
toSet to remove the duplicated
and mkString to join the set with comma

Detect letter where number was expected [closed]

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How can I detect and show an error if someone put a letter and the program expects a number?
A regexp match makes this easy. Searching for any character which isn't a numeral or an arithmetic symbol:
if ( $input =~ /[^0-9+*/-]/ ) {
print "Incorrect character detected!\n"
}
Literally anything which is a letter:
if ( $input =~ /[A-Za-z]/ ) {
print "Incorrect character detected!\n"
}

Converting single line to multiple lines in perl [closed]

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I have file as below.
ID || DATE || AMOUNT
XX||20130801##20130901##20131001##20131101||100##200##300##400
and I want the output as below using perl.
xx||20130801||100
xx||20130901||200
xx||20131001||300
xx||20131101||400
Please help me how to convert using perl.
perl -F'\|\||##' -lanE'$.>1 or $" ="||",next; say "#F[0,$_,$_+4]" for 1..4' file
perl -F'\|\|' -lane '#a=split(/##/,$F[1]); #b=split(/##/,$F[2]); print "$F[0]||$a[$_]||$b[$_]" foreach 0..$#a;' file
Output:
ID || DATE || AMOUNT
XX||20130801||100
XX||20130901||200
XX||20131001||300
XX||20131101||400