I am trying to escape the spaces in a Linux path. However, whenever I try to escape my backslash I end up with a double slash.
Example path:
/mnt/drive/site/usa/1201 East/1201 East Invoice.pdf
So that I can use this in Linux I want to escape it as:
/mnt/drive/site/usa/1201\ East/1201\ East\ Invoice.pdf
So I'm trying this:
backup_item.gsub("\s", "\\\s")
But I get an unexpected output of
/mnt/drive/site/usa/1201\\ East/1201\\ East\\ Invoice.pdf
Stefan is right; I just want to point out that if you have to escape strings for shell use you should check Shellwords::shellescape:
require 'shellwords'
puts Shellwords.shellescape "/mnt/drive/site/usa/1201 East/1201 East Invoice.pdf"
# prints /mnt/drive/site/usa/1201\ East/1201\ East\ Invoice.pdf
# or
puts "/mnt/drive/site/usa/1201 East/1201 East Invoice.pdf".shellescape
# prints /mnt/drive/site/usa/1201\ East/1201\ East\ Invoice.pdf
# or (as reported by #hagello)
puts shellwords.escape "/mnt/drive/site/usa/1201 East/1201 East Invoice.pdf"
# prints /mnt/drive/site/usa/1201\ East/1201\ East\ Invoice.pdf
That is the string's inspect value, "a printable version of str, surrounded by quote marks, with special characters escaped":
quoted = "path/to/file with spaces".gsub(/ /, '\ ')
=> "path/to/file\\ with\\ spaces"
Just print the string:
puts quoted
Output:
path/to/file\ with\ spaces
Related
s = "#main= 'quotes'
s.gsub "'", "\\'" # => "#main= quotes'quotes"
This seems to be wrong, I expect to get "#main= \\'quotes\\'"
when I don't use escape char, then it works as expected.
s.gsub "'", "*" # => "#main= *quotes*"
So there must be something to do with escaping.
Using ruby 1.9.2p290
I need to replace single quotes with back-slash and a quote.
Even more inconsistencies:
"\\'".length # => 2
"\\*".length # => 2
# As expected
"'".gsub("'", "\\*").length # => 2
"'a'".gsub("'", "\\*") # => "\\*a\\*" (length==5)
# WTF next:
"'".gsub("'", "\\'").length # => 0
# Doubling the content?
"'a'".gsub("'", "\\'") # => "a'a" (length==3)
What is going on here?
You're getting tripped up by the specialness of \' inside a regular expression replacement string:
\0, \1, \2, ... \9, \&, \`, \', \+
Substitutes the value matched by the nth grouped subexpression, or by the entire match, pre- or postmatch, or the highest group.
So when you say "\\'", the double \\ becomes just a single backslash and the result is \' but that means "The string to the right of the last successful match." If you want to replace single quotes with escaped single quotes, you need to escape more to get past the specialness of \':
s.gsub("'", "\\\\'")
Or avoid the toothpicks and use the block form:
s.gsub("'") { |m| '\\' + m }
You would run into similar issues if you were trying to escape backticks, a plus sign, or even a single digit.
The overall lesson here is to prefer the block form of gsub for anything but the most trivial of substitutions.
s = "#main = 'quotes'
s.gsub "'", "\\\\'"
Since \it's \\equivalent if you want to get a double backslash you have to put four of ones.
You need to escape the \ as well:
s.gsub "'", "\\\\'"
Outputs
"#main= \\'quotes\\'"
A good explanation found on an outside forum:
The key point to understand IMHO is that a backslash is special in
replacement strings. So, whenever one wants to have a literal
backslash in a replacement string one needs to escape it and hence
have [two] backslashes. Coincidentally a backslash is also special in a
string (even in a single quoted string). So you need two levels of
escaping, makes 2 * 2 = 4 backslashes on the screen for one literal
replacement backslash.
source
I'm trying to do a regex with lookbehind that changes \n to but not if it's a \\n.
My closest attempt has no effect:
text.gsub /(?<!\\)\n/, ''
Unfortunately, no number of backslashes in the lookbehind seem to fix the problem. How can I address this?
You need to double the backslash before the n in the regex, otherwise it's looking for a newline instead of a literal backslash followed by n:
irb(main):001:0> puts "hello\\nthere\\\\n".gsub(/(?<!\\)\\n/, ' ')
hello there\\n
You don't need anything special. "\n" is a single character. It does not include a "\" or "n" character.
text.gsub(/\n/, "")
But instead of that, you should do:
text.gsub("\n", "")
or
text.tr("\n", "")
But I would do:
text.tr($/, "")
s = "#main= 'quotes'
s.gsub "'", "\\'" # => "#main= quotes'quotes"
This seems to be wrong, I expect to get "#main= \\'quotes\\'"
when I don't use escape char, then it works as expected.
s.gsub "'", "*" # => "#main= *quotes*"
So there must be something to do with escaping.
Using ruby 1.9.2p290
I need to replace single quotes with back-slash and a quote.
Even more inconsistencies:
"\\'".length # => 2
"\\*".length # => 2
# As expected
"'".gsub("'", "\\*").length # => 2
"'a'".gsub("'", "\\*") # => "\\*a\\*" (length==5)
# WTF next:
"'".gsub("'", "\\'").length # => 0
# Doubling the content?
"'a'".gsub("'", "\\'") # => "a'a" (length==3)
What is going on here?
You're getting tripped up by the specialness of \' inside a regular expression replacement string:
\0, \1, \2, ... \9, \&, \`, \', \+
Substitutes the value matched by the nth grouped subexpression, or by the entire match, pre- or postmatch, or the highest group.
So when you say "\\'", the double \\ becomes just a single backslash and the result is \' but that means "The string to the right of the last successful match." If you want to replace single quotes with escaped single quotes, you need to escape more to get past the specialness of \':
s.gsub("'", "\\\\'")
Or avoid the toothpicks and use the block form:
s.gsub("'") { |m| '\\' + m }
You would run into similar issues if you were trying to escape backticks, a plus sign, or even a single digit.
The overall lesson here is to prefer the block form of gsub for anything but the most trivial of substitutions.
s = "#main = 'quotes'
s.gsub "'", "\\\\'"
Since \it's \\equivalent if you want to get a double backslash you have to put four of ones.
You need to escape the \ as well:
s.gsub "'", "\\\\'"
Outputs
"#main= \\'quotes\\'"
A good explanation found on an outside forum:
The key point to understand IMHO is that a backslash is special in
replacement strings. So, whenever one wants to have a literal
backslash in a replacement string one needs to escape it and hence
have [two] backslashes. Coincidentally a backslash is also special in a
string (even in a single quoted string). So you need two levels of
escaping, makes 2 * 2 = 4 backslashes on the screen for one literal
replacement backslash.
source
So I'm having an issue replacing \" in a string.
My Objective:
Given a string, if there's an escaped quote in the string, replace it with just a quote
So for example:
"hello\"74" would be "hello"74"
simp"\"sons would be simp"sons
jump98" would be jump98"
I'm currently trying this: but obviously that doesn't work and messes everything up, any assistance would be awesome
str.replace "\\"", "\""
I guess you are being mistaken by how \ works. You can never define a string as
a = "hello"74"
Also escape character is used only while defining the variable its not part of the value. Eg:
a = "hello\"74"
# => "hello\"74"
puts a
# hello"74
However in-case my above assumption is incorrect following example should help you:
a = 'hello\"74'
# => "hello\\\"74"
puts a
# hello\"74
a.gsub!("\\","")
# => "hello\"74"
puts a
# hello"74
EDIT
The above gsub will replace all instances of \ however OP needs only to replace '" with ". Following should do the trick:
a.gsub!("\\\"","\"")
# => "hello\"74"
puts a
# hello"74
You can use gsub:
word = 'simp"\"sons';
print word.gsub(/\\"/, '"');
//=> simp""sons
I'm currently trying str.replace "\\"", "\"" but obviously that doesn't work and messes everything up, any assistance would be awesome
str.replace "\\"", "\"" doesn't work for two reasons:
It's the wrong method. String#replace replaces the entire string, you are looking for String#gsub.
"\\"" is incorrect: " starts the string, \\ is a backslash (correctly escaped) and " ends the string. The last " starts a new string.
You have to either escape the double quote:
puts "\\\"" #=> \"
Or use single quotes:
puts '\\"' #=> \"
Example:
content = <<-EOF
"hello\"74"
simp"\"sons
jump98"
EOF
puts content.gsub('\\"', '"')
Output:
"hello"74"
simp""sons
jump98"
There is something mysterious to me about the escape status of a backslash within a single quoted string literal as argument of String#tr. Can you explain the contrast between the three examples below? I particularly do not understand the second one. To avoid complication, I am using 'd' here, which does not change the meaning when escaped in double quotation ("\d" = "d").
'\\'.tr('\\', 'x') #=> "x"
'\\'.tr('\\d', 'x') #=> "\\"
'\\'.tr('\\\d', 'x') #=> "x"
Escaping in tr
The first argument of tr works much like bracket character grouping in regular expressions. You can use ^ in the start of the expression to negate the matching (replace anything that doesn't match) and use e.g. a-f to match a range of characters. Since it has control characters, it also does escaping internally, so you can use - and ^ as literal characters.
print 'abcdef'.tr('b-e', 'x') # axxxxf
print 'abcdef'.tr('b\-e', 'x') # axcdxf
Escaping in Ruby single quote strings
Furthermore, when using single quotes, Ruby tries to include the backslash when possible, i.e. when it's not used to actually escape another backslash or a single quote.
# Single quotes
print '\\' # \
print '\d' # \d
print '\\d' # \d
print '\\\d' # \\d
# Double quotes
print "\\" # \
print "\d" # d
print "\\d" # \d
print "\\\d" # \d
The examples revisited
With all that in mind, let's look at the examples again.
'\\'.tr('\\', 'x') #=> "x"
The string defined as '\\' becomes the literal string \ because the first backslash escapes the second. No surprises there.
'\\'.tr('\\d', 'x') #=> "\\"
The string defined as '\\d' becomes the literal string \d. The tr engine, in turn uses the backslash in the literal string to escape the d. Result: tr replaces instances of d with x.
'\\'.tr('\\\d', 'x') #=> "x"
The string defined as '\\\d' becomes the literal \\d. First \\ becomes \. Then \d becomes \d, i.e. the backslash is preserved. (This particular behavior is different from double strings, where the backslash would be eaten alive, leaving only a lonesome d)
The literal string \\d then makes tr replace all characters that are either a backslash or a d with the replacement string.