I am trying to search a keyword/pattern match in a file, where the lines will be starting with date.
Line will be like below
11/02/15 02:28:49%%PROGRAM$$SUCCESS$$End.
So i tried with below command,
Select-String -Path C:\Path\To\File.txt -Pattern $(Get-Date -format d) | Select-String -Pattern SUCCESS
So that i can get lines which contain SUCCESS with a starting of current date.
Its working on my test box and when i tried the same on a big file (~200 MB), its not giving any results. Tried below too,
Get-Content -Path C:\Path\To\File.txt | Select-String -Pattern $(Get-Date -format d) | Select-String -Pattern SUCCESS
Any help any help would be greatly appreciated!
Some things to consider here. As PetSerAl brings to light, Get-Date -Format d depends on the culture, so you need to be careful about relying on the output of that.
If the files you're searching are generated using Get-Date -Format d then it makes sense to do the search that way as long as the files will always be searched on a machine with the same culture they were generated with.
By the way on my machine it's 11/2/2015 not 11/02/15 and I am in the US.
Also, when you use Select-String -Pattern it's a regular expression, so you need to make sure that there are no special characters in the string. In the case of PetSerAl's date, the dots . would be interpreted as special characters. To avoid that use [RegEx]::Escape().
Select-String returns a match object (or objects), so piping it directly into another Select-String may not work. Consider making a single pattern out of it.
Just a guess here, but it kind of seems like the pattern you want is to match the current date string at the beginning of the line and then find SUCCESS anywhere after that in the line.
I think for that you could use a pattern like this: 11/02/15.+?SUCCESS
So code like this:
Get-Content -Path C:\Path\To\File.txt | Select-String -Pattern "$([RegEx]::Escape((Get-Date -Format d))).+?SUCCESS"
Would do the trick I think, again assuming culture issues don't mess you up.
Related
I am trying to read lines in a file and search for a pattern that spans two lines. Looking at the file in notepad++ I see a LF char in the file.
Example log.txt:
I want to find this
value here: OK
My simple code does not work and returns nothing:
select-string -Path "log.txt" -Pattern "find this\n*value here: OK"
I have tried many combos of various things here including .+ and \r that I found posted on various threads. I can get the first line by using:
select-string -Path "log.txt" -Pattern "find this\n*"
Result of above is: I want to find this
Adding anything more to the line above results in nothing being returned. Any ideas how to do this using select-string? I was trying to avoid using get content due to the potential size of the files I am working with.
So I think I understand your question. If you have a file that has a line that you want to key off of then the next line is the line that you want to look at:
(Select-String -Path "Log.txt" -Pattern "find this" -Context 1).Context.PostContext
I wasn't sure if that carriage return was an artifact of your formatting or not. If it is not then this would work better:
(Select-String -Path "Log.txt" -Pattern "find this" -Context 2).Context.PostContext[1]
Here is a way to do it if you don't know how many lines will be between the two bits:
$file = Get-Content 'Log.txt' -Raw
$file -match '(?smi)I want to find this.*(value here: OK)'
$matches[1]
Since you might want a multi line regex solution you need to read in the text file as one string.
Using the test file:
Stfuf
Bagel
I want to find this
value here: OK
Things
I was able to get the result using a simple matching pattern that satisfies your example text.
(Get-Content -Raw c:\temp\test.txt | Select-String -Pattern "([\w ]+)\s*value here: OK").Matches.Groups[1].Value
regex101.com
Basically gets the text preceding invariant spaces, including newlines and the static text "value here: OK". Could be made better with positive look aheads but this seems to work fine.
I have a text file that could contain up to 1000 lines of data in the following format:
14410:3012669|EU14410|20/01/2017||||1|6|4|OUT FROM UNDER||22/02/2017 04:01:47|22/02/2017 21:19:52
14:3012670|EU016271751|20/01/2017||||2|6|4|BLOCK BET|\\acis-prod\Pictures\Entry\EU01627.jpg|22/02/2017 04:02:02|22/02/2017 21:19:52
301111:3012671|EU016275|20/01/2017||||2|6|4|VITAE MEDICAL CLINIC|\\tm-prod\Pictures\Entry\EU01.jpg|22/02/2017 04:02:11|22/02/2017 21:19:53
each line will start with the following format
"set of characters up to max of 8":"set of characters unlimited max"
I want to search the characters ONLY up until the first colon. Those characters could contain any amount up to a maximum of 8. (hopefully shown well in my examples above) I'm trying to search those first characters, up to the ":" of each line to see if it contains a string, and return the whole line. still new to powershell so I've only tried a simple select:
$path = "C:\Users\ME\Desktop\acsep22\acsnic-20170222_233324.done"
Get-ChildItem $path -recurse | Select-String -pattern ("14410","3011981","3011982",) | out-file $logfile |format-list
which works - but I didn't take into account that the string could also appear twice in the same line ( though unrelated to the first 7 characters)
for example:
14410:3012669|EU14410|
contains 14410 twice, they're unrelated in terms of their significance and I only want to search and return based on the first number
could somebody help me achieve this or could some one point me toward the cmdlet that would help?
I've tried various searches online (and via the Microsoft online resource) but a lot of results are more to do with "return the first X amount of characters" rather than "search using only the first X amount and return line"
Kind Regards
You could use a simple Where-Object filter to check whether the string before the : is one of the strings you expect:
$strings = '14410','3011981','3011982'
Get-Content $path |Where-Object {$strings -contains ($_ -split ':')[0]}
This is probably the most PowerShell-idiomatic approach.
If you want to use Select-String, you'll have to construct a regex pattern that will match on strings that start with one of the strings and then a colon:
$strings = '14410','3011981','3011982'
$pattern = '^(?:{0}):' -f ($strings -join '|') # pattern is now '^(?:14410|3011981|3011982):'
Select-String -Path $path -Pattern $pattern
If you just want the bare string itself from the output, grab the Line property from the objects returned by Select-String:
Select-String -Path $path -Pattern $pattern |Select-Object -Expand Line
or
Select-String -Path $path -Pattern $pattern |ForEach-Object Line
The pattern above uses a non-capturing group (?:pattern-goes-here) to match any one of the strings, at the start ^ of a string, followed by :.
Both solutions will work with an arbitrary number of strings
I've looked all around this site and can't quite seem to find anything that fits my situation. Basically, I am trying to write an addition to the NETLOGON file that will replace text in a text file on all of our users' desktops. The current text is static across the board.
The text I want it changed to will be unique to each user. I want to change the current text (user1) to the users AD username (i.e. johnd, janed, etc.). I am using Windows Server 2008 R2 and all the workstations are Windows 7 Professional SP1 64 bit.
Here's what I have tried so far (with a few variables, which none have worked for one reason or the other):
gc c:\Users\%USERNAME%\desktop\VPN.txt' -replace "user1",$env:username | out-file c:\Users\%USERNAME%\desktop\VPN.txt
I didn't get an error, but it also did not go back to the normal "PS C:>" prompt, just ">>>" and the file did not change as anticipated.
If that is how you have the code exactly then I suppose it is because you have an opening single quote without a closing one. You are still going to have two other problems and you have one answer in your code. The >>> is the line continuation characters because the parser knows that the code is not complete and giving you the option to continue with the code. If you were purposely coding a single line on multiple lines you would consider this a feature.
$path = "c:\Users\$($env:username)\desktop\VPN.txt"
(Get-Content $path) -replace "user1",$env:username | out-file $path
Closed the path in quotes and used a variable since you called the path twice.
%name% is used in command prompt. Environment variables in PowerShell use the $env: provider which you did you once in your snippet.
-replace is a regex replaced tool that can work against Get-Content but you need to capture the result in a sub expression first.
Secondly with -replace is for regex and your string is not regex based you could just use .Replace() as well.
Set-Content is generally preferred over Out-File for performance reasons.
All that being said...
you could also try something like this.
$path = "c:\Users\$($env:username)\desktop\VPN.txt"
(Get-Content $path).Replace("user1",$env:username) | Set-Content $path
Do you want to only replace the first occurrence?
You could use a little regex here with a tweak in how you get the use Get-Content
$path = "c:\Users\$($env:username)\desktop\VPN.txt"
(Get-Content $path | Out-String) -replace "(.*?)user1(.*)",('$1{0}$2' -f $env:username) | out-file $path
Regex will match the entire file. There are two groups which it captures.
(.*?) - Up until the first "user1"
(.*) - Everything after that
Then we use the format operator to sandwich the new username in between those capture groups.
Use:
(Get-Content $fileName) | % {
if ($_.ReadCount -eq 1) {
$_ -replace "$original", "$content"
}
else {
$_
}
} | Set-Content $fileName
I need to do something similar to Unix's ls | grep 'my[rR]egexp?' in Powershell. The similar expression ls | Select-String -Pattern 'my[rR]egexp?' seems to go through contents of the listed files, rather than simply filtering the filenames themselves.
The Select-String documentation hasn't been of much help either.
Very simple:
ls | where Name -match 'myregex'
There are other options, though:
(ls) -match 'myregex'
Or, depending on how complex your regex is, you could maybe also solve it with a simple wildcard match:
ls wild[ck]ard*.txt
which is faster than above options. And if you can get it into a wildcard match without character classes you can also just use the -Filter parameter to Get-ChildItem (ls), which performs filtering on the file system level and thus is even faster. Note also that PowerShell is case-insensitive by default, so a character class like [rR] is unnecessary.
While researching based on #Joey's answer, I stumbled upon another way to achieve the same (based on Select-String itself):
ls -Name | Select-String -Pattern 'my[Rr]egexp?'
The -Name argument seems to make ls return the result as a plain string rather than FileInfo object, so Select-String treats it as the string to be searched in rather than a list of files to be searched.
I have the text of a couple hundred Word documents saved into individual .txt files in a folder. I am having an issue where a MergeField in the Word document wasn't formatted correctly, and now I need to find all the instances in the folder where the incorrect formatting occurs. the incorrect formatting is the string \#,$##,##0.00\* So, I'm trying to use PowerShell as follows:
select-string -path MY_PATH\.*txt -pattern '\#,$##,##0.00\*'
select-string -path MY_PATH\.*txt -pattern "\#`,`$##`,##0.00\*"
But neither of those commands finds any results, even though I'm sure the string exists in at least one file. I feel like the error is occurring because there are special characters in the parameter (specifically $ and ,) that I'm not escaping correctly, but I'm not sure how else to format the pattern. Any suggestions?
If you are actually looking for \#,$##,##0.00\* then you need to be aware that Select-String uses regex and you have a lot of control characters in there. Your string should be
\\\#,\$\#\#,\#\#0\.00\\\*
Or you can use the static method Escape of regex to do the dirty work for you.
[regex]::Escape("\#,$##,##0.00\*")
To put this all together you would get the following:
select-string -path MY_PATH\.*txt -pattern ([regex]::Escape("\#,$##,##0.00\*"))
Or even simpler would be to use the parameter -SimpleMatch since it does not interpet the string .. just searches as is. More here
select-string -path MY_PATH\.*txt -SimpleMatch "\#,$##,##0.00\*"
My try, similar to Matts:
select-string -path .\*.txt -pattern '\\#,\$##,##0\.00\\\*'
result:
test.txt:1:\#,$##,##0.00\*