Regsubbing simple matches - match

I'm looking for a regsub example that does the following:
123tcl456TCL789 => 123!tcl!456!TCL!789
This is an Tcl example => This is an !Tcl! example
Yes, I could use string first to find a position and mash things but I saw in past a regsub command that does what I want but I can't recall. What would be the regsub command that allows that? I would guess regsub -all -nocase is a start.
I am bad at regsub and regexps. I wonder if there is a site or tool/script that we can supply a string, the final result and then we get the regsub form.


You're looking at the right tool, but there are various options, depending on exactly what the conditions are when faced with other text. Here's one that wraps each occurrence of "Tcl" (any capitalisation) with exclamation marks:
set inputString "123tcl456TCL789"
set replaced [regsub -all -nocase {tcl} $inputString {!&!}]
puts $replaced
That's using a very simple regular expression with the -nocase option, and the replacement means "put ! on either side of the substring matched".
Another (more generally applicable... perhaps) might be to put ! after any letter or number sequence that is followed by a number or letter.
set replaced [regsub -all {[A-Za-z]+(?=[0-9])|[0-9]+(?=[A-Za-z])} $inputString {&!}]
Note that doing things correctly typically requires understanding the real input data fairly well. For example, whether the numbers include floating point numbers in scientific notation, or whether the substrings to delimit are of fixed length.

Related

match string pattern by certain characters but exclude combinations of those characters

I have the following sample string:
'-Dparam="x" -f hello-world.txt bye1.txt foo_bar.txt -Dparam2="y"'
I am trying to use RegEx (PowerShell, .NET flavor) to extract the filenames hello-world.txt, bye1.txt, and foo_bar.txt.
The real use case could have any number of -D parameters, and the -f <filenames> argument could appear in any position between these other parameters. I can't easily use something like split to extract it as the delimiter positioning could change, so I thought RegEx might be a good proposition here.
My attempt is something like this in PowerShell (can be opened on any Windows system and copy pasted into it):
'-Dparam="x" -f hello-world.txt bye1.txt foo_bar.txt -Dparam2="y"' -replace '^.* -f ([a-zA-Z0-9_.\s-]+).*$','$1'
Desired output:
hello-world.txt bye1.txt foo_bar.txt
My problem is that I either only take hello-world.txt, or I get hello-world.txt all the way to the end of the string or next = symbol (as in the example above).
I am having trouble expressing that \s is allowed, since I need to capture multiple space-delimited filenames, but that the combination of \s-[a-zA-Z] is not allowed, as that indicates the start of the next argument.

Partial String Replacement using PowerShell

Problem
I am working on a script that has a user provide a specific IP address and I want to mask this IP in some fashion so that it isn't stored in the logs. My problem is, that I can easily do this when I know what the first three values of the IP typically are; however, I want to avoid storing/hard coding those values into the code to if at all possible. I also want to be able to replace the values even if the first three are unknown to me.
Examples:
10.11.12.50 would display as XX.XX.XX.50
10.12.11.23 would also display as XX.XX.XX.23
I have looked up partial string replacements, but none of the questions or problems that I found came close to doing this. I have tried doing things like:
# This ended up replacing all of the numbers
$tempString = $str -replace '[0-9]', 'X'
I know that I am partway there, but I aiming to only replace only the first 3 sets of digits so, basically every digit that is before a '.', but I haven't been able to achieve this.
Question
Is what I'm trying to do possible to achieve with PowerShell? Is there a best practice way of achieving this?

Here's an example of how you can accomplish this:
Get-Content 'File.txt' |
ForEach-Object { $_ = $_ -replace '\d{1,3}\.\d{1,3}\.\d{1,3}','xx.xx.xx' }
This example matches a digit 1-3 times, a literal period, and continues that pattern so it'll capture anything from 0-999.0-999.0-999 and replace with xx.xx.xx

TheIncorrigible1's helpful answer is an exact way of solving the problem (replacement only happens if 3 consecutive .-separated groups of 1-3 digits are matched.)
A looser, but shorter solution that replaces everything but the last .-prefixed digit group:
PS> '10.11.12.50' -replace '.+(?=\.\d+$)', 'XX.XX.XX'
XX.XX.XX.50
(?=\.\d+$) is a (positive) lookahead assertion ((?=...)) that matches the enclosed subexpression (a literal . followed by 1 or more digits (\d) at the end of the string ($)), but doesn't capture it as part of the overall match.
The net effect is that only what .+ captured - everything before the lookahead assertion's match - is replaced with 'XX.XX.XX'.
Applied to the above example input string, 10.11.12.50:
(?=\.\d+$) matches the .-prefixed digit group at the end, .50.
.+ matches everything before .50, which is 10.11.12.
Since the (?=...) part isn't captured, it is therefore not included in what is replaced, so it is only substring 10.11.12 that is replaced, namely with XX.XX.XX, yielding XX.XX.XX.50 as a result.

PCRE Regex - How to return matches with multiline string looking for multiple strings in any order

I need to use Perl-compatible regex to match several strings which appear over multiple lines in a file.
The matches need to appear in any order (server servernameA.company.com followed by servernameZ.company.com followed by servernameD.company.com or any order combination of the three). Note: All matches will appear at the beginning of each line.
In my testing with grep -P, I haven't even been able to produce a match on simple string terms that appear in any order over new lines (even when using the /s and /m modifiers). I am pretty sure from reading I need a look-ahead assertion but the samples I used didn't produce a match for me even after analyzing each bit of the regex to make sure it was relevant to my scenario.
Since I need to support this in Production, I would like an answer that is simple and relatively straight-forward to interpret.
Sample Input
irrelevant_directive = 0
# Comment
server servernameA.company.com iburst
additional_directive = yes
server servernameZ.company.com iburst
server servernameD.company.com iburst
# Additional Comment
final_directive = true
Expectation
The regex should match and return the 3 lines beginning with server (that appear in any order) if and only if there is a perfect match for strings'serverA.company.com', 'serverZ.company.com', and 'serverD.company.com' followed by iburst. All 3 strings must be included.
Finally, if the answer (or a very similar form of the answer) can address checking for strings in any order on a single line, that would be very helpful. For example, if I have a single-line string of: preauth param audit=true silent deny=5 severe=false unlock_time=1000 time=20ms and I want to ensure the terms deny=5 and time=20ms appear in any order and if so match.
Thank you in advance for your assistance.

Regarding the main issue [for the secondary question see Casimir et Hippolyte answer] (using x modifier): https://regex101.com/r/mkxcap/5
(?:
(?<a>.*serverA\.company\.com\s+iburst.*)
|(?<z>.*serverZ\.company\.com\s+iburst.*)
|(?<d>.*serverD\.company\.com\s+iburst.*)
|[^\n]*(?:\n|$)
)++
(?(a)(?(z)(?(d)(*ACCEPT))))(*SKIP)(*F)
The matches are now all in the a, z and d capturing groups.
It's not the most efficient (it goes three times over each line with backtracking...), but the main takeaway is to register the matches with capturing groups and then checking for them being defined.

You don't need to use the PCRE features, you can simply write in ERE:
grep -E '.*(\bdeny=5\b.*\btime=20ms\b|\btime=20ms\b.*\bdeny=5\b).*' file
The PCRE approach will be different: (however you can also use the previous pattern)
grep -P '^(?=.*\bdeny=5\b).*\btime=20ms\b.*' file

Substitute only one part of a string using perl

I have an array that have some symbols that I want to remove and even thought I find a solution, I will like to know if this is the right way because I'm afraid if I use it with array will remove the character that I might need on future arrays.
Here is an example item on my array:
$string1='22 | logging monitor informational';
so I try the following:
$string1=~ s/\s{6}\|(?=\s{6})//;
So my output is:
22 logging monitor informational
Is the other way that best match "|". I just want to remove the pipe character.
Thanks in advance

"I want to remove just the pipe character."
OK, then do this:
$string1 =~ s/\|//;
This will remove the first pipe character in the string. (You said in another comment that you don't want to remove any additional pipe characters.) If that's not what you want, then I'd suggest telling us exactly what you do want. We can't read minds, you know.
In the mean time, I'd also strongly recommend reading the Perl regular expressions tutorial.

How does this Perl one-liner actually work?

So, I happened to notice that last.fm is hiring in my area, and since I've known a few people who worked there, I though of applying.
But I thought I'd better take a look at the current staff first.
Everyone on that page has a cute/clever/dumb strapline, like "Is life not a thousand times too short for us to bore ourselves?". In fact, it was quite amusing, until I got to this:
perl -e'print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34'
Which I couldn't resist pasting into my terminal (kind of a stupid thing to do, maybe), but it printed:
Just another Last.fm hacker,
I thought it would be relatively easy to figure out how that Perl one-liner works. But I couldn't really make sense of the documentation, and I don't know Perl, so I wasn't even sure I was reading the relevant documentation.
So I tried modifying the numbers, which got me nowhere. So I decided it was genuinely interesting and worth figuring out.
So, 'how does it work' being a bit vague, my question is mainly,
What are those numbers? Why are there negative numbers and positive numbers, and does the negativity or positivity matter?
What does the combination of operators +=$_ do?
What's pack+q,c*,, doing?

This is a variant on “Just another Perl hacker”, a Perl meme. As JAPHs go, this one is relatively tame.
The first thing you need to do is figure out how to parse the perl program. It lacks parentheses around function calls and uses the + and quote-like operators in interesting ways. The original program is this:
print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34
pack is a function, whereas print and map are list operators. Either way, a function or non-nullary operator name immediately followed by a plus sign can't be using + as a binary operator, so both + signs at the beginning are unary operators. This oddity is described in the manual.
If we add parentheses, use the block syntax for map, and add a bit of whitespace, we get:
print(+pack(+q,c*,,
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
The next tricky bit is that q here is the q quote-like operator. It's more commonly written with single quotes:
print(+pack(+'c*',
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
Remember that the unary plus is a no-op (apart from forcing a scalar context), so things should now be looking more familiar. This is a call to the pack function, with a format of c*, meaning “any number of characters, specified by their number in the current character set”. An alternate way to write this is
print(join("", map {chr($.+=$_)} (74, …, -34)))
The map function applies the supplied block to the elements of the argument list in order. For each element, $_ is set to the element value, and the result of the map call is the list of values returned by executing the block on the successive elements. A longer way to write this program would be
#list_accumulator = ();
for $n in (74, …, -34) {
$. += $n;
push #list_accumulator, chr($.)
}
print(join("", #list_accumulator))
The $. variable contains a running total of the numbers. The numbers are chosen so that the running total is the ASCII codes of the characters the author wants to print: 74=J, 74+43=117=u, 74+43-2=115=s, etc. They are negative or positive depending on whether each character is before or after the previous one in ASCII order.
For your next task, explain this JAPH (produced by EyesDrop).
''=~('(?{'.('-)#.)#_*([]#!#/)(#)#-#),#(##+#)'
^'][)#]`}`]()`#.#]#%[`}%[#`#!##%[').',"})')
Don't use any of this in production code.

The basic idea behind this is quite simple. You have an array containing the ASCII values of the characters. To make things a little bit more complicated you don't use absolute values, but relative ones except for the first one. So the idea is to add the specific value to the previous one, for example:
74 -> J
74 + 43 -> u
74 + 42 + (-2 ) -> s
Even though $. is a special variable in Perl it does not mean anything special in this case. It is just used to save the previous value and add the current element:
map($.+=$_, ARRAY)
Basically it means add the current list element ($_) to the variable $.. This will return a new array with the correct ASCII values for the new sentence.
The q function in Perl is used for single quoted, literal strings. E.g. you can use something like
q/Literal $1 String/
q!Another literal String!
q,Third literal string,
This means that pack+q,c*,, is basically pack 'c*', ARRAY. The c* modifier in pack interprets the value as characters. For example, it will use the value and interpret it as a character.
It basically boils down to this:
#!/usr/bin/perl
use strict;
use warnings;
my $prev_value = 0;
my #relative = (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34);
my #absolute = map($prev_value += $_, #relative);
print pack("c*", #absolute);