Replace the separator between pairs of numbers - regex

I want to replace all strings like [0-9][0-9]-[0-9][0-9] with [0-9][0-9]/[0-9][0-9] using sed.
In other words, I want to replace - with /.
If I have somewhere in my text:
09-36
32-43
54-65
I want this change:
09/36
32/43
54/65

Using GNU sed:
$ echo '09-36 32-43 54-65' | sed -r 's|\<([0-9]{2})-([0-9]{2})\>|\1/\2|g'
09/36 32/43 54/65
-r turns on extended regular expressions, which:
doesn't require \-escaping ( ) { } char.
enables use of \< and /> to only match at word boundaries (if the expression should only match full lines, use ^ and $ instead, and omit the g option)
| is used as an alternative regex delimiter so that / can be used without \-escaping.
A BSD/macOS sed solution would look slightly different:
echo '09-36 32-43 54-65' | sed -E 's|[[:<:]]([0-9]{2})-([0-9]{2})[[:>:]]|\1/\2|g'

sed -e 's/\([0-9]\{2\}\)-\([0-9]\{2\}\)/\1\/\2/g'
Might not be the most elegant version, but works for me. The gazillion backslashes make this rather unreadable in my opinion. You might improve the readability by not using / to separate the pattern and the replacement maybe?

perl -C -npe 's/(?<!\d)(\d\d)-(\d\d)(?!\d)/\1\/\2/g' file
Input
维基 1-11 22-33 444-44 55-555 66-66百科
77-77
8 88-88
Output
维基 1-11 22/33 444-44 55-555 66/66百科
77/77
8 88/88
In the command above
-C enables Unicode;
-n causes Perl to process the script for each input line;
-p causes Perl to print the result of the script to the standard output;
-e accepts a Perl expression (particularly, it is a substitution).
In this mode (-npe), Perl works just like sed. The script substitutes each pair of digits separated with - to the same pair separated with a slash.
(?<!\d) and (?!\d) are negative lookaround expressions.
To edit the file in place use -i option: perl -C -i.backup -npe ....
If the input is not a file, you can pass the input to Perl via pipe, e.g.:
echo '维基 1-11 22-33 444-44 55-555 66-66百科' | \
perl -C -npe 's/(?<!\d)(\d\d)-(\d\d)(?!\d)/\1\/\2/g'

Related

Why does this regex work in grep but not sed?

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

Is it possible to escape regex metacharacters reliably with sed

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" <<<''

Retrieve value of attribute in bash

I have a list of lines:
<some_random_text="someval" my_val_="0.4" some_random_text_1="someval_">
<some_random_text="someval" my_val_="0.8" some_random_text_1="someval_">
<some_random_text="someval" my_val_="1.2" some_random_text_1="someval_">
and so on.
From each line, I want to return the numeric value given after my_val_. How can I do this in bash?
Within this very rigid structure, what you want to do is quite easy using sed:
sed 's/.*my_val_="\([0-9.]\{1,\}\)".*/\1/' file
or using extended regular expressions:
sed -r 's/.*my_val_="([0-9.]+)".*/\1/' file
This captures the part you're interested in (the digits and dots between the quotes) and uses them to replace the contents of the line.
As mentioned in the comments (thanks), the switch to enable extended regular expressions differs between versions of sed. Out of habit, I tend to use -r but some implementations (such as BSD sed on OSX) work with -E instead. Others work with either -r or -E but neither option is defined by the standard.
This could also be done in native bash (although I wouldn't recommend it...):
re='my_val_="([0-9.]+)"'
while read -r line; do
[[ $line =~ $re ]] && echo "${BASH_REMATCH[1]}"
done < file
=~ is the regex match operator. The captured digits and dots are stored in element 1 of the special array BASH_REMATCH.
The sed and bash approaches are subtly different, as the sed version will print all lines in the file, even if they don't match the pattern. If this is a problem, you can add the -n switch and a p at the end of the command to print matching lines:
sed -nr 's/.*my_val_="([0-9.]+)".*/\1/p' file
With grep:
grep -oP 'my_val_="\K[^"]*' filename
-o so that grep only prints only the match, -P so that Perl-compatible regexes are used.
The \K in the regex removes from the match everything that was matched by the part of the regex that came before it; this has the effect of a lookbehind: only non-quote characters that come directly after my_val_=" are matched.

Difference between grep and perl regex?

I have a problem with what I think is a difference in grep's regex and perl's regex. Consider the following little test:
$ cat testfile.txt
A line of text
SOME_RULE = $(BIN)
Another line of text
$ grep "SOME_RULE\s*=\s*\$(BIN)" testfile.txt
SOME_RULE = $(BIN)
$ perl -p -e "s/SOME_RULE\s*=\s*\$(BIN)/Hello/g" testfile.txt
A line of text
SOME_RULE = $(BIN)
Another line of text
As you can see, using the regex "SOME_RULE\s*=\s*$(BIN)", grep could find the match, but perl was unable to update the file using the same expression. How should I solve this problem?
Perl wants the '(' and ')' to be escaped. Also, the shell eats the '\' on the '$', so you need:
$ perl -p -e "s/SOME_RULE\s*=\s*\\$\(BIN\)/Hello/g" testfile.txt
(or use single quotes--which is highly advisable in any case.)
You need to escape ( and )(Capturing group).
perl -p -e 's/SOME_RULE\s*=\s*\$\(BIN\)/Hello/g' testfile.txt
Actually you need it in Extended Regular Expression(ERE):
grep -E "SOME_RULE\s*=\s*\$\(BIN\)" testfile.txt
perl -ne '(/SOME_RULE\s*?=\s*?\$\(BIN\)/) && print' testfile.txt
If you want to modify use
perl -pe 's/SOME_RULE\s*?=\s*?\$\(BIN\)/Hello/' testfile.txt
Perl's regex syntax is different to the POSIX regexes used by grep. In this case, you're falling foul of parentheses being metacharacters in Perl's regexes - they denote a capturing group.
You should have more success by altering the Perl regex:
s/SOME_RULE\s*=\s*\$\(BIN\)/Hello/g
which will then match the literal parentheses in the source text.

put regular expression in variable

output=`grep -R -l "${images}" *`
new_output=`regex "slide[0-9]" $output`
Basically $output is a string like this:
slides/_rels/slide9.xml.rels
The number in $output will change. I want to grab "slide9" and put that in a variable. I was hoping new_output would do that but I get a command not found for using regex. Any other options? I'm using a bash shell script.
Well, regex is not a program like grep. ;)
But you can use
grep -Eo "(slide[0-9]+)"
as a simple approach. -o means: show only the matching part, -E means: extended regex (allows more sophisticated patterns).
Reading I want to grab "slide9" and put that in a variable. I assume you want what matches your regexp to be the only thing put in $new_output? If so, then you can change that to:
new_output=`egrep -R -l "${images}" * | sed 's/.*\(slide[0-9]+\).*/\1/'`
Note no setting of output= is required (unless you use that for something else)
If you need $output to use elsewhere then instead use:
output=`grep -R -l "${images}" *`
new_output=`echo ${ouput} | sed 's/.*\(slide[0-9]+\).*/\1/'`
sed's s/// command is similar to perls s// command and has an equivalent in most languages.
Here I'm matching zero or more characters .* before and after your slide[0-9]+ and then remembering (backrefrencing) the result \( ... \) in sed (the brackets may or may not need to be escaped depending on the version of sed). We then replace that whole match (i.e the whole line) with \1 which expands to the first captured result in this case your slide[0-9]+ match.
In these situations using awk is better :
output="`grep -R -l "main" codes`"
echo $output
tout=`echo $output | awk -F. '{for(i=1;i<=NF;i++){if(index($i,"/")>0){n=split($i,ar,"/");print ar[n];}}}'`
echo $tout
This prints the filename without the extension. If you want to grab only slide9 than use the solutions provided by others.
Sample output :
A#A-laptop ~ $ bash try.sh
codes/quicksort_iterative.cpp codes/graham_scan.cpp codes/a.out
quicksort_iterative graham_scan a