I have text that prints out like this:
mdbAppText_Arr: [0]: The cover is open. {goes to next line here}
Please close the cover. and [1] Backprinter cover open
46
I tried getting rid of the newline after open., and it's still there. Any idea of a better way or fix for what I'm doing? I need to get rid of the newline because it's going to a csv file, and messing up formatting (going to newline there).
This is my code:
$mdbAppText_Arr = $mdbAppText.Split("|")
$mdbAppText_Arr[0].replace("`r",";").replace("`n",";").replace("`t",";").replace("&",";")
#replace newline/carriage return/tab with semicolon
if($alarmIdDef -eq "12-7")
{
Write-Host "mdbAppText_Arr: [0]: $($mdbAppText_Arr[0]) and [1] $($mdbAppText_Arr[1]) "
[byte] $mdbAppText_Arr[0][31]
}
I've been looking at:
replace
replace - this one has a link reference to lookup in the asci table, but it's unclear to me what column the byte equivalent is in the table/link.
I'm using PowerShell 5.1.
-replace is a regex operator, so you need to supply a valid regular expression pattern as the right-hand side operand.
You can replace most newline sequences with a pattern describing a substring consisting of:
an optional carriage return (\r? in regex), followed by
a (non-optional) newline character (\n in regex):
$mdbAppText_Arr = $mdbAppText_Arr -replace '\r?\n'
Related
I am trying to remove all lines of text after a single line of text "[info]" Here is an example:
Top=1266
[info]
name=tod
space=456
number=221,441,111,0
[version]
version=1
I only need the top, the other text will be replaced later on in the script. Here is all that I have tried
$Content -replace '\[Info\]*',''
Only removes the Info line and not anything past that. I have tried to loop, but I can't seem to find the line with a where object search.
What is a quick and easy way to remove all lines of code after a single line of set text?
To make the -replace operator treat it as one string, add (?s) to the pattern.
$Content -replace '(?s)\[Info\].*'
You also needed to match any character so .* works in this case. The second part is optional. Since you're replacing it with nothing you can simply omit it.
Read more about regular expression in powershell
https://learn.microsoft.com/en-us/powershell/module/microsoft.powershell.core/about/about_regular_expressions?view=powershell-7.1
and operators
https://learn.microsoft.com/en-us/powershell/module/microsoft.powershell.core/about/about_operators?view=powershell-7.1
I'm using the .Replace() function to replace line feeds in the file I'm working on with a carriage return and a line feed but I would also like to match any number of spaces preceding the line feed. Can this be done in the same operation using a regular expression?
I've tried various combinations of "\s +*" but none have worked, except with a fixed number of manually typed spaces.
This version works for the one space case:
.Replace(" `n","`r`n")
For example, a file like this:
...end of line one\n
...end of line two \n
would look like:
...end of line one\r\n
...end of line two\r\n
The .Replace() method of the .NET [string] type performs literal string replacements.
By contrast, PowerShell's -replace operator is based on regexes (regular expressions), so it allows you to match a variable number of spaces (including none) with *:
"...end of line two `n" -replace ' *\n', "`r`n"
Get-Content $user| Foreach-Object{
$user = $_.Split('=')
New-Variable -Name $user[0] -Value $user[1]}
Im trying to work on a script and have it split a text file into an array, splitting the file based on each new line
What should I change the "=" sign to
It depends on the exact encoding of the textfile, but [Environment]::NewLine usually does the trick.
"This is `r`na string.".Split([Environment]::NewLine)
Output:
This is
a string.
The problem with the String.Split method is that it splits on each character in the given string. Hence, if the text file has CRLF line separators, you will get empty elements.
Better solution, using the -Split operator.
"This is `r`na string." -Split "`r`n" #[Environment]::NewLine, if you prefer
You can use the String.Split method to split on CRLF and not end up with the empty elements by using the Split(String[], StringSplitOptions) method overload.
There are a couple different ways you can use this method to do it.
Option 1
$input.Split([string[]]"`r`n", [StringSplitOptions]::None)
This will split on the combined CRLF (Carriage Return and Line Feed) string represented by `r`n. The [StringSplitOptions]::None option will allow the Split method to return empty elements in the array, but there should not be any if all the lines end with a CRLF.
Option 2
$input.Split([Environment]::NewLine, [StringSplitOptions]::RemoveEmptyEntries)
This will split on either a Carriage Return or a Line Feed. So the array will end up with empty elements interspersed with the actual strings. The [StringSplitOptions]::RemoveEmptyEntries option instructs the Split method to not include empty elements.
The answers given so far consider only Windows as the running environment. If your script needs to run in a variety of environments (Linux, Mac and Windows), consider using the following snippet:
$lines = $input.Split(
#("`r`n", "`r", "`n"),
[StringSplitOptions]::None)
There is a simple and unusual way to do this.
$lines = [string[]]$input
This will split $input like:
$input.Split(#("`r`n", "`n"))
This is undocumented at least in docs for Conversions.
Beware, this will not remove empty entries.
And it doesn't work for Carriage Return (\r) line ending at least on Windows.
Experimented in Powershell 7.2.
This article also explains a lot about how it works with carriage return and line ends. https://virot.eu/powershell-and-newlines/
having some issues with additional empty lines and such i found the solution to understanding the issue. Excerpt from virot.eu:
So what makes up a new line. Here comes the tricky part, it depends.
To understand this we need to go to the line feed the character.
Line feed is the ASCII character 10. It in most programming languages
escaped by writing \n, but in powershell it is `n. But Windows is not
content with just one character, Windows also uses carriage return
which is ASCII character 13. Escaped \r. So what is the difference?
Line feed advances the pointer down one row and carriage return
returns it to the left side again. If you store a file in Windows by
default are linebreaks are stored as first a carriage return and then
a line feed (\r\n). When we aren’t using any parameters for the
split() command it will split on all white-space characters, that is
both carriage return, linefeed, tabs and a few more. This is why we
are getting 5 results when there is both carriage return and line
feeds.
I have some trailing characters at the end of a string peregrinevwap^_^_
print "JH 4 - app: $application \n";
app: peregrinevwap^_^_
Do you know why they are there and how I can remove them. I tried the chomp command but this hasn't worked.
Check out the tr//cd operator to get rid of unwanted characters.
It's documented in "perldoc perlop"
$application =~ tr/a-zA-Z//cd;
Will remove everything except letters from the string and
$application =~ tr/^_//d;
Will remove all "^" and "_" characters.
If you only want to remove certain characters when they at the end of the string, use the s// search/replace operator with regular expressions and the $ anchor to match the end of the string.
Here's an example:
s/[\^_]*$//;
Let's hope the underscores do not occur at the end of your strings, otherwise you can't automatically separate them from these unwanted characters.
Are you sure these characters are actually ^ and _ characters?
^_ could also indicate Ctrl-Underscore, ASCII character 0x1F (Unit Separator). (Not a character I've ever seen used, but you never know.)
If this is in fact the case, you can remove them with something like:
$application =~ s/\x1F//g;
Here is a piece of code
while($l=~/(\\\s*)$/) {
statements;
}
$l contains a line of text taken form file, in effect this code is for go through lines in file.
Questions:
I don't clearly understand what the condition in while is doing. I think it is trying to match group of \ followed by some number of white spaces at the end of line and loop should stop whenever a line ends with \ and may be some white spaces. I am not sure of it.
I came across statement $a ~= s/^(.*$)/$1/ . What I understand that ^ will force matching at the beginning of string, but in (.*$) would mean match all the characters at the end of string . Dose it mean that the statement is trying to find if any group of character at the end is same as group of character in the beginning of text ?
It is interesting to note that this statement:
while ( $l =~ /(\\\s*)$/ ) {
Is an infinite loop unless $l is altered inside the loop so that the regex no longer matches. As has already been mentioned by others, this is what it matches:
( ... ) a capture group, captures string to $1 (that's the number one, not lower case L)
\\ matches a literal backslash
\s* matches 0 or more whitespace characters.
$ matches end of line with optional newline.
Since you do not have the /g modifier, this regex will not iterate through matches, it will simply check if there is a match, resetting the regex each iteration, thereby causing an endless loop.
The statement
$a ~= s/^(.*$)/$1/
Looks rather pointless. It captures a string of characters up until end of string, then replaces it with itself. The captured text is stored in $1 and is simply replaced. The only marginally useful thing about this regex is that:
It matches up until newline \n, and nothing further, which may be of some use to a parser. A period . matches any character except newline, unless the /s modifier is present on the regex.
It captures the line in $1 for future use. However, a simple /^(.*$)/ would do the same.
1. the while
Usually while (regex) is used with the /g modifier, otherwise, if it matches, you get an infinite loop (unless you exit the loop, like using last).
statements would be executed continuously in an infinite loop.
In your case, adding the g
while($l=~/(\\\s*)$/g)
will have the while make only one loop, due to the $ - making a match unique (whatever matches up to the end of string is unique, as $ marks the end, and there is nothing after...).
2. $a ~= s/^(.*$)/$1/
This is a substitution. If the string ^.*$ matches (and it will, since ^.*$ matches (almost, see comment) anything) it is replaced with... $1 or what's inside the (), ie itself, since the match occurs from 1st char to the end of string
^ means beginning of string
(.*) means all chars
$ end of string
so that will replace $a with itself - probably not what you want.
it matches a literal backslash followed by 0 or more spaces followed by the end of the line.
it executes statements for all the lines in that text file that contain a \, followed by zero or more spaces ( \s* ), at the end of the line ($).
It matches lines that end with a backslash character, ignoring any trailing whitespace characters.
Ending a line with a backslash is used in some languages and data files to indicate that the line is being continued on the next line. So I suspect this is part of a parser that merges these continuation lines.
If you enter a regular expression at RegExr and hover your mouse over the pieces, it displays the meaning of each piece in a tooltip.
(\\\s*)$ this regex means --- a \ followed by zero or more number of white space characters which is followed by end of the line. Since you have your regex in (...), you can extract what you matched using $1, if you need.
http://rubular.com/r/dtHtEPh5DX
EDIT -- based on your update
$a ~= s/^(.$)/$1/ --- this is search and replace. So your regex matches a line which contains exactly one character (since you use . http://www.regular-expressions.info/dot.html), except a new-line character. Since you use (...), the character which matched the regex is extracted and stored in variable a
EDIT -- you changed your regex so here is the updated answer
$a ~= s/^(.*$)/$1/ -- same as above except now it matches zero or more characters (except new-line)