Say I have a function that accepts a string and interpolates it inside quotes.
print_str() (
printf '"%s"\n' "$1"
)
Since the function encloses the input in double quotes, I want to escape all double quotes, if any, in the argument. But, critically, double quotes that are already escaped must be preceded by a backslash. In short, the function must add one layer of escaping to all double quotes in the argument, be that the first layer or an additional layer.
Examples:
'abc"def' -> 'abc\"def'
'{ "json": "{ \"key\": \"value\" }" }' -> '{ \"json\": \"{ \\\"key\\\": \\\"value\\\" }\" }'
The tricky part for me is determining whether a backslash preceding a double quote is actually escaping the double quote, because, for example, in '\\"' the backslash does not escape the double quote even though it immediately precedes the double quote.
I've thought about using sed and checking whether the number of backslashes before a double quote is odd or even, but that seems pretty convoluted. I managed to get the solution I wanted by piping the input through jq -R ., but I am wondering if there is another way to do it. In particular, I am looking for something that works in POSIX shell.
EDIT:
For further clarification, I am seeking some function that will add an additional layer of escaping to double quotes in a given string.
some_function() {
# IDK what the implementation would be
}
Then, some_function in conjunction with print_str would producing the following output.
string='{ "json": "{ \"key\": \"value\" }" }'
print_str "$(some_function "$string")"
# -> prints "{ \"json\": \"{ \\\"key\\\": \\\"value\\\" }\" }"
You don't need to count anything
sed 's/["\]/\\&/g; s/.*/"&"/'
Related
In perl suppose I have a string like 'hello\tworld\n', and what I want is:
'hello world
'
That is, "hello", then a literal tab character, then "world", then a literal newline. Or equivalently, "hello\tworld\n" (note the double quotes).
In other words, is there a function for taking a string with escape sequences and returning an equivalent string with all the escape sequences interpolated? I don't want to interpolate variables or anything else, just escape sequences like \x, where x is a letter.
Sounds like a problem that someone else would have solved already. I've never used the module, but it looks useful:
use String::Escape qw(unbackslash);
my $s = unbackslash('hello\tworld\n');
You can do it with 'eval':
my $string = 'hello\tworld\n';
my $decoded_string = eval "\"$string\"";
Note that there are security issues tied to that approach if you don't have 100% control of the input string.
Edit: If you want to ONLY interpolate \x substitutions (and not the general case of 'anything Perl would interpolate in a quoted string') you could do this:
my $string = 'hello\tworld\n';
$string =~ s#([^\\A-Za-z_0-9])#\\$1#gs;
my $decoded_string = eval "\"$string\"";
That does almost the same thing as quotemeta - but exempts '\' characters from being escaped.
Edit2: This still isn't 100% safe because if the last character is a '\' - it will 'leak' past the end of the string though...
Personally, if I wanted to be 100% safe I would make a hash with the subs I specifically wanted and use a regex substitution instead of an eval:
my %sub_strings = (
'\n' => "\n",
'\t' => "\t",
'\r' => "\r",
);
$string =~ s/(\\n|\\t|\\n)/$sub_strings{$1}/gs;
I am unsure why this doesn't work:
string.replaceAll('\\"','"')
I want to replace all \" with "
Any idea?
I have also tried
string.replaceAll("[\"]","\"")
The first argument to the replaceAll method is a regular expression, so the backslash character has significance there and needs to be escaped. You could use the forward-slash string delimiter to avoid double-escaping.
assert (/Hello, \"Joe\"/.replaceAll(/\\"/, '"') == 'Hello, "Joe"')
I have a variable from which I have to grep the which in middle of %% adn the word which starts with $$. I used split it works... but for only some scenarios.
Example:
#!/usr/bin/perl
my $lastline ="%Filters_LN_RESS_DIR%\ARC\Options\Pega\CHF_Vega\$$(1212_GV_DATE_LDN)";
my #lastline_temp = split(/%/,$lastline);
print #lastline_temp;
my #var=split("\\$\\$",$lastline_temp[2]);
print #var;
I get the o/p as expected. But can i get the same using Grep command. I mean I dont want to use the array[2] or array[1]. So that I can replace the values easily.
I don't really see how you can get the output you expect. Because you put your data in "busy" quotes (interpolating, double, ...), it comes out being stored as:
'%Filters_LN_RESS_DIR%ARCOptionsPegaCHF_Vega$01212_GV_DATE_LDN)'
See Quote and Quote-like Operators and perhaps read Interpolation in Perl
Notice that the backslashes are gone. A backslash in interpolating quotes simply means "treat the next character as literal", so you get literal 'A', literal 'O', literal 'P', ....
That '0' is the value of $( (aka $REAL_GROUP_ID) which you unwittingly asked it to interpolate. So there is no sequence '$$' to split on.
Can you get the same using a grep command? It depends on what "the same" is. You save the results in arrays, the purpose of grep is to exclude things from the arrays. You will neither have the arrays, nor the output of the arrays if you use a non-trivial grep: grep {; 1 } #data.
Actually you can get the exact same result with this regular expression, assuming that the single string in #vars is the "result".
m/%([^%]*)$/
Of course, that's no more than
substr( $lastline, rindex( $lastline, '%' ) + 1 );
which can run 8-10 times faster.
First, be very careful in your use of quotes, I'm not sure if you don't mean
'%Filters_LN_RESS_DIR%\ARC\Options\Pega\CHF_Vega\$$(1212_GV_DATE_LDN)'
instead of
"%Filters_LN_RESS_DIR%\ARC\Options\Pega\CHF_Vega\$$(1212_GV_DATE_LDN)"
which might be a different string. For example, if evaluated, "$$" means the variable $PROCESS_ID.
After trying to solve riddles (not sure about that), and quoting your string
my $lastline =
'%Filters_LN_RESS_DIR%\ARC\Options\Pega\CHF_Vega\$$(1212_GV_DATE_LDN)'
differently, I'd use:
my ($w1, $w2) = $lastline =~ m{ % # the % char at the start
([^%]+) # CAPTURE everything until next %
[^(]+ # scan to the first brace
\( # hit the brace
([^)]+) # CAPTURE everything up to closing brace
}x;
print "$w1\n$w2";
to extract your words. Result:
Filters_LN_RESS_DIR
1212_GV_DATE_LDN
But what do you mean by replace the values easily. Which values?
Addendum
Now lets extract the "words" delimited by '\'. Using a simple split:
my #words = split /\\/, # use substr to start split after the first '\\'
substr $lastline, index($lastline,'\\');
you'll get the words between the backslashes if you drop the last entry (which is the $$(..) string):
pop #words; # remove the last element '$$(..)'
print join "\n", #words; # print the other elements
Result:
ARC
Options
Pega
CHF_Vega
Does this work better with grep? Seems to:
my #words = grep /^[^\$%]+$/, split /\\/, $lastline;
and
print join "\n", #words;
also results in:
ARC
Options
Pega
CHF_Vega
Maybe that is what you are after? What do you want to do with these?
Regards
rbo
Say I have the following string:
"abcdefghijklmnopqrstuvwxyz"
And I think its too long for one line in my YAML file, is there some way to split that over several lines?
>-
abcdefghi
jklmnopqr
stuvwxyz
Would result in "abcdefghi jklmnopqr stuvwxyz" which is close, but it shouldn't have any spaces.
Use double-quotes, and escape the newline:
"abcdefghi\
jklmnopqr\
stuvwxyz"
There are some subtleties that Jesse's answer will miss.
YAML (like many programming languages) treats single and double quotes differently. Consider this document:
regexp: "\d{4}"
This will fail to parse with an error such as:
found unknown escape character while parsing a quoted scalar at line 1 column 9
Compare that to:
regexp: '\d{4}'
Which will parse correctly. In order to use backslash character inside double-quoted strings you would need to escape them, as in:
regexp: "\\d{4}"
I'd also like to highlight Steve's comment about single-quoted strings. Consider this document:
s1: "this\
is\
a\
test"
s2: 'this\
is\
a\
test'
When parsed, you will find that it is equivalent to:
s1: thisisatest
s2: "this\\ is\\ a\\ test"
This is a direct result of the fact that YAML treats single-quoted strings as literals, while double-quoted strings are subject to escape character expansion.
Is there an easy way, using a subroutine maybe, to print a string in Perl without escaping every special character?
This is what I want to do:
print DELIMITER <I don't care what is here> DELIMITER
So obviously it will great if I can put a string as a delimiter instead of special characters.
perldoc perlop, under "Quote and Quote-like Operators", contains everything you need.
While we usually think of quotes as literal values, in Perl they function as operators, providing various kinds of interpolating and pattern matching
capabilities. Perl provides customary quote characters for these behaviors, but also provides a way for you to choose your quote character for any of
them. In the following table, a "{}" represents any pair of delimiters you choose.
Customary Generic Meaning Interpolates
'' q{} Literal no
"" qq{} Literal yes
`` qx{} Command yes*
qw{} Word list no
// m{} Pattern match yes*
qr{} Pattern yes*
s{}{} Substitution yes*
tr{}{} Transliteration no (but see below)
<<EOF here-doc yes*
* unless the delimiter is ''.
$str = q(this is a "string");
print $str;
if you mean quotes and apostrophes with 'special characters'
You can use the __DATA__ directive which will treat all of the following lines as a file that can be accessed from the DATA handle:
while (<DATA>) {
print # or do something else with the lines
}
__DATA__
#!/usr/bin/perl -w
use Some::Module;
....
or you can use a heredoc:
my $string = <<'END'; #single quotes prevent any interpolation
#!/usr/bin/perl -b
use Some::Module;
....
END
The printing is not doing special things to the escapes, double quoted strings are doing it. You may want to try single quoted strings:
print 'this is \n', "\n";
In a single quoted string the only characters that must be escaped are single quotes and a backslash that occurs immediately before the end of the string (i.e. 'foo\\').
It is important to note that interpolation does not work with single quoted strings, so
print 'foo is $foo', "\n";
Will not print the contents of $foo.
You can pretty much use any character you want with q or qq. For example:
#!/usr/bin/perl
use utf8;
use strict; use warnings;
print q∞This is a test∞;
print qq☼\nThis is another test\n☼;
print q»But, what is the point?»;
print qq\nYou are just making life hard on yourself!\n;
print qq¿That last one is tricky\n¿;
You cannot use qq DELIMITER foo DELIMITER. However, you could use heredocs for a similar effect:
print <<DELIMITER
...
DELIMETER
;
or
print <<'DELIMETER'
...
DELIMETER
;
but your source code would be really ugly.
If you want to print a string literally and you have Perl 5.10 or later then
say 'This is a string with "quotes"' ;
will print the string with a newline.. The importaning thing is to use single quotes ' ' rather than double ones " "