Can you please help me to get the desired output, where SIT is the environment and type of file is properties, i need to remove the environment and the extension of the string.
#$string="<ENV>.<can have multiple period>.properties
*$string ="SIT.com.local.test.stack.properties"
$b=$string.split('.')
$b[0].Substring(1)*
Required output : com.local.test.stack //can have multiple period
This should do.
$string = "SIT.com.local.test.stack.properties"
# capture anything up to the first period, and in between first and last period
if($string -match '^(.+?)\.(.+)\.properties$') {
$environment = $Matches[1]
$properties = $Matches[2]
# ...
}
You may use
$string -replace '^[^.]+\.|\.[^.]+$'
This will remove the first 1+ chars other than a dot and then a dot, and the last dot followed with any 1+ non-dot chars.
See the regex demo and the regex graph:
Details
^ - start of string
[^.]+ - 1+ chars other than .
\. - a dot
| - or
\. - a dot
[^.]+ - 1+ chars other than .
$ - end of string.
You can use -match to capture your desired output using regex
$string ="SIT.com.local.test.stack.properties"
$string -match "^.*?\.(.+)\.[^.]+$"
$Matches.1
You can do this with the Split operator also.
($string -split "\.",2)[1]
Explanation:
You split on the literal . character with regex \.. The ,2 syntax tells PowerShell to return 2 substrings after the split. The [1] index selects the second element of the returned array. [0] is the first substring (SIT in this case).
I'm using regular expressions in a custom text editor to in effect whitelist certain modules (assert and crypto). I'm close to what I need but not quite there. Here it is:
/require\s*\(\s*'(?!(\bassert\b|\bcrypto\b)).*'\s*\)/
I want the regular expression to match any line with require('foo'); where foo is anything except for 'assert' or 'crypto'. The case I'm failing is require('assert '); which is not being matched with my regex however require(' assert'); is correctly being matched.
https://regexr.com/4i6ot
If you don't want to match assert or crypto between ', you could change the lookahead to assert exactly that. You can omit the word boundaries matching the words right after the '.
If what follows should match until the first occurrence of ', you could use a negated character class [^'\r\n]* to match any char except ' or a newline.
require\s*\(\s*'(?!(assert|crypto)')[^'\r\n]*'\s*\)
^
Regex demo
You can use: require\s*\(\s*'(?!(\bassert'|\bcrypto')).*'\s*\)
Online demo
The difference is that I replaced word boundary \b with ' at the end of the module names. With \b a module name of 'assert ' was matched by negative lookahead, because t was matched by \b. In the new version, we require ' at the end of the name of the module.
EDIT
As Cary Swoveland advised, leading \b are not required:
require\s*\(\s*'(?!(assert'|crypto')).*'\s*\)
Demo
I assume from the flawed regex that if there is a match the string between "('" and "')" is to be captured. One way to do that follows.
r = /
require # match word
\ * # match zero or more spaces (note escaped space)
\( # match a left paren
(?! # begin a negative lookahead
' # match a single quote
(?:assert|crypto) # match either word
' # match a single quote
(?=\)) # match a right paren in a forward lookahead
) # end negative lookahead
' # match a single quote
(.*?) # match any number of characters lazily in a capture group 1
' # match a single quote
\) # match a right paren
/x # free-spacing regex definition mode
As the capture group is followed by a single quote, matching characters in the capture group lazily ensures that a single quote is not matched in the capture group. I could have instead written ([^']*). In conventional form this regex is written as follows:
r = /require *\((?!'(?:assert|crypto)'(?=\)))'(.*?)'\)/
Note that in free-spacing regex definition mode spaces will be removed unless they are escaped, put in a character class ([ ]), replaced with \p{Space} and so on.
"require ('victory')" =~ r #=> 0
$1 #=> "victory"
"require (' assert')" =~ r #=> 0
$1 #=> " assert"
"require ('assert ')" =~ r #=> 0
$1 #=> "assert "
"require ('crypto')" =~ r #=> nil
"require ('assert')" =~ r #=> nil
"require\n('victory')" =~ r #=> nil
Notice that had I replace the space character in the regex with "\s" in the last example I would have obtained:
"require\n('victory')" =~ r #=> 0
$1 #=> "victory"
I don't think you need anything remotely that complicated, this simple pattern will work just fine:
require\((?!'crypto'|'assert')'.*'\);
regex101 demo
I have the following string:
my $string = "Ethernet FlexNIC (NIC 1) LOM1:1-a FC:15:B4:13:6A:A8";
I want to extract the number that is in brackets (1) in another variable.
The following statement does not work:
my ($NAdapter) = $string =~ /\((\d+)\)/;
What is the correct syntax?
\d+(?=[^(]*\))
You can use this.See demo.Yours will not work as inside () there is more data besides \d+.
https://regex101.com/r/fM9lY3/57
You could try something like
my ($NAdapter) = $string =~ /\(.*(\d+).*\)/;
After that, $NAdapter should include the number that you want.
my $string = "Ethernet FlexNIC (NIC 1) LOM1:1-a FC:15:B4:13:6A:A8";
I want to extract the number that is in brackets (1) in another
variable
Your regex (with some spaces for clarity):
/ \( (\d+) \) /x;
says to match:
A literal opening parenthesis, immediately followed by...
A digit, one or more times (captured in group 1), immediately followed by...
A literal closing parenthesis.
Yet, the substring you want to match:
(NIC 1)
is of the form:
A literal opening parenthesis, immediately followed by...
Some capital letters
STOP EVERYTHING! NO MATCH!
As an alternative, your substring:
(NIC 1)
could be described as:
Some digits, immediately followed by...
A literal closing parenthesis.
Here's the regex:
use strict;
use warnings;
use 5.020;
my $string = "Ethernet FlexNIC (NIC 1234) LOM1:1-a FC:15:B4:13:6A:A8";
my ($match) = $string =~ /
(\d+) #Match any digit, one or more times, captured in group 1, followed by...
\) #a literal closing parenthesis.
#Parentheses have a special meaning in a regex--they create a capture
#group--so if you want to match a parenthesis in your string, you
#have to escape the parenthesis in your regex with a backslash.
/xms; #Standard flags that some people apply to every regex.
say $match;
--output:--
1234
Another description of your substring:
(NIC 1)
could be:
A literal opening parenthesis, immediately followed by...
Some non-digits, immediately followed by...
Some digits, immediately followed by..
A literal closing parenthesis.
Here's the regex:
use strict;
use warnings;
use 5.020;
my $string = "Ethernet FlexNIC (ABC NIC789) LOM1:1-a FC:15:B4:13:6A:A8";
my ($match) = $string =~ /
\( #Match a literal opening parethesis, followed by...
\D+ #a non-digit, one or more times, followed by...
(\d+) #a digit, one or more times, captured in group 1, followed by...
\) #a literal closing parentheses.
/xms; #Standard flags that some people apply to every regex.
say $match;
--output:--
789
If there might be spaces on some lines and not others, such as:
spaces
||
VV
(NIC 1 )
(NIC 2)
You can insert a \s* (any whitespace, zero or more times) in the appropriate place in the regex, for instance:
my ($match) = $string =~ /
#Parentheses have special meaning in a regex--they create a capture
#group--so if you want to match a parenthesis in your string, you
#have to escape the parenthesis in your regex with a backslash.
\( #Match a literal opening parethesis, followed by...
\D+ #a non-digit, one or more times, followed by...
(\d+) #a digit, one or more times, captured in group 1, followed by...
\s* #any whitespace, zero or more times, followed by...
\) #a literal closing parentheses.
/xms; #Standard flags that some people apply to every regex.
So I have little problem, I need to check in perl if last character in string is "a". I know, that I can do it in this way:
$test = "mama";
$test2 = substr $test, -1
And now just check if $test2 not equal "a". But how can I do this with regex?
The $ matches the end of the string:
my $test = "mama";
print "Terminal 'a' in $test\n" if $test =~ /a$/;
In Perl, $ does not necessarily match the end of the string:
^ Match string start (or line, if /m is used)
$ Match string end (or line, if /m is used) or before newline
\b Match word boundary (between \w and \W)
\B Match except at word boundary (between \w and \w or \W and \W)
\A Match string start (regardless of /m)
\Z Match string end (before optional newline)
\z Match absolute string end
\G Match where previous m//g left off
\K Keep the stuff left of the \K, don't include it in $&
Therefore, to check if the last character of $s really is 'a', you must use:
if ($s =~ /a\z/) { ...
because
$ perl -E 'say "yes" if "a\n" =~ /a$/'
yes
I have an SQL Select dump with many lines each looks like this:
07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',,,,'text',,,0,0,
I want to do 2 things to each line:
Replace all dates with Oracle's sysdate function. Dates can also come without hour (like 07/11/2011).
Replace all null values with null string
Here's my attempt:
$_ =~ s/,(,|\n)/,null$1/g; # Replace no data by "null"
$_ =~ s/\d{2}\/\d{2}\/d{4}.*?,/sysdate,/g; # Replace dates by "sysdate"
But this would transform the string to:
07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',null,,null,'text',null,,0,0,null
while I expect it to be
sysdate,sysdate,'YD','MANUAL',0,1,'text','text','text','text',null,null,null,'text',null,null,0,0,null
I don't understand why dates do not match and why some ,, are not replaced by null.
Any insights welcome, thanks in advance.
\d{2}\/\d{2}\/d{4}.*?, didn't work because the last d wasn't escaped.
If a , can be on either side, or begin/end of string, you could do it in 2 steps:
step 1
s/(?:^|(?<=,))(?=,|\n)/null/g
expanded:
/
(?: ^ # Begining of line, ie: nothing behind us
| (?<=,) # Or, a comma behind us
)
# we are HERE!, this is the place between characters
(?= , # A comma in front of us
| \n # Or, a newline in front of us
)
/null/g
# The above regex does not consume, it just inserts 'null', leaving the
# same search position (after the insertion, but before the comma).
# If you want to consume a comma, it would be done this way:
s/(?:^|(?<=,))(,|\n)/null$1/xg
# Now the search position is after the 'null,'
step 2
s/(?:^|(?<=,))\d{2}\/\d{2}\/\d{4}.*?(?=,|\n)/sysdate/g
Or, you could combine them into a single regex, using the eval modifier:
$row =~ s/(?:^|(?<=,))(\d{2}\/\d{2}\/\d{4}.*?|)(?=,|\n)/ length $1 ? 'sysdate' : 'null'/eg;
Broken down it looks like this
s{
(?: ^ | (?<=,) ) # begin of line or comma behind us
( # Capt group $1
\d{2}/\d{2}/\d{4}.*? # date format and optional non-newline chars
| # Or, nothing at all
) # End Capt group 1
(?= , | \n ) # comma or newline in front of us
}{
length $1 ? 'sysdate' : 'null'
}eg
If there is a chance of non-newline whitespace padding, it could be written as:
$row =~ s/(?:^|(?<=,))(?:([^\S\n]*\d{2}\/\d{2}\/\d{4}.*?)|[^\S\n]*)(?=,|\n)/ defined $1 ? 'sysdate' : 'null'/eg;
You could do this:
$ cat perlregex.pl
use warnings;
use strict;
my $row = "07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',,,,'text',,,0,0,\n";
print( "$row\n" );
while ( $row =~ /,([,\n])/ ) { $row =~ s/,([,\n])/,null$1/; }
print( "$row\n" );
$row =~ s/\d{2}\/\d{2}\/\d{4}.*?,/sysdate,/g;
print( "$row\n" );
Which results in this:
$ ./perlregex.pl
07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',,,,'text',,,0,0,
07/11/2011 16:48:08,07/11/2011 16:48:08,'YD','MANUAL',0,1,'text','text','text','text',null,null,null,'text',null,null,0,0,null
sysdate,sysdate,'YD','MANUAL',0,1,'text','text','text','text',null,null,null,'text',null,null,0,0,null
This could certainly be optimized, but it gets the point across.
You want to replace something. Usually lookaheads are a better option for this :
$subject =~ s/(?<=,)(?=,|$)/null/g;
Explanation :
"
(?<= # Assert that the regex below can be matched, with the match ending at this position (positive lookbehind)
, # Match the character “,” literally
)
(?= # Assert that the regex below can be matched, starting at this position (positive lookahead)
# Match either the regular expression below (attempting the next alternative only if this one fails)
, # Match the character “,” literally
| # Or match regular expression number 2 below (the entire group fails if this one fails to match)
\$ # Assert position at the end of the string (or before the line break at the end of the string, if any)
)
"
Secodnly you wish to replace the dates :
$subject =~ s!\d{2}/\d{2}/\d{4}.*?(?=,)!sysdate!g;
That's almost the same with your original regex. Just replace the last , with lookahead. (If you don't want to replace it , don't match it.)
# \d{2}/\d{2}/\d{4}.*?(?=,)
#
# Match a single digit 0..9 «\d{2}»
# Exactly 2 times «{2}»
# Match the character “/” literally «/»
# Match a single digit 0..9 «\d{2}»
# Exactly 2 times «{2}»
# Match the character “/” literally «/»
# Match a single digit 0..9 «\d{4}»
# Exactly 4 times «{4}»
# Match any single character that is not a line break character «.*?»
# Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
# Assert that the regex below can be matched, starting at this position (positive lookahead) «(?=,)»
# Match the character “,” literally «,»
Maybe .*? is too greedy, try:
$_ =~ s/\d{2}\/\d{2}\/d{4}[^,]+,/sysdate,/g;