I would like to take part of a string to use it elsewhere. For example, I have the following strings:
Project XYZ is the project name - 20-12-11
I would like to get the value "XYZ is the project name" from the string. The word "Project" and character "-" before the number will always be there.
I think a lookaround regular expression would work here since "Project" and "-" are always there:
(?<=Project ).+?(?= -)
A lookaround can be useful for cases that deal with getting a sub string.
Explanation:
(?<= = negative lookbehind
Project = starting string (including space)
) = closing negative lookbehind
.+? = matches anything in between
(?= = positive lookahead
- = ending string
) = closing positive lookahead
Example in PowerShell:
Function GetProjectName($InputString) {
$regExResult = $InputString | Select-String -Pattern '(?<=Project ).+?(?= -)'
$regExResult.Matches[0].Value
}
$projectName = GetProjectName -InputString "Project XYZ is the project name - 20-12-11"
Write-Host "Result = '$($projectName)'"
here is yet another regex version. [grin] it may be easier to understand since it uses somewhat basic regex patterns.
what it does ...
defines the input string
defines the prefix to match on
this will keep only what comes after it.
defines the suffix to match on
this part will keep only what is before it.
trigger the replace
the part in the () is what will be placed into the 1st capture group.
show what was kept
the code ...
$InString = 'Project XYZ is the project name - 20-12-11'
# "^" = start of string
$Prefix = '^project '
# ".+' = one or more of any character
# "$" = end of string
$Suffix = ' - .+$'
# "$1" holds the content of the 1st [and only] capture group
$OutString = $InString -replace "$Prefix(.+)$Suffix", '$1'
$OutString
# define the input string
$str = 'Project XYZ is the project name - 20-12-11'
# use regex (-match) including the .*? regex pattern
# this patterns means (.)any char, (*) any times, (?) maximum greed
# to capture (into brackets) the desired pattern substring
$str -match "(Project.*?is the project name)"
# show result (the first capturing group)
$matches[1]
Related
I have a block of text I need to parse (saved in a variable) but I'm unsure how to go about it. This block of text, saved in a variable we can call $block for simplicity's sake, includes all the whitespace shown below.
I would like the result to be an iterable list, the first value being Health_AEPOEP_Membership_Summary - Dev and the second one being Health_AEPOEP_YoY_Comparison_Summary - Dev. Assume this list of workbooks can be longer (up to 50) or shorter (minimum 1 workbook), and all workbooks are formatted similarly (in terms of name_with_underscores - Dev. I'd try the $block.split(" ") method, but this method gives many spaces which may be hard to enumerate and account for.
Workbooks : Health_AEPOEP_Membership_Summary - Dev [Project: Health - Dev]
Health_AEPOEP_YoY_Comparison_Summary - Dev [Project: Health - Dev]
Any help is much appreciated!
You could write a multi-line regex pattern and try to extract the names, but it might be easier to reason about if you just breaking it into simple(r) steps:
$string = #'
Workbooks : Health_AEPOEP_Membership_Summary - Dev [Project: Health - Dev]
Health_AEPOEP_YoY_Comparison_Summary - Dev [Project: Health - Dev]
'#
# Split into one string per line
$strings = $string -split '\r?\n'
# Remove leading whitespace
$strings = $strings -replace '^\s*'
# Remove `Workbooks : ` prefix (strings that don't match will be left untouched)
$strings = $strings -replace '^Workbooks :\s*'
# Remove `[Project $NAME]` suffix
$strings = $strings -replace '\s*\[Project: [^\]]+\]'
# Get rid of empty lines
$strings = $strings |Where-Object Length
$strings now contains the two project names
If the text is in a file it would make this a little easier, and I would recommend this approach
switch -Regex -file ($file){
'(\w+_.+- Dev)' {$matches.1}
}
Regex details
() - capture group
\w+ - match one or more letter characters
_ - match literal underscore
.+ - match one or more of any character
- Dev - literal match of dash space Dev
If it's already in a variable, it would depend if it's a string array or a single string. Assuming it's a single string, I'd recommend this approach
$regex = [regex]'(\w+_.+)(?=(\s\[.+))'
$regex.Matches($block).value
Health_AEPOEP_Membership_Summary - Dev
Health_AEPOEP_YoY_Comparison_Summary - Dev
Regex details
Same as above but added the following
(?=) - Look ahead
\s\[.+ - match a space, a left square bracket, one or more characters
Simply add a variable assignment $strings = before either of these to capture the output. Either would work on one or 500 workbooks.
Think I've found the worst way to do this:
$ip = "192.168.13.1"
$a,$b,$c,$d = $ip.Split(".")
[int]$c = $c
$c = $c+1
[string]$c = $c
$newIP = $a+"."+$b+"."+$c+"."+$d
$newIP
But what is the best way? Has to be string when completed. Not bothered about validating its a legit IP.
Using your example for how you want to modify the third octet, I'd do it pretty much the same way, but I'd compress some of the steps together:
$IP = "192.168.13.1"
$octets = $IP.Split(".") # or $octets = $IP -split "\."
$octets[2] = [string]([int]$octets[2] + 1) # or other manipulation of the third octet
$newIP = $octets -join "."
$newIP
You can simply use the -replace operator of PowerShell and a look ahead pattern. Look at this script below
Set-StrictMode -Version "2.0"
$ErrorActionPreference="Stop"
cls
$ip1 = "192.168.13.123"
$tests=#("192.168.13.123" , "192.168.13.1" , "192.168.13.12")
foreach($test in $tests)
{
$patternRegex="\d{1,3}(?=\.\d{1,3}$)"
$newOctet="420"
$ipNew=$test -replace $patternRegex,$newOctet
$msg="OLD ip={0} NEW ip={1}" -f $test,$ipNew
Write-Host $msg
}
This will produce the following:
OLD ip=192.168.13.123 NEW ip=192.168.420.123
OLD ip=192.168.13.1 NEW ip=192.168.420.1
OLD ip=192.168.13.12 NEW ip=192.168.420.12
How to use the -replace operator?
https://powershell.org/2013/08/regular-expressions-are-a-replaces-best-friend/
Understanding the pattern that I have used
The (?=) in \d{1,3}(?=.\d{1,3}$) means look behind.
The (?=.\d{1,3}$ in \d{1,3}(?=.\d{1,3}$) means anything behind a DOT and 1-3 digits.
The leading \d{1,3} is an instruction to specifically match 1-3 digits
All combined in plain english "Give me 1-3 digits which is behind a period and 1-3 digits located towards the right side boundary of the string"
Look ahead regex
https://learn.microsoft.com/en-us/dotnet/standard/base-types/regular-expression-language-quick-reference
CORRECTION
The regex pattern is a look ahead and not look behind.
If you have PowerShell Core (v6.1 or higher), you can combine -replace with a script block-based replacement:
PS> '192.168.13.1' -replace '(?<=^(\d+\.){2})\d+', { 1 + $_.Value }
192.168.14.1
Negative look-behind assertion (?<=^(\d+\.){2}) matches everything up to, but not including, the 3rd octet - without considering it part of the overall match to replace.
(?<=...) is the look-behind assertion, \d+ matches one or more (+) digits (\d), \. a literal ., and {2} matches the preceding subexpression ((...)) 2 times.
\d+ then matches just the 3rd octet; since nothing more is matched, the remainder of the string (. and the 4th octet) is left in place.
Inside the replacement script block ({ ... }), $_ refers to the results of the match, in the form of a [MatchInfo] instance; its .Value is the matched string, i.e. the 3rd octet, to which 1 can be added.
Data type note: by using 1, an implicit [int], as the LHS, the RHS (the .Value string) is implicitly coerced to [int] (you may choose to use an explicit cast).
On output, whatever the script block returns is automatically coerced back to a string.
If you must remain compatible with Windows PowerShell, consider Jeff Zeitlin's helpful answer.
For complete your method but shortly :
$a,$b,$c,$d = "192.168.13.1".Split(".")
$IP="$a.$b.$([int]$c+1).$d"
function Replace-3rdOctet {
Param(
[string]$GivenIP,
[string]$New3rdOctet
)
$GivenIP -match '(\d{1,3}).(\d{1,3}).(\d{1,3}).(\d{1,3})' | Out-Null
$Output = "$($matches[1]).$($matches[2]).$New3rdOctet.$($matches[4])"
Return $Output
}
Copy to a ps1 file and dot source it from command line, then type
Replace-3rdOctet -GivenIP '100.201.190.150' -New3rdOctet '42'
Output: 100.201.42.150
From there you could add extra error handling etc for random input etc.
here's a slightly different method. [grin] i managed to not notice the answer by JeffZeitlin until after i finished this.
[edit - thanks to JeffZeitlin for reminding me that the OP wants the final result as a string. oops! [*blush*]]
what it does ...
splits the string on the dots
puts that into an [int] array & coerces the items into that type
increments the item in the targeted slot
joins the items back into a string with a dot for the delimiter
converts that to an IP address type
adds a line to convert the IP address to a string
here's the code ...
$OriginalIPv4 = '1.1.1.1'
$TargetOctet = 3
$OctetList = [int[]]$OriginalIPv4.Split('.')
$OctetList[$TargetOctet - 1]++
$NewIPv4 = [ipaddress]($OctetList -join '.')
$NewIPv4
'=' * 30
$NewIPv4.IPAddressToString
output ...
Address : 16908545
AddressFamily : InterNetwork
ScopeId :
IsIPv6Multicast : False
IsIPv6LinkLocal : False
IsIPv6SiteLocal : False
IsIPv6Teredo : False
IsIPv4MappedToIPv6 : False
IPAddressToString : 1.1.2.1
==============================
1.1.2.1
I'm not very experienced in Regex. Can you tell me how to get a string value from between two strings?
The subject will always be in this format : //subject/some_other_stuff
I need to get the string found between // and /.
For example:
Full String = //Manhattan/Project
Output = Manhattan
Any help will be very much appreciated.
You can use a negated character class and reference capturing group #1 for your match result.
//([^/]+)/
Explanation:
// # '//'
( # group and capture to \1:
[^/]+ # any character except: '/' (1 or more times)
) # end of \1
/ # '/'
You could use the below regex which uses lookarounds.
(?<=\/\/)[^\/]+(?=\/)
Since the strings are always of the same format, you can simply split them on / and then retrieve the element at index 2 (the third element):
PS > $str = "//Manhattan/Project"
PS > $str.split('/')[2]
Manhattan
PS > $str = "//subject/some_other_stuff"
PS > $str.split('/')[2]
subject
PS >
I'm trying to learn something about regular expressions.
Here is what I'm going to match:
/parent/child
/parent/child?
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/
/parent/child/?
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789
My expression should "grabs" abc123 and def456.
And now just an example about what I'm not going to match ("question mark" is missing):
/parent/child/firstparam=abc123&secondparam=def456
Well, I built the following expression:
^(?:/parent/child){1}(?:^(?:/\?|\?)+(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?)?
But that doesn't work.
Could you help me to understand what I'm doing wrong?
Thanks in advance.
UPDATE 1
Ok, I made other tests.
I'm trying to fix the previous version with something like this:
/parent/child(?:(?:\?|/\?)+(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?)?$
Let me explain my idea:
Must start with /parent/child:
/parent/child
Following group is optional
(?: ... )?
The previous optional group must starts with ? or /?
(?:\?|/\?)+
Optional parameters (I grab values if specified parameters are part of querystring)
(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?
End of line
$
Any advice?
UPDATE 2
My solution must be based just on regular expressions.
Just for example, I previously wrote the following one:
/parent/child(?:[?&/]*(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*))*$
And that works pretty nice.
But it matches the following input too:
/parent/child/firstparam=abc123&secondparam=def456
How could I modify the expression in order to not match the previous string?
You didn't specify a language so I'll just usre Perl. So basically instead of matching everything, I just matched exactly what I thought you needed. Correct me if I am wrong please.
while ($subject =~ m/(?<==)\w+?(?=&|\W|$)/g) {
# matched text = $&
}
(?<= # Assert that the regex below can be matched, with the match ending at this position (positive lookbehind)
= # Match the character “=” literally
)
\\w # Match a single character that is a “word character” (letters, digits, and underscores)
+? # Between one 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 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 (attempting the next alternative only if this one fails)
\\W # Match a single character that is a “non-word character”
| # Or match regular expression number 3 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)
)
Output:
This regex will work as long as you know what your parameter names are going to be and you're sure that they won't change.
\/parent\/child\/?\?(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)?
Whilst regex is not the best solution for this (the above code examples will be far more efficient, as string functions are way faster than regexes) this will work if you need a regex solution with up to 3 parameters. Out of interest, why must the solution use only regex?
In any case, this regex will match the following strings:
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789
It will now only match those containing query string parameters, and put them into capture groups for you.
What language are you using to process your matches?
If you are using preg_match with PHP, you can get the whole match as well as capture groups in an array with
preg_match($regex, $string, $matches);
Then you can access the whole match with $matches[0] and the rest with $matches[1], $matches[2], etc.
If you want to add additional parameters you'll also need to add them in the regex too, and add additional parts to get your data. For example, if you had
/parent/child/?secondparam=def456&firstparam=abc123&fourthparam=jkl01112&thirdparam=ghi789
The regex will become
\/parent\/child\/?\?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?
This will become a bit more tedious to maintain as you add more parameters, though.
You can optionally include ^ $ at the start and end if the multi-line flag is enabled. If you also need to match the whole lines without query strings, wrap this whole regex in a non-capture group (including ^ $) and add
|(?:^\/parent\/child\/?\??$)
to the end.
You're not escaping the /s in your regex for starters and using {1} for a single repetition of something is unnecessary; you only use those when you want more than one repetition or a range of repetitions.
And part of what you're trying to do is simply not a good use of a regex. I'll show you an easier way to deal with that: you want to use something like split and put the information into a hash that you can check the contents of later. Because you didn't specify a language, I'm just going to use Perl for my example, but every language I know with regexes also has easy access to hashes and something like split, so this should be easy enough to port:
# I picked an example to show how this works.
my $route = '/parent/child/?first=123&second=345&third=678';
my %params; # I'm going to put those URL parameters in this hash.
# Perl has a way to let me avoid escaping the /s, but I wanted an example that
# works in other languages too.
if ($route =~ m/\/parent\/child\/\?(.*)/) { # Use the regex for this part
print "Matched route.\n";
# But NOT for this part.
my $query = $1; # $1 is a Perl thing. It contains what (.*) matched above.
my #items = split '&', $query; # Each item is something like param=123
foreach my $item (#items) {
my ($param, $value) = split '=', $item;
$params{$param} = $value; # Put the parameters in a hash for easy access.
print "$param set to $value \n";
}
}
# Now you can check the parameter values and do whatever you need to with them.
# And you can add new parameters whenever you want, etc.
if ($params{'first'} eq '123') {
# Do whatever
}
My solution:
/(?:\w+/)*(?:(?:\w+)?\?(?:\w+=\w+(?:&\w+=\w+)*)?|\w+|)
Explain:
/(?:\w+/)* match /parent/child/ or /parent/
(?:\w+)?\?(?:\w+=\w+(?:&\w+=\w+)*)? match child?firstparam=abc123 or ?firstparam=abc123 or ?
\w+ match text like child
..|) match nothing(empty)
If you need only query string, pattern would reduce such as:
/(?:\w+/)*(?:\w+)?\?(\w+=\w+(?:&\w+=\w+)*)
If you want to get every parameter from query string, this is a Ruby sample:
re = /\/(?:\w+\/)*(?:\w+)?\?(\w+=\w+(?:&\w+=\w+)*)/
s = '/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789'
if m = s.match(re)
query_str = m[1] # now, you can 100% trust this string
query_str.scan(/(\w+)=(\w+)/) do |param,value| #grab parameter
printf("%s, %s\n", param, value)
end
end
output
secondparam, def456
firstparam, abc123
thirdparam, ghi789
This script will help you.
First, i check, is there any symbol like ?.
Then, i kill first part of line (left from ?).
Next, i split line by &, where each value splitted by =.
my $r = q"/parent/child
/parent/child?
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/
/parent/child/?
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789";
for my $string(split /\n/, $r){
if (index($string,'?')!=-1){
substr($string, 0, index($string,'?')+1,"");
#say "string = ".$string;
if (index($string,'=')!=-1){
my #params = map{$_ = [split /=/, $_];}split/\&/, $string;
$"="\n";
say "$_->[0] === $_->[1]" for (#params);
say "######next########";
}
else{
#print "there is no params!"
}
}
else{
#say "there is no params!";
}
}
What is the best way to ignore the white space in a target string when searching for matches using a regular expression pattern, but only if the whitespace comes after a newline (\n)? For example, if my search is for "cats", I would want "c\n ats" or "ca\n ts" to match but not "c ats" since the whitespace doesn't come after a newline. I can't strip out the whitespace beforehand because I need to find the begin and end index of the match (including any whitespace) in order to highlight that match and any whitespace needs to be there for formatting purposes.
If the regex engine you're using supports lookaround assertions, use a positive lookbehind assertion to check for the presence of a preceding newline:
(?<=\n)\s
"What is the best way to ignore the white space in a target string when searching for matches using a regular expression pattern"
I would construct a regex dynamically, inserting a (?:\n\s)? between each character.
use strict;
use warnings;
my $needed = 'cats';
my $regex = join '(?:\n\s)?' , split ( '',$needed );
print "\nRegex = $regex\n", '-'x40, "\n\n";
my $target = "
cats
c ats
c\n ats
ca ts
ca\n ts
cat s
cat\n s
";
while ( $target =~ /($regex)/g)
{
print "Found - '$1'\n\n";
}
The output:
Regex = c(?:\n\s)?a(?:\n\s)?t(?:\n\s)?s
----------------------------------------
Found - 'cats'
Found - 'c
ats'
Found - 'ca
ts'
Found - 'cat
s'
I have made a small ruby snippet based on the rules you have listed. Is this what you are looking for?
data = <<DATA
test1c\n atsOKexpected
test2ca\n tsOKexpected
test3catsOKexpected
test5ca tsBADexpected
test6 catsOKexpected
test7cats OKexpected
DATA
tests = data.split(/\n\n/)
regex = /c(\n )?a(\n )?t(\n )?s/
tests.each do |s|
if s =~ regex
puts "OK\n#{s}\n\n"
else
puts "BAD\n#{s}\n\n"
end
end
# RESULTS
# OK
# test1c
# atsOKexpected
#
# OK
# test2ca
# tsOKexpected
#
# OK
# test3catsOKexpected
#
# BAD
# test5ca tsBADexpected
#
# OK
# test6 catsOKexpected
#
# OK
# test7cats OKexpected