I have to replace following String
//#Config(manifest
with below string,
#Config(manifest
So this i created following regex
\/\/#Config\(manifest
And tried
grep -rl \/\/#Config\(manifest . | xargs sed -i "\/\/#Config\(manifest#Config\(manifest/g"
But i am getting following error:
sed: -e expression #1, char 38: Unmatched ( or \(
I have to search recursively and do this operation, though i am stuck with above error.
grep -rl '//#Config(manifest' | xargs sed -i 's|//#Config(manifest|#Config(manifest|g'
Specifying . for current directory is optional for grep -r
sed allows Any character other than backslash or newline to be used as delimiter
Edit
If file name contains spaces, use
grep -rlZ '//#Config(manifest' | xargs -0 sed -i 's|//#Config(manifest|#Config(manifest|g'
Explanation (assumes GNU version of commands)
grep
-r performs recursive search
-l option outputs only filenames instead of matched patterns
-Z outputs a zero byte (ASCII NUL character) after each file name instead of usual newline
'pattern' by default, grep uses BRE (basic regular expression) where characters like ( do not have special meaning and hence need not be escaped
xargs -0 tells xargs to separate arguments by the ASCII NUL character
sed
-i inplace edit, use -i.bkp if you want to create backup of original files
s|pattern|replace|g the g flag tells sed to search and replace all occurrences. sed also defaults to BRE and so no need to escape (. Using \( would mean start of capture groups and hence the error when it doesn't find the closing \)
Related
I have two regular expressions:
$ grep -E '\-\- .*$' *.sql
$ sed -E '\-\- .*$' *.sql
(I am trying to grep lines in sql files that have comments and remove lines in sql files that have comments)
The grep command works using this regex; however, the sed returns the following error:
sed: -e expression #1, char 7: unterminated address regex
What am I doing incorrectly with sed?
(The space after the two hyphens is required for sql comments if you are unfamiliar with MySql comments of this type)
You're trying to use:
sed -E '\-\- .*$' *.sql
Here sed command is not correct because you're not really telling sed to do something.
It should be:
sed -n '/-- /p' *.sql
and equivalent grep would be:
grep -- '-- ' *.sql
or even better with a fixed string search:
grep -F -- '-- ' *.sql
Using -- to separate pattern and arguments in grep command.
There is no need to escape - in a regex if it is outside bracket expression (or character class) i.e. [...].
Based on comments below it seems OP's intent is to remove commented section in all *.sql files that start with 2 hyphens.
You may use this sed for that:
sed -i 's/-- .*//g' *.sql
The problem here is not the regex, the problem is that sed requires a command. The equivalent of your grep would be:
sed -n '/\-\- .*$/p'
You suppress output for non-matching lines -n ... you search (wrap your regex in slashes) and you print p (after the last slash).
P.S.: As Anub pointed out, escaping the hyphens - inside the regex is unnecessary.
You are trying to use sed's \cregexpc syntax where with \-<...> you are telling sed the delimiter character you want use is a dash -, but you didn't terminate it where it should be: \-<...>- also add d command to delete those lines.
sed '\-\-\-.*$-d' infile
see man sed about that:
\cregexpc
Match lines matching the regular expression regexp. The c may be any character.
if default / was used this was not required so:
sed '/--.*$/d' infile
or simply:
sed '/^--/d' infile
and more accurately:
sed '/^[[:blank:]]*--/d' infile
I have the example text:
"are "insulin sensitizers" "
I am trying to use the below command to find and replace the xml quote command with a single quote but it only works for the first one and the second that follows "sensitizers" is left unchanged.
command:
grep -rl """ ./ | xargs sed -i "s/"/'/"
result:
"are 'insulin sensitizers" "
desired result:
"are 'insulin sensitizers' "
You will need to use g flag in sed for global substitution:
grep -Zirl """ . | xargs -0 sed -i "s/"/'/g"
Also note use of -Z option in grep and -0 in xargs to take care of files with special characters and whitespaces.
As per man grep:
-Z, --null
Output a zero byte (the ASCII NUL character) instead of the character that normally follows a file name.
I have the following lines in an apache access log
/sms/receiveHLRLookup?Ported=No&Status=Success&MSISDN=647930229655&blah
/sms/receiveHLRLookup?Ported=No&Status=Success&MSISDN=647930229656&blah
/sms/receiveHLRLookup?Ported=No&Status=Success&MSISDN=647930229657&blah
/sms/receiveHLRLookup?Ported=No&Status=Success&MSISDN=647930229658&blah
and i want to extract the MSISDN value only, so expected output would be
647930229655
647930229656
647930229657
647930229658
I'm using the following sed command but i can't get it to stop capturing at &
sed 's/.*MSISDN=\(.*\)/\1/'
sed solution:
sed -E 's/.*&MSISDN=([^&]+).*/\1/' file
& - is key/value pair separator in URL syntax, so you should rely on it
([^&]+) - 1st captured group containing any character sequence except &
\1 - backreference to the 1st captured group
The output:
647930229655
647930229656
647930229657
647930229658
-o : means print only matching string not the whole line.
-P: To enable pcre regex.
\K: means ignore everything on the left. But should be part of actual input string.
\d: means digit, + means one or more digit.
grep -oP 'MSISDN=\K\d+' input
647930229655
647930229656
647930229657
647930229658
Following simple sed may help you on same.
sed 's/.*MSISDN=//;s/&.*//' Input_file
Explanation:
s/.*MSISDN=//: s means substitute .*MSISDN= string with // NULL in current line.
; semi colon tells sed that there is 1 more statement to be executed.
s/&.*//g': s/&.*// means substitute &.* from & to everything with NULL.
$ grep -oP '(?<=&MSISDN=)\d+' file
647930229655
647930229656
647930229657
647930229658
-o option is meant to show only matched output
-P option is meant to enable PCRE (Perl Compatible Regex)
(?<=regex) this is to enforce positive look behind assertion. You can read more about them over here. Lookarounds dont consume any characters while matching unlike normal regex. Hence the only matched output you get it \d+ which is 1 or more digits.
or using sed:
$ sed -r 's/^.*MSISDN=([0-9]+).*$/\1/' file
647930229655
647930229656
647930229657
647930229658
you can also pipe cut to cut
cut -d '&' -f3 Input_file |cut -d '=' -f2
$ cat file
anna
amma
kklks
ksklaii
$ grep '\`' file
anna
amma
kklks
ksklaii
Why? How is that match working ?
This appears to be a GNU extension for regular expressions. The backtick ('\`') anchor matches the very start of a subject string, which explains why it is matching all lines. OS X apparently doesn't implement the GNU extensions, which would explain why your example doesn't match any lines there. See http://www.regular-expressions.info/gnu.html
If you want to match an actual backtick when the GNU extensions are in effect, this works for me:
grep '[`]' file
twm's answer provides the crucial pointer, but note that it is the sequence \`, not ` by itself that acts as the start-of-input anchor in GNU regexes.
Thus, to match a literal backtick in a regex specified as a single-quoted shell string, you don't need any escaping at all, neither with GNU grep nor with BSD/macOS grep:
$ { echo 'ab'; echo 'c`d'; } | grep '`'
c`d
When using double-quoted shell strings - which you should avoid for regexes, for reasons that will become obvious - things get more complicated, because you then must escape the ` for the shell's sake in order to pass it through as a literal to grep:
$ { echo 'ab'; echo 'c`d'; } | grep "\`"
c`d
Note that, after the shell has parsed the "..." string, grep still only sees `.
To recreate the original command with a double-quoted string with GNU grep:
$ { echo 'ab'; echo 'c`d'; } | grep "\\\`" # !! BOTH \ and ` need \-escaping
ab
c`d
Again, after the shell's string parsing, grep sees just \`, which to GNU grep is the start-of-the-input anchor, so all input lines match.
Also note that since grep processes input line by line, \` has the same effect as ^ the start-of-a-line anchor; with multi-line input, however - such as if you used grep -z to read all lines at once - \` only matches the very start of the whole string.
To BSD/macOS grep, \` simply escapes a literal `, so it only matches input lines that contain that character.
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.
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.
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 [ 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.
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.
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" <<<''