Was hoping someone could help me out a one-liner in bash, using anything standard like sed, awk, etc. to take a string and insert a backslash before any characters that are matched to a given list of characters.
For example, the input string abj:"si8'h4# should become abj\:\"si8\'h4\# if the list of characters I want to escape is "':#
Any help is appreciated!
Here's a simple perl version.
echo 'abj:"si8'\''h4#' | perl -pe 's/(["'\'':#])/\\\1/g'
abj\:\"si8\'h4\#
The problem is how you want to input the list of characters to escape. If you're inputting them yourself, this solution will work fine, though you have to be careful with escaping single quotes. If you want to avoid escaping issues, perhaps you'd want to read the list of characters to escape from a file?
sed
echo 'abj:"si8'"'h4#" | sed 's/["\x27:#]/\\&/g'
abj\:\"si8\'h4\#
The set of characters that you want to escape as a variable:
$ esc="\"':#"
The string that you want to escape:
$ str=abj:\"si8\'h4\#
Escaping with sed:
$ echo "$str" | sed 's/['"$esc"']/\\&/g'
Output:
abj\:\"si8\'h4\# <--- Output
abj\:\"si8\'h4\# <--- Correct string for comparison
Yup, that seems to do it.
Related
Having the following string inside of a text file.
{"_job":"delete","query":{"query":{"bool":{"must":[{"term":{"_id":"28381"}}],"should":[]}}},"script":{"inline":"ctx._source.meta='This
is a ' test string Peedr'"},"timestamp":1518165383,"host":"","port":"9200","index":"","docType":"","customIndexer":""}
I would like to replace all the ' that are inside the ctx._source.meta='' part with \' using sed.
In the example above I've This is a ' test string Peedr which I would like to convert to This is a \' test string Peedr, so the desired output would be:
{"_job":"delete","query":{"query":{"bool":{"must":[{"term":{"_id":"28381"}}],"should":[]}}},"script":{"inline":"ctx._source.meta='This
is a \' test string
Peedr'"},"timestamp":1518165383,"host":"","port":"9200","index":"","docType":"","customIndexer":""}
I'm using the following regex to get the ' that is inside the ctx._source.meta string (3rd capture group).
(meta=')(.*?)(')(.*?)(')
I've the regex, but I dont know how to use the sed comand in order to replace the 3rd capture group with \'.
Can someone give me a hand and tell me the sed comand I have to use?
Thanks in advance
sed generally does not support the Perl regex extensions, so the non-greedy .*? will probably not do what you hope. If you want to use Perl regex, use Perl!
perl -pe "s/(meta='.*?)(')(.*?')/\$1\\\\\$2\$3/"
This will still not necessarily work if the input is malformed; a better approach would be to specifically exclude single quotes from the match, and then you don't need the non-greedy matching.
sed "s/\\(meta='[^']*\\)'\\([^']*'\\)/\\1\\\\'\\2/"
In both cases, the number of backslashes required to escape the backslashes inside the shell's double quotes is staggering.
You put back-references to groups except one you want to replace. There is a better way to accomplish same task:
sed -E "s/(ctx\._source\.meta=')([^']*)(')([^']*')/\1\2\\'\4/"
You may use:
sed "s/ ' / \\\' /g" sample.txt
The first part will instruct sed to only look for a single quote between 2 spaces, as such ctx._source.meta='This and string Peedr'"} will not match, hence will not be changed.
Edit:
At the poster's request, I edited my sed command to apply to extra use cases:
sed "s/\(ctx._source.meta='.*\)'\(.*Peedr'\"\)/\1\\\'\2/g"
I have a text file with a line that reads:
<div id="page_footer"><div><? print('Any phrase's characters can go here!'); ?></div></div>
And I'm wanting to use sed or awk to extract the substring above between the single quotes so it just prints ...
Any phrase's characters can go here!
I want the phrase to be delimited as I have above, starting after the single quote and ending at the single-quote immediately followed by a parenthesis and then semicolon. The following sed command with a capture group doesn't seem to be working for me. Suggestions?
sed '/^<div id="page_footer"><div><? print(\'\(.\+\)\');/ s//\1/p' /home/foobar/testfile.txt
Incorrect would be using cut like
grep "page_footer" /home/foobar/testfile.txt | cut -d "'" -f2
It will go wrong with single quotes inside the string. Counting the number of single quotes first will change this from a simple to an over-complicated solution.
A solution with sed is better: remove everything until the first single quote and everything after the last one. A single quote in the string becomes messy when you first close the sed parameter with a single quote, escape the single quote and open a sed string again:
grep page_footer /home/foobar/testfile.txt | sed -e 's/[^'\'']*//' -e 's/[^'\'']*$//'
And this is not the full solution, you want to remove the first/last quotes as well:
grep page_footer /home/foobar/testfile.txt | sed -e 's/[^'\'']*'\''//' -e 's/'\''[^'\'']*$//'
Writing the sed parameters in double-quoted strings and using the . wildcard for matching the single quote will make the line shorter:
grep page_footer /home/foobar/testfile.txt | sed -e "s/^[^\']*.//" -e "s/.[^\']*$//"
Using advanced grep (such as in Linux), this might be what you are looking for
grep -Po "(?<=').*?(?='\);)"
The general form of the substitution command in sed is:
s/regexp/replacement/flags
where the '/' characters may be uniformly replaced by any other single character. But how do you choose this separator character when the replacement string is being fed in by an environment variable and might contain any printable character? Is there a straightforward way to escape the separator character in the variable using bash?
The values are coming from trusted administrators so security is not my main concern. (In other words, please don't answer with: "Never do this!") Nevertheless, I can't predict what characters will need to appear in the replacement string.
You can use control character as regex delimiters also like this:
s^Aregexp^Areplacement^Ag
Where ^A is CTRLva pressed together.
Or else use awk and don't worry about delimiters:
awk -v s="search" -v r="replacement" '{gsub(s, r)} 1' file
Here isn't (easy) solution for the following using the sed.
while read -r string from to wanted
do
echo "in [$string] want replace [$from] to [$to] wanted result: [$wanted]"
final=$(echo "$string" | sed "s/$from/$to/")
[[ "$final" == "$wanted" ]] && echo OK || echo WRONG
echo
done <<EOF
=xxx= xxx === =====
=abc= abc /// =///=
=///= /// abc =abc=
EOF
what prints
in [=xxx=] want replace [xxx] to [===] wanted result: [=====]
OK
in [=abc=] want replace [abc] to [///] wanted result: [=///=]
sed: 1: "s/abc/////": bad flag in substitute command: '/'
WRONG
in [=///=] want replace [///] to [abc] wanted result: [=abc=]
sed: 1: "s/////abc/": bad flag in substitute command: '/'
WRONG
Can't resists: Never do this! (with sed). :)
Is there a straightforward way to escape the separator character in
the variable using bash?
No, because you passing the strings from variables, you can't easily escape the separator character, because in "s/$from/$to/" the separator can appear not only in the $to part but in the $from part too. E.g. when you escape the separator it in the $from part it will not do the replacement at all, because will not find the $from.
Solution: use something other as sed
1.) Using pure bash. In the above script instead of the sed use the
final=${string//$from/$to}
2.) If the bash's substitutions are not enough, use something to what you can pass the $from and $to as variables.
as #anubhava already said, can use: awk -v f="$from" -v t="$to" '{gsub(f, t)} 1' file
or you can use perl and passing values as environment variables
final=$(echo "$string" | perl_from="$from" perl_to="$to" perl -pe 's/$ENV{perl_from}/$ENV{perl_to}/')
or passing the variables to perl via the command line arguments
final=$(echo "$string" | perl -spe 's/$f/$t/' -- -f="$from" -t="$to")
2 options:
1) take a char not in the string (need a pre process on content check and possible char without warranty that a char is available)
# Quick and dirty sample using `'/_##|!%=:;,-` arbitrary sequence
Separator="$( printf "%sa%s%s" '/_##|!%=:;,-' "${regexp}" "${replacement}" \
| sed -n ':cycle
s/\(.\)\(.*a.*\1.*\)\1/\1\2/g;t cycle
s/\(.\)\(.*a.*\)\1/\2/g;t cycle
s/^\(.\).*a.*/\1/p
' )"
echo "Separator: [ ${Separator} ]"
sed "s${Separator}${regexp}${Separator}${replacement}${Separator}flag" YourFile
2) escape the wanted char in the string patterns (need a pre process to escape char).
# Quick and dirty sample using # arbitrary with few escape security check
regexpEsc="$( printf "%s" "${regexp}" | sed 's/#/\\#/g' )"
replacementEsc"$( printf "%s" "${replacement}" | sed 's/#/\\#/g' )"
sed 's#regexpEsc#replacementEsc#flags' YourFile
From man sed
\cregexpc
Match lines matching the regular expression regexp. The c may be any
character.
When working with paths i often use # as separator:
sed s\#find/path#replace/path#
No need to escape / with ugly \/.
I use this classic perl one liner to replace strings in multiple files recursively
perl -pi -e 's/oldstring/newstring/g' `grep -irl oldstring *`
But this has failed me as I want to find the string:
'$user->primaryorganisation->id'
and replace with
$user->primaryorganisation->id
I can't seem to escape the string correctly for the line to run successfully.
Any help gratefully received!
Try this one. Lots of escapes. Go with TLPs suggestion and use a source file.
perl -pi -e "s/'\\\$user->primaryorganisation->id'/\\\$user->primaryorganisation->id/g" `grep -irl "'\$user->primaryorganisation->id'" *`
Explanation:
three backslashes: the first two tell the shell to produce a literal backslash; the thrid one escapes the $ for the shell; that makes \$ for Perl, which needs the backslash to escape the variable interpolation
double quotes " to put single quotes ' inside them
one backslash and a dollar \$ for grep so the shell passes on a literal dollar sign
When you want to represent a single quote in a perl but can't because the one-liner uses single quotes itself, you can use \047, the octal code for single quote. So, this should work:
s/\047(\$user->primaryorganisation->id)\047/$1/g
I recommend Minimal Perl by Maher for more-than-you-wanted-to-know about the art of one-lining perl.
To produce
...'...
you can generically use
'...'\''...'
As such,
s/'(\$user->primaryorganisation->id)'/$1/g
becomes
's/'\''(\$user->primaryorganisation->id)'\''/$1/g'
so
find -type f \
-exec perl -i -pe's/'\''(\$user->primaryorganisation->id)'\''/$1/g' {} +
I'm trying to use sed with perl to replace ^[(s3B with an empty string in several files.
s/^[(s3B// isn't working though, so I'm wondering what else I could try.
You need to quote the special characters:
$ echo "^[(s3B AAA ^[(s3B"|sed 's/\^\[[(]s3B//g'
AAA
$ echo "^[(s3B AAA ^[(s3B" >file.txt
$ perl -p -i -e 's/\^\[[(]s3B//g' file.txt
$ cat file.txt
AAA
The problem is that there are several characters that have a special meaning in regular expressions. ^ is a start-of-line anchor, [ opens a character class, and ( opens a capture.
You can escape all non-alphanumerics in a Perl string by preceding it with \Q, so you can safely use
s/\Q^[(s3B//
which is equivalent to, and more readable than
s/\^\[\(s3B//
If you're dealing with ANSI sequences (xterm color sequences, escape sequences), then ^[ is not '^' followed by '[' but rather an unprintable character ESC, ASCII code 0x1B.
To put that character into a sed expression you need to use \x1B in GNU sed, or see http://www.cyberciti.biz/faq/unix-linux-sed-ascii-control-codes-nonprintable/ . You can also insert special characters directly into your command line using ctrl+v in Bash line editing.
In regex "^", "[" and "(" (and many others) are special characters used for special regex features, if you are referencing the characters themselves you should preceed them with "\".
The correct substitution reges would be:
$string =~ s/\^\[\(3B//g
if you want to replace all occurences.