Is it possible to escape regex metacharacters reliably with sed

Question

I'm wondering whether it is possible to write a 100% reliable sed command to escape any regex metacharacters in an input string so that it can be used in a subsequent sed command. Like this:

#!/bin/bash
# Trying to replace one regex by another in an input file with sed

search="/abc\n\t[a-z]\+\([^ ]\)\{2,3\}\3"
replace="/xyz\n\t[0-9]\+\([^ ]\)\{2,3\}\3"

# Sanitize input
search=$(sed 'script to escape' <<< "$search")
replace=$(sed 'script to escape' <<< "$replace")

# Use it in a sed command
sed "s/$search/$replace/" input

I know that there are better tools to work with fixed strings instead of patterns, for example awk, perl or python. I would just like to prove whether it is possible or not with sed. I would say let's concentrate on basic POSIX regexes to have even more fun! :)

I have tried a lot of things but anytime I could find an input which broke my attempt. I thought keeping it abstract as script to escape would not lead anybody into the wrong direction.

Btw, the discussion came up here. I thought this could be a good place to collect solutions and probably break and/or elaborate them.

Aggress

Why do you need to escape them? Do you want to find a literal copy of the regexp in input? – Erskine

@Erskine Yes. It is just theoretical. – Aggress

Find all the special characters that can appear in regular expressions, and then use s/[chars]/\\&/' – Erskine

@Erskine Sometimes they are sequences of characters like \1 (back reference) or \( (start of capturing group) – Aggress

The list of special characters will include backslash. So you'll convert \1 to \\1 with this. – Erskine

How close (or not) does bash's printf %q come? – Swerve

Yes, it should. He says he wants to search for the regexp literally, not treat it as a regular expression. – Erskine

@EtanReisner printf "%q" "\n" would destroy the \n .. – Aggress

@Aggress What do you want \n in the input search string to match? For it to match a literal \n in the file it needs to be \\n in the pattern (which is what printf does). – Swerve

@EtanReisner It should stay a newline, keeping it's special meaning. – Aggress

sed can't match newlines without extra work (pulling multiple lines into the pattern area) so I'm not sure how useful that is, but ok. Does this caveat apply to all the escapes then? You just want to handle the regex metacharacters themselves? Because mkelement0's solution doesn't preserve the escapes either. – Swerve

@EdMorton it should be treated a literal backslash followed by a literal n - yes, I meant this. – Aggress

@EtanReisner No, I meant they should not preserved meaning \n should stay \n and not getting \\n.. (Same with \t, \0 ... ) – Aggress

don't forget that there are several standard regex used depending the sed version and option.In this case, some escaped character become regex special meaning for other version like the { or ( between posix and gnu native. – Adynamia

For the LHS if you don't know what data will show up its best to pay attention to the delimiters, the answers I see still use slash but a slash is somewhat common, you can use a non-printable character like SOH \001 - see my answer to another question that uses this here: https://mcmap.net/q/48795/-replace-multiple-lines-identifying-end-character On the RHS there is no perfect solution that I know of but you can scan the string quickly and find a unique replacement value at runtime, you can see this technique here: github.com/AdamDanischewski/r-n-f-bash-rename-script – Mopes

Answer 1

Note:

• If you're looking for prepackaged functionality based on the techniques discussed in this answer:

  • bash functions that enable robust escaping even in multi-line substitutions can be found at the bottom of this post (plus a perl solution that uses perl's built-in support for such escaping).
  • @EdMorton's answer contains a tool (bash script) that robustly performs single-line substitutions.
    • Ed's answer now has an improved version of the sed command used below, corrected in calestyo's answer, which is needed if you want to escape string literals for potential use with other regex-processing tools, such as awk and perl. In short: for cross-tool use, \ must be escaped as \\ rather than as [\], which means: instead of the
      sed 's/[^^]/[&]/g; s/\^/\\^/g' command used below, you must use
      sed 's/[^^\]/[&]/g; s/[\^]/\\&/g;'

• All snippets below assume bash as the shell (POSIX-compliant reformulations are possible):

---

SINGLE-line Solutions

---

Escaping a string literal for use as a regex in sed:

^{To give credit where credit is due: I found the regex used below in this answer.}

Assuming that the search string is a single-line string:

search='abc\n\t[a-z]\+\([^ ]\)\{2,3\}\3'  # sample input containing metachars.

searchEscaped=$(sed 's/[^^]/[&]/g; s/\^/\\^/g' <<<"$search") # escape it.

sed -n "s/$searchEscaped/foo/p" <<<"$search" # Echoes 'foo'

• Every character except ^ is placed in its own character set [...] expression to treat it as a literal.
  • Note that ^ is the one char. you cannot represent as [^], because it has special meaning in that location (negation).
• Then, ^ chars. are escaped as \^.
  • Note that you cannot just escape every char by putting a \ in front of it because that can turn a literal char into a metachar, e.g. \< and \b are word boundaries in some tools, \n is a newline, \{ is the start of a RE interval like \{1,3\}, etc.

The approach is robust, but not efficient.

The robustness comes from not trying to anticipate all special regex characters - which will vary across regex dialects - but to focus on only 2 features shared by all regex dialects:

• the ability to specify literal characters inside a character set.
• the ability to escape a literal ^ as \^

---

Escaping a string literal for use as the replacement string in sed's s/// command:

The replacement string in a sed s/// command is not a regex, but it recognizes placeholders that refer to either the entire string matched by the regex (&) or specific capture-group results by index (\1, \2, ...), so these must be escaped, along with the (customary) regex delimiter, /.

Assuming that the replacement string is a single-line string:

replace='Laurel & Hardy; PS\2' # sample input containing metachars.

replaceEscaped=$(sed 's/[&/\]/\\&/g' <<<"$replace") # escape it

sed -n "s/.*/$replaceEscaped/p" <<<"foo" # Echoes $replace as-is

---

---

MULTI-line Solutions

---

Escaping a MULTI-LINE string literal for use as a regex in sed:

Note: This only makes sense if multiple input lines (possibly ALL) have been read before attempting to match.
Since tools such as sed and awk operate on a single line at a time by default, extra steps are needed to make them read more than one line at a time.

# Define sample multi-line literal.
search='/abc\n\t[a-z]\+\([^ ]\)\{2,3\}\3
/def\n\t[A-Z]\+\([^ ]\)\{3,4\}\4'

# Escape it.
searchEscaped=$(sed -e 's/[^^]/[&]/g; s/\^/\\^/g; $!a\'$'\n''\\n' <<<"$search" | tr -d '\n')           #'

# Use in a Sed command that reads ALL input lines up front.
# If ok, echoes 'foo'
sed -n -e ':a' -e '$!{N;ba' -e '}' -e "s/$searchEscaped/foo/p" <<<"$search"

• The newlines in multi-line input strings must be translated to '\n' strings, which is how newlines are encoded in a regex.
• $!a\'$'\n''\\n' appends string '\n' to every output line but the last (the last newline is ignored, because it was added by <<<)
• tr -d '\n then removes all actual newlines from the string (sed adds one whenever it prints its pattern space), effectively replacing all newlines in the input with '\n' strings.

• -e ':a' -e '$!{N;ba' -e '}' is the POSIX-compliant form of a sed idiom that reads all input lines a loop, therefore leaving subsequent commands to operate on all input lines at once.

  • If you're using GNU sed (only), you can use its -z option to simplify reading all input lines at once:
    sed -z "s/$searchEscaped/foo/" <<<"$search"

---

Escaping a MULTI-LINE string literal for use as the replacement string in sed's s/// command:

# Define sample multi-line literal.
replace='Laurel & Hardy; PS\2
Masters\1 & Johnson\2'

# Escape it for use as a Sed replacement string.
IFS= read -d '' -r < <(sed -e ':a' -e '$!{N;ba' -e '}' -e 's/[&/\]/\\&/g; s/\n/\\&/g' <<<"$replace")
replaceEscaped=${REPLY%$'\n'}

# If ok, outputs $replace as is.
sed -n "s/\(.*\) \(.*\)/$replaceEscaped/p" <<<"foo bar" 

• Newlines in the input string must be retained as actual newlines, but \-escaped.
• -e ':a' -e '$!{N;ba' -e '}' is the POSIX-compliant form of a sed idiom that reads all input lines a loop.
• 's/[&/\]/\\&/g escapes all &, \ and / instances, as in the single-line solution.
• s/\n/\\&/g' then \-prefixes all actual newlines.
• IFS= read -d '' -r is used to read the sed command's output as is (to avoid the automatic removal of trailing newlines that a command substitution ($(...)) would perform).
• ${REPLY%$'\n'} then removes a single trailing newline, which the <<< has implicitly appended to the input.

---

---

bash functions based on the above (for sed):

• quoteRe() quotes (escapes) for use in a regex
• quoteSubst() quotes for use in the substitution string of a s/// call.
• both handle multi-line input correctly
  • Note that because sed reads a single line at at time by default, use of quoteRe() with multi-line strings only makes sense in sed commands that explicitly read multiple (or all) lines at once.
  • Also, using command substitutions ($(...)) to call the functions won't work for strings that have trailing newlines; in that event, use something like IFS= read -d '' -r escapedValue <(quoteSubst "$value")

# SYNOPSIS
#   quoteRe <text>
quoteRe() { sed -e 's/[^^]/[&]/g; s/\^/\\^/g; $!a\'$'\n''\\n' <<<"$1" | tr -d '\n'; }

# SYNOPSIS
#  quoteSubst <text>
quoteSubst() {
  IFS= read -d '' -r < <(sed -e ':a' -e '$!{N;ba' -e '}' -e 's/[&/\]/\\&/g; s/\n/\\&/g' <<<"$1")
  printf %s "${REPLY%$'\n'}"
}

Example:

from=$'Cost\(*):\n$3.' # sample input containing metachars. 
to='You & I'$'\n''eating A\1 sauce.' # sample replacement string with metachars.

# Should print the unmodified value of $to
sed -e ':a' -e '$!{N;ba' -e '}' -e "s/$(quoteRe "$from")/$(quoteSubst "$to")/" <<<"$from" 

Note the use of -e ':a' -e '$!{N;ba' -e '}' to read all input at once, so that the multi-line substitution works.

---

---

perl solution:

Perl has built-in support for escaping arbitrary strings for literal use in a regex: the quotemeta() function or its equivalent \Q...\E quoting.
The approach is the same for both single- and multi-line strings; for example:

from=$'Cost\(*):\n$3.' # sample input containing metachars.
to='You owe me $1/$& for'$'\n''eating A\1 sauce.' # sample replacement string w/ metachars.

# Should print the unmodified value of $to.
# Note that the replacement value needs NO escaping.
perl -s -0777 -pe 's/\Q$from\E/$to/' -- -from="$from" -to="$to" <<<"$from" 

• Note the use of -0777 to read all input at once, so that the multi-line substitution works.

• The -s option allows placing -<var>=<val>-style Perl variable definitions following -- after the script, before any filename operands.

Unlade

FWIW, newer sed allow sed -z to match NUL separated lines, so the matches can include \n. Example use: find -print0 | sed -z ... | xargs --null script etc. Multiline regex with \n come in very handy, as Linux (or Ubuntu for Windows) allows linefeeds in filenames (like: echo help me world > $'\n\nminime\nwas here\n') – Shady

@Tino: Thanks, I've added a -z-based variant to the answer, but note that it's not about older or newer per se, it's about GNU sed, which defines -z as a nonstandard option, vs. other sed implementations, such as the BSD sed found on macOS, which do not. – Unlade

This is an excellent answer. In my case, I needed to escape the string for input to sed, and need it to work in POSIX sh (not bash), so I ended up with: fixed=`printf '%s\n' "${val}" | sed 's#\/#\\\/#g' | sed 's/\&/\\\&/g' ` – Latterll

Ah shoot, it worked for my original big test case, but further experiments show that it's still finicky with backslashes in the substitution string, so I'll have to take down my answer. Et tu, sed... – Gastrotomy

Ah, good point, @Gastrotomy - I was only focused on the search pattern - answer updated. – Unlade

FYI, something about zsh interprets this differently – Barfuss

@WilliamEntriken, which of the several commands in this answer are you referring to? – Unlade

@Unlade I think quoteSubst is wrong in at least one case: printf 'a\n' | sed -e ':a; $!{N; ba;}; s%[&/\]%\\&%g; s/\n/\\&/g' causes a<newline> to be printed, but POSIX mandates that in sed replacements the <newline> is prefixed by <backslash>. This works however for e.g. the a\nb case. – Technicolor

@calestyo, quoteSubst is designed to receive its input as an argument, so try quoteSubst $'a\n' – Unlade

@mklement0, ah, I see why it works with that, because the <<< adds a <newline>. – Technicolor

Answer 2

Building upon @mklement0's answer in this thread, the following tool will replace any single-line string (as opposed to regexp) with any other single-line string using sed and bash:

$ cat sedstr
#!/bin/bash
old="$1"
new="$2"
file="${3:--}"
escOld=$(sed 's/[^^\\]/[&]/g; s/\^/\\^/g; s/\\/\\\\/g' <<< "$old")
escNew=$(sed 's/[&/\]/\\&/g' <<< "$new")
sed "s/$escOld/$escNew/g" "$file"

To illustrate the need for this tool, consider trying to replace a.*/b{2,}\nc with d&e\1f by calling sed directly:

$ cat file
a.*/b{2,}\nc
axx/bb\nc

$ sed 's/a.*/b{2,}\nc/d&e\1f/' file  
sed: -e expression #1, char 16: unknown option to `s'
$ sed 's/a.*\/b{2,}\nc/d&e\1f/' file
sed: -e expression #1, char 23: invalid reference \1 on `s' command's RHS
$ sed 's/a.*\/b{2,}\nc/d&e\\1f/' file
a.*/b{2,}\nc
axx/bb\nc
# .... and so on, peeling the onion ad nauseum until:
$ sed 's/a\.\*\/b{2,}\\nc/d\&e\\1f/' file
d&e\1f
axx/bb\nc

or use the above tool:

$ sedstr 'a.*/b{2,}\nc' 'd&e\1f' file  
d&e\1f
axx/bb\nc

The reason this is useful is that it can be easily augmented to use word-delimiters to replace words if necessary, e.g. in GNU sed syntax:

sed "s/\<$escOld\>/$escNew/g" "$file"

whereas the tools that actually operate on strings (e.g. awk's index()) cannot use word-delimiters.

NOTE: the reason to not wrap \ in a bracket expression is that if you were using a tool that accepts [\]] as a literal ] inside a bracket expression (e.g. perl and most awk implementations) to do the actual final substitution (i.e. instead of sed "s/$escOld/$escNew/g") then you couldn't use the approach of:

sed 's/[^^]/[&]/g; s/\^/\\^/g'

to escape \ by enclosing it in [] because then \x would become [\][x] which means ] or [ or x. Instead you'd need:

sed 's/[^^\\]/[&]/g; s/\^/\\^/g; s/\\/\\\\/g'

So while [\] is probably OK for all current sed implementations, we know that \\ will work for all sed, awk, perl, etc. implementations and so use that form of escaping.

Teenybopper

Answer 3

It should be noted that the regular expression used in some answers above among this and that one:

's/[^^\\]/[&]/g; s/\^/\\^/g; s/\\/\\\\/g'

seems to be wrong:

• Doing first s/\^/\\^/g followed by s/\\/\\\\/g is an error, as any ^ escaped first to \^ will then have its \ escaped again.

A better way seems to be: 's/[^\^]/[&]/g; s/[\^]/\\&/g;'.

• [^^\\] with sed (BRE/ERE) should be just [^\^] (or [^^\]). \ has no special meaning inside a bracket expression and needs not to be quoted.

Technicolor

Good point, but instead of (only) posting a separate answer, you could have (also) left a comment on those answers. I've just updated mine based on your suggestion, but I suggest you let Ed know too. – Unlade

Answer 4

Bash parameter expansion can be used to escape a string for use as a Sed replacement string:

# Define a sample multi-line literal. Includes a trailing newline to test corner case
replace='a&b;c\1
d/e
'

# Escape it for use as a Sed replacement string.
: "${replace//\\/\\\\}"
: "${_//&/\\\&}"
: "${_//\//\\\/}"
: "${_//$'\n'/\\$'\n'}"
replaceEscaped=$_

# Output should match "$replace"
sed -n "s/.*/$replaceEscaped/p" <<<''

In bash 5.2+, it can be simplified further:

# Define a sample multi-line literal. Includes a trailing newline to test corner case
replace='a&b;c\1
d/e
'

# Escape it for use as a Sed replacement string.
shopt -s extglob
shopt -s patsub_replacement # An & in the replacement will expand to what matched. bash 5.2+
: "${replace//@(&|\\|\/|$'\n')/\\&}"
replaceEscaped=$_

# Output should match "$replace"
sed -n "s/.*/$replaceEscaped/p" <<<''

Encapsulate it in a bash function:

##
# escape_replacement -v var replacement
#
# Escape special characters in _replacement_ so that it can be
# used as the replacement part in a sed substitute command.
# Store the result in _var_.
escape_replacement() {
  if ! [[ $# = 3 && $1 = '-v' ]]; then
    echo "escape_replacement: invalid usage" >&2
    echo "escape_replacement: usage: escape_replacement -v var replacement" >&2
    return 1
  fi
  local -n var=$2 # nameref (requires Bash 4.3+)
  # We use the : command (true builtin) as a dummy command as we  
  # trigger a sequence of parameter expansions
  # We exploit that the $_ variable (last argument to the previous command
  # after expansion) contains the result of the previous parameter expansion
  : "${3//\\/\\\\}" # Backslash-escape any existing backslashes
  : "${_//&/\\\&}"  # Backslash-escape &
  : "${_//\//\\\/}" # Backslash-escape the delimiter (we assume /)
  : "${_//$'\n'/\\$'\n'}" # Backslash-escape newline
  var=$_ # Assign to the nameref
  # To support Bash older than 4.3, the following can be used instead of nameref
  #eval "$2=\$_" # Use eval instead of nameref https://mywiki.wooledge.org/BashFAQ/006
}

# Test the function
# =================

# Define a sample multi-line literal. Include a trailing newline to test corner case
replace='a&b;c\1
d/e
'

escape_replacement -v replaceEscaped "$replace"

# Output should match "$replace"
sed -n "s/.*/$replaceEscaped/p" <<<''

Stylet