Perl's q function or single quote is supposed to return the string literal as typed (except \'). But it doesn't work as expected for the following scenario.
I want to print the following UNC path
\\dir1\dir2\dir3
So I have used
my $path = q(\\dir1\dir2\dir3);
OR
my $path = '\\dir1\dir2\dir3';
But this skips one backslash at the front.
So if I print it i.e. print $path; it prints
\dir1\dir2\dir3
I want to know why? I have to type 3 or 4 backslashes at the beginning of the UNC path to make it work as expected. What am I missing?
From perldoc perlop:
q/STRING/
'STRING'
A single-quoted, literal string. A backslash represents a backslash unless followed by the delimiter or another backslash, in which case the delimiter or backslash is interpolated.
Change:
my $path = q(\\dir1\dir2\dir3);
to:
my $path = q(\\\dir1\dir2\dir3);
As for why, it's because Perl lets you include the quote delimiter in your string by escaping it with a backslash:
my $single_quote = 'This is a single quote: \'';
But if a backslash before the delimiter always escaped the delimiter, there would be no way to end a string with a backslash:
my $backslash = 'This is a backslash: \'; # nope
Allowing backslashes to be escaped too takes care of that:
my $backslash = 'This is a backslash: \\';
Interestingly enough, there is only one way to type in double backslashes in a perl string without it being interpolated as a single backslash.
As all the other answers showed, any of the quote operators treat backslashes as a backslash unless there is another one following it directly.
The only way to get the double backslashes to display exactly as you have typed them is to use a single quote here-doc.
my $path = <<'VISTA';
\\dir1\dir2\dir3
VISTA
chomp $path;
print $path."\n";
Would print it exactly as you've typed it in.
Related
I have this folder in my OneDrive and Just wondering how should I pass the path. Right now I'm getting "Folder not Found" when I try to do it like this.
$ServerRelativeUrl= "Documents/PC-OFFBOARD_USMPGWNCAT15C61-GZJY8T-01-Dec-0257/C$/'$Windows.~WS'"
With double-quoted " strings, you must escape characters with special meanings if you want them to be processed literally. PowerShell's escape character is the backtick `. The dollar-symbol $ must be prefixed with a backtick like this to be part of a literal file path:
"Documents/PC-OFFBOARD_USMPGWNCAT15C61-GZJY8T-01-Dec-0257/C`$/'`$Windows.~WS'"
Alternatively, you can use a single-quoted ' string instead, making sure to escape the literal single-quotes with two single-quotes '' (backticks won't escape in a literal string):
'Documents/PC-OFFBOARD_USMPGWNCAT15C61-GZJY8T-01-Dec-0257/C$/''$Windows.~WS'''
This loses your ability to insert actual intended variables though. You can however rely on the format operator in this case. To insert the literal string '$Windows.~WS' into the path, for example:
$folderName = '''$Windows.~WS'''
$fullPath = 'Documents/PC-OFFBOARD_USMPGWNCAT15C61-GZJY8T-01-Dec-0257/C$/{0}' -f $folderName
The following example prints "SAME":
if (q/\\a/ eq q/\a/) {
print "SAME\n";
}
else {
print "DIFFERENT\n";
}
I understand this is consistent with the documentation. But I think this behavior is undesirable. Is there a need to escape a backlash lilteral in single-quoted string? If I wanted 2 backlashes, I'd need to specify 4; this does not seem convenient.
Shouldn't Perl detect whether a backslash serves as an escape character or not? For instance, when a backslash does not precede a delimiter, it should be treated as a literal; and if that were the case, I wouldn't need 3 backslashes to express two, e.g.,
q<a\\b>
instead of
q<a\\\b>.
Is there a need to escape a backlash in single-quoted string?
Yes, if the backslash is followed by another backslash, or is the last character in the string:
$ perl -e'print q/C:\/'
Can't find string terminator "/" anywhere before EOF at -e line 1.
$ perl -e'print q/C:\\/'
C:\
This makes it possible to include any character in a single-quoted string, including the delimiter and the escape character.
If I wanted 2 backlashes, I'd need to specify 4; this does not seem convenient.
Actually, you only need three (because the second backslash isn't followed by another backslash). But as an alternative, if your string contains a lot of backslashes you can use a single-quoted heredoc, which requires no escaping:
my $path = <<'END';
C:\a\very\long\path
END
chomp $path;
print $path; # C:\a\very\long\path
Note that the heredoc adds a newline to the end, which you can remove with chomp.
In single-quoted string literals,
A backslash represents a backslash unless followed by the delimiter or another backslash, in which case the delimiter or backslash is interpolated.
In other words,
You must escape delimiters.
You must escape \ that are followed by \ or the delimiter.
You may escape \ that aren't followed by \ or the delimiter.
So,
q/\// ⇒ /
q/\\\\a/ ⇒ \\a
q/\\\a/ ⇒ \\a
q/\\a/ ⇒ \a
q/\a/ ⇒ \a
Is there a need to escape a backlash in single-quoted string?
Yes, if it's followed by another backslash or the delimiter.
If I wanted 2 backlashes, I'd need to specify 4
Three would suffice.
this does not seem convenient.
It's more convenient than double-quoted strings, where backslashes must always be escaped. Single-quoted string require the minimum amount of escaping possible without losing the ability to produce the delimiter.
I am parsing a text doc and replacing some text. Lines of text without the "\" seem to be found and replaced no issues.
By the way this is to be done in Perl
I have a string like below:
Path=S:\2014 March\Test Scenarios\load\2014 March
that contains "\" that slash is an issue. I am using a simple search and replace line of code
$nExit =~ s/$sMatchPattern/$sFullReplacementString/;
How should I do it?
I suspect that you're trying to match a literal string, and therefore need to escape regex special characters.
You can use quotemeta or the escape codes \Q ... \E to do that:
$nExit = s/\Q$sMatchPattern/$sFullReplacementString/;
The above variable $sMatchPattern will be interpolated, but then any special characters will be escaped before the regex is compiled. Therefore the value of $sMatchPattern will be treated like a literal string.
Is this string inputed, or is it embedded in your program. You could do this to get rid of the backslash character:
my $path = "S:/2014 March/Test Scenarios/load/2014 March";
By the way, it's best not to have spaces in file and path names. They can be a bit problematic in certain situations. If you can't eliminate them, it's understandable.
Two things you should look at:
Use quotemeta which can help quote special characters in strings and allow you to use them in substitutions. Even if you had backslashes in your strings, quotemeta will handle them.
You don't have to use / as separators in match and substitutions. Instead, you can substitute various other characters.
These are all the same:
$string =~ s/$regex/$replace/;
$string =~ s#$regex#$replace#;
$string =~ s|$regex|$replace|;
You can also use parentheses, square braces, or curly brackets:
$string =~ s($regex)($replace);
$string =~ s[$regex][$replace]; # Not really recommended because `[...]` is a common regex
$string =~ s{$regex}{$replace};
The advantage of these as regular expression quote-like characters is that they must be balanced, so if I had this:
my $string = "I have (parentheses) in my string";
my $regex = "(parentheses}";
my $replace = "{curly braces}";
$string = s($regex)($replace);
print "$string\n"; # Still works. This will be "I have {curly braces} in my string"
Even if my string contains these types of characters, as long as they're balanced, everything will still work.
For yours:
my $Path = 'S:\2014 March\Test Scenarios\load\2014 March';
$nExit = quotemeta $string; #Quotes all meta characters...
$nExit =~ s($sMatchPattern)($sFullReplacementString);
That should work for you.
if you want to have a \ in your replacement string or match string dont forget to put another backslash in front of the backslash you want, as its an operator...
$sFullReplacementString = "\\";
That would turn the string into a single \
I want to replace something with a path like C:\foo, so I:
s/hello/c:\foo
But that is invalid.
Do I need to escape some chars?
Two problems that I can see.
Your first problem is that your s/// replacement is not terminated:
s/hello/c:\foo # fatal syntax error: "Substitution replacement not terminated"
s/hello/c:\foo/ # syntactically okay
s!hello!c:\foo! # also okay, and more readable with backslashes (IMHO)
Your second problem, the one you asked about, is that the \f is taken as a form feed escape sequence (ASCII 0x0C), just as it would be in double quotes, which is not what you want.
You may either escape the backslash, or let variable interpolation "hide" the problem:
s!hello!c:\\foo! # This will do what you want. Note double backslash.
my $replacement = 'c:\foo' # N.B.: Using single quotes here, not double quotes
s!hello!$replacement!; # This also works
Take a look at the treatment of Quote and Quote-like Operators in perlop for more information.
If I understand what you're asking, then this might be something like what you're after:
$path = "hello/there";
$path =~ s/hello/c:\\foo/;
print "$path\n";
To answer your question, yes you do need to double the backslash because \f is an escape sequence for "form feed" in a Perl string.
The problem is that you are not escaping special characters:
s/hello/c:\\foo/;
would solve your problem. \ is a special character so you need to escape it. {}[]()^$.|*+?\ are meta (special) characterss which you need to escape.
Additional reference: http://perldoc.perl.org/perlretut.html
Is there some way to replace a string such as #or * or ? or & without needing to put a "\" before it?
Example:
perl -pe 'next if /^#/; s/\#d\&/new_value/ if /param5/' test
In this example I need to replace a #d& with new_value but the old value might contain any character, how do I escape only the characters that need to be escaped?
You have several problems:
You are using \b incorrectly
You are replacing code with shell variables
You need to quote metacharacters
From perldoc perlre
A word boundary ("\b") is a spot between two characters that has a "\w" on one side of it
Neither of the characters # or & are \w characters. So your match is guaranteed to fail. You may want to use something like s/(^|\s)\#d\&(\s|$)/${1}new text$2/
(^|\s) says to match either the start of the string (^)or a whitespace character (\s).
(\s|$) says to match either the end of the string ($) or a whitespace character (\s).
To solve the second problem, you should use %ENV.
To solve the third problem, you should use the \Q and \E escape sequences to escape the value in $ENV{a}.
Putting it all together we get:
#!/bin/bash
export a='#d&'
export b='new text'
echo 'param5 #d&' |
perl -pe 'next if /^#/; s/(^|\s)\Q$ENV{a}\E(\s|$)/$1$ENV{b}$2/ if /param5/'
Which prints
param5 new text
As discussed at perldoc perlre:
...Today it is more common to use the quotemeta() function or the "\Q" metaquoting
escape sequence to disable all metacharacters' special meanings like this:
/$unquoted\Q$quoted\E$unquoted/
Beware that if you put literal backslashes (those not inside interpolated variables) between "\Q" and "\E", double-quotish backslash interpolation may
lead to confusing results. If you need to use literal backslashes within "\Q...\E", consult "Gory details of parsing quoted constructs" in perlop.
You can also use a ' as the delimiter in the s/// operation to make everything be parsed literally:
my $text = '#';
$text =~ s'#'1';
print $text;
In your example, you can do (note the single quotes):
perl -pe 's/\b\Q#f&\E\b/new_value/g if m/param5/ and not /^ *#/'
The other answers have covered the question, now here's your meta-problem: Leaning Toothpick Syndrome. Its when the delimiter and escapes start to blur together:
s/\/foo\/bar\\/\/bar\/baz/
The solution is to use a different delimiter. You can use just about anything, but balanced braces work best. Most editors can parse them and you generally don't have to worry about escaping.
s{/foo/bar\\}{/bar/baz}
Here's your regex with braced delimiters.
s{\#d\&}{new_value}
Much easier on the eyeholes.
If you really want to avoid typing the \s, put your search string into a variable and then use that in your regex instead. You don't need quotemeta or \Q ... \E in that case. For example:
my $s = '#d&';
s/$s/new_value/g;
If you must use this in a one-liner, bear in mind that you will have to escape the $s if you use "s to contain your perl code, or escape the 's if you use 's to contain your perl code.
If you have a string like
my $var1 = abc$123
and you want to replace it with abcd then you have to use \Q \E. If you don't then no matter what perl doesn't replace the string.
This is the only thing that worked for me.
my $var2 = s/\Q$var1\E/abcd/g;