From the example, the name of the processor is returned but with a space at the end, how can I make it return a value without a space in the end?
[MeasureRun]
Measure=Plugin
Plugin=RunCommand
Parameter=wmic cpu get Name
OutputType=ANSI
RegExpSubstitute=1
Substitute="Name.*#CRLF#":"","#CRLF#":""
ClipString=1
IfCondition=1
IfTrueAction=[!CommandMeasure MeasureRun "Run"]
[MeterResult]
Meter=String
MeasureName=MeasureRun
FontSize=14
FontColor=255,255,255,255
AntiAlias=1
Text=%1!
I think this can be done through regular expression, but I'm not strong at it.
Just add another Substitute in the substitution list to replace the occurence of one or more whitespaces from the end of the string (the pattern "\s+$") with an empty string like below:
Substitute="Name.*#CRLF#":"","#CRLF#":"","\s+$":""
Related
I have this regex that's worked well so far that splits 'name=value' pairs separated by a given character.
(?s)([^\s=]+)=(.*?)(?=\s+[^\s=]+=|\Z)
I know the separator, but the problem is in the example below (tab separated):
usrName=Wilma sev=4 cat=Detection CommandLine="C:\powershell.exe" -Enc 0ATQBpAG0AAcABDAHIAZQBkAHMAIgA= IOCValue= ProcessEndTime=2023-01-18 15:51:05
https://regex101.com/r/1wgVxs/5
Some values can have no value in the case of 'IOCValue' which works as expected, however some values like the CommandLine are giving me up to -Enc as one match and the remainder to the next pair as another.
What I'm hoping to get out from the above is:
usrName=Wilma
sev=4
cat=Detection
CommandLine="C:\powershell.exe" -Enc 0ATQBpAG0AAcABDAHIAZQBkAHMAIgA=
IOCValue=
ProcessEndTime=2023-01-18 15:51:05
But I'm getting:
usrName=Wilma
sev=4
cat=Detection
CommandLine="C:\powershell.exe" -Enc
0ATQBpAG0AAcABDAHIAZQBkAHMAIgA=
IOCValue=
ProcessEndTime=2023-01-18 15:51:05
Given I know the separator is a tab I think what I need is to only look for name=value pairs when they are at the start of the line or proceeded by the separator (tab). Is this possible?
Note, I can expect a space separator too, but I have a less performant and messy non-regex version I can send these too, so presume tab.
You may use this simplified regex:
(?s)([^\s=]+)=(.*?)(?=\t|\Z)
Updated RegEx Demo
Here, lookahead (?=\t|\Z) will make sure that value part is followed by either a tab character or end position.
Here is a sample string.
"BLAH, blah, going to the store &^5, light Version 12.7(2)L6, anyway
plus other stuff Version 3.3.4.6. Then goes on an on for several lines..."
I want to capture only the first version number without including the word version if possible but not include the periods and parenthesis. The result would stop when it encounters a comma. The result would be:
"1272L6"
I don't want it to include other instances of version in the text. Can this be done?
I've tried (?<=version)[^,]* I know it does not address removing the periods and parens and does not address the subsequent versions.
This exact RegEx, maybe not the best solution, but it might help you to get 1272L6:
([0-9]{2})\.([0-9]{1})\(([0-9]{1})\)([A-Z]{1}[0-9]{1})
It creates four groups (where $1$2$3$4 is your target 1272L6) and passes ., ) and (.
You might change {1} to other numbers of repetitions, such as {1,2}.
Assuming the version number is fixed on format but not on the specific digits or letters, you could do this.
String s = "this is a test 12.7(2)L6, 13.7(2)L6, 14.7(2)L6";
String reg = "(\\d\\d\\.\\d\\(\\d\\)[A-Z]\\d),";
Matcher m = Pattern.compile(reg).matcher(s);
if (m.find()) { // should only find first one
System.out.println(m.group(1).replaceAll("[.()]", ""));
}
I made a regex which inserts a space where ever there is any of the characters
-:\*_/;, present for example JET*AIRWAYS\INDIA/858701/IDBI 05/05/05;05:05:05 a/c should beJET* AIRWAYS\ INDIA/ 858701/ IDBI 05/05/05; 05:05:05 a/c
The regex I used is (?!a\/c|w\/d|m\/s|s\/w|m\/o)(\D-|\D:|\D\*|\D_|\D\\|\D\/|\D\;)
I have added some words exceptions like a/c w/d etc. \D conditions given to avoid date/time values getting separated, but this created an issue, the numbers followed by the above mentioned characters never get split.
My requirement is
1. Insert a space after characters -:\*_/;,
2. but date and time should not get split which may have / :
3. need exception on words like a/c w/d
The following is the full code
Private Function formatColon(oldString As String) As String
Dim reg As New RegExp: reg.Global = True: reg.Pattern = "(?!a\/c|w\/d|m\/s|s\/w|m\/o)(\D-|\D:|\D\*|\D_|\D\\|\D\/|\D\;)" '"(\D:|\D/|\D-|^w/d)"
Dim newString As String: newString = reg.Replace(oldString, "$1 ")
formatColon = XtraspaceKill(newString)
End Function
I would use 3 replacements.
Replace all date and time special characters with a special macro that should never be found in your text, e.g. for 05/15/2018 4:06 PM, something based on your name:
05MANUMOHANSLASH15MANUMOHANSLASH2018 4MANUMOHANCOLON06 PM
You can encode exceptions too, like this:
aMANUMOHANSLASHc
Now run your original regex to replace all special characters.
Finally, unreplace the macros MANUMOHANSLASH and MANUMOHANCOLON.
Meanwhile, let me tell you why this is complicated in a single regex.
If trying to do this in a single regex, you have to ask, for each / or :, "Am I a part of a date or time?"
To answer that, you need to use lookahead and lookbehind assertions, the latter of which Microsoft has finally added support for.
But given a /, you don't know if you're between the first and second, or second and third parts of the date. Similar for time.
The number of cases you need to consider will render your regex unmaintainably complex.
So please just use a few separate replacements :-)
Okay so I wanted a regex to parse uncontracted(if that's what it is called) ipv6 adresses
Example ipv6 adress: 1050:::600:5:1000::
What I want returned: 1050:0000:0000:600:5:1000:0000:0000
My try at this:
ip:gsub("%:([^0-9a-zA-Z])", ":0000")
The first problem with this: It replaces the first and second :
So :: gets replaced with :0000
Replacing it with :0000: wouldn't work because then it will end with a :. Also this would note parse the newly added : resulting in: 1050:0000::600:5:1000:0000:
So what would I need this regex to do?
Replace every : by :0000 if it isn't followed by a number or letter
Main problem: :: gets replaced instead of 1 :
gsub and other functions from Lua's string library use Lua Patterns which are much simpler than regex. Using the pattern more than once will handle the cases where the pattern overlaps the replacement text. The pattern only needs to be applied twice since the first time will catch even pairings and the second will catch the odd/new pairings of colons. The trailing and leading colons can be handled separately with their own patterns.
ip = "1050:::600:5:1000::"
ip = ip:gsub("^:", "0000:"):gsub(":$", ":0000")
ip = ip:gsub("::", ":0000:"):gsub("::", ":0000:")
print(ip) -- 1050:0000:0000:600:5:1000:0000:0000
There is no single statement pattern to do this but you can use a function to do this for any possible input:
function fill_ip(s)
local ans = {}
for s in (s..':'):gmatch('(%x*):') do
if s == '' then s = '0000' end
ans[ #ans+1 ] = s
end
return table.concat(ans,':')
end
--examples:
print(fill_ip('1050:::600:5:1000::'))
print(fill_ip(':1050:::600:5:1000:'))
print(fill_ip('1050::::600:5:1000:1'))
print(fill_ip(':::::::'))
My regex is (?<![\u0410-\u042F])[.!?](?=(\s)?(\s)?[\u0410-\u042F]|[\u04E8]|["]|[\u201C]|![0-9])
I want to split a paragraph into sentences.
I do the regex with re.split() and I print the array
This is a sample input I did:
Мамлекеттик айыптоочу Биринчи май райондук сотуна берген бул сунушун диний кастыкты ырбатпоо аракети менен негиздеди. Мусулмандарга акаарат келтирип жатат деген кайрылуу каттын негизинде УКМК Тезекбаевге каршы кылмыш ишин козгоп, сотко өткөргөн. Бул ишти бүгүн Биринчи май райондук соту карап бүттү жана өкүм эртең чыгарыларын маалымдады. Тараптар мунаса тапты;
Ишти карап жаткан мамлекеттик айыптоочу Кудайберди Чаргынов Кубанычбек Тезекбаевдин диний кастыкты козутууга болгон аракети толугу менен далилденгенин билдирүүдө. Бирок мамлекеттик айыптоочу диний кастыкты ырбатпоо максатында Кыргызстандын Кылмыш кодексинин 65-беренесине ылайык, иш өз маанисин жоготконуна байланыштуу кылмыш ишин Т.У. кыскартып салууну сунуштады.
It prints out fine except that the last character (in this case a period, ?, or !) gets removed!
I searched online and it says to surround the punctuation with lookahead tags, but it doesn't work.
I'm using Python 3.
Put a capturing group around the character(s) you want to preserve in the split:
(?<![\u0410-\u042F])([.!?])(?=(\s)?(\s)?[\u0410-\u042F]|[\u04E8]|["]|[\u201C]|![0-9])
The periods will be added as new elements in the resulting list. From the documentation:
If capturing parentheses are used in pattern, then the text of all groups in the pattern are also returned as part of the resulting list.
If you don't want this, you'll have to split on the space itself, by ensuring that the space is preceded by a matching period of other punctuation marks (using a look-behind assertion).