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.
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 '/--.*$/d' infile
or simply:
sed '/^--/d' infile
and more accurately:
sed '/^[[:blank:]]*--/d' infile
I can't seem to get a substring correctly.
declare BRANCH_NAME="bugfix/US3280841-something-duh";
# Trim it down to "US3280841"
TRIMMED=$(echo $BRANCH_NAME | sed -e 's/\(^.*\)\/[a-z0-9]\|[A-Z0-9]\+/\1/g')
That still returns bugfix/US3280841-something-duh.
If I try an use perl instead:
declare BRANCH_NAME="bugfix/US3280841-something-duh";
# Trim it down to "US3280841"
TRIMMED=$(echo $BRANCH_NAME | perl -nle 'm/^.*\/([a-z0-9]|[A-Z0-9])+/; print $1');
That outputs nothing.
What am I doing wrong?
Using bash parameter expansion only:
$: # don't use caps; see below.
$: declare branch="bugfix/US3280841-something-duh"
$: tmp="${branch##*/}"
$: echo "$tmp"
US3280841-something-duh
$: trimmed="${tmp%%-*}"
$: echo "$trimmed"
US3280841
Which means:
$: tmp="${branch_name##*/}"
$: trimmed="${tmp%%-*}"
does the job in two steps without spawning extra processes.
In sed,
$: sed -E 's#^.*/([^/-]+)-.*$#\1#' <<< "$branch"
This says "after any or no characters followed by a slash, remember one or more that are not slashes or dashes, followed by a not-remembered dash and then any or no characters, then replace the whole input with the remembered part."
Your original pattern was
's/\(^.*\)\/[a-z0-9]\|[A-Z0-9]\+/\1/g'
This says "remember any number of anything followed by a slash, then a lowercase letter or a digit, then a pipe character (because those only work with -E), then a capital letter or digit, then a literal plus sign, and then replace it all with what you remembered."
GNU's manual is your friend. I look stuff up all the time to make sure I'm doing it right. Sometimes it still takes me a few tries, lol.
An aside - try not to use all-capital variable names. That is a convention that indicates it's special to the OS, like RANDOM or IFS.
You may use this sed:
sed -E 's~^.*/|-.*$~~g' <<< "$BRANCH_NAME"
US3280841
Ot this awk:
awk -F '[/-]' '{print $2}' <<< "$BRANCH_NAME"
US3280841
sed 's:[^/]*/\([^-]*\)-.*:\1:'<<<"bugfix/US3280841-something-duh"
Perl version just has + in wrong place. It should be inside the capture brackets:
TRIMMED=$(echo $BRANCH_NAME | perl -nle 'm/^.*\/([a-z0-9A-Z]+)/; print $1');
Just use a ^ before A-Z0-9
TRIMMED=$(echo $BRANCH_NAME | sed -e 's/\(^.*\)\/[a-z0-9]\|[^A-Z0-9]\+/\1/g')
in your sed case.
Alternatively and briefly, you can use
TRIMMED=$(echo $BRANCH_NAME | sed "s/[a-z\/\-]//g" )
too.
type on shell terminal
$ BRANCH_NAME="bugfix/US3280841-something-duh"
$ echo $BRANCH_NAME| perl -pe 's/.*\/(\w\w[0-9]+).+/\1/'
use s (substitute) command instead of m (match)
perl is a superset of sed so it'd be identical 'sed -E' instead of 'perl -pe'
Another variant using Perl Regular Expression Character Classes (see perldoc perlrecharclass).
echo $BRANCH_NAME | perl -nE 'say m/^.*\/([[:alnum:]]+)/;'
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'
Currently I'm trying to use sed with regex on Solaris but it doesn't work.
I need to show only lines matching to my regex.
sed -n -E '/^[a-zA-Z0-9]*$|^a_[a-zA-Z0-9]*$/p'
input file:
grtad
a_pitr
_aupa
a__as
baman
12353
ai345
ki_ag
-MXx2
!!!23
+_)#*
I want to show only lines matching to above regex:
grtad
a_pitr
baman
12353
ai345
Is there another way to use alternative? Is it possible in perl?
Thanks for any solutions.
With Perl
perl -ne 'print if /^(a_)?[a-zA-Z0-9]*$/' input.txt
The (a_)? matches a_ one-or-zero times, so optionally. It may or may not be there.
The (a_) also captures the match, what is not needed. So you can use (?:a_)? instead. The ?: makes () only group what is inside (so ? applies to the whole thing), but not remember it.
with grep
$ grep -xiE '(a_)?[a-z0-9]*' ip.txt
grtad
a_pitr
baman
12353
ai345
-x match whole line
-i ignore case
-E extended regex, if not available, use grep -xi '\(a_\)\?[a-z0-9]*'
(a_)? zero or one time match a_
[a-z0-9]* zero or more alphabets or numbers
With sed
sed -nE '/^(a_)?[a-zA-Z0-9]*$/p' ip.txt
or, with GNU sed
sed -nE '/^(a_)?[a-z0-9]*$/Ip' ip.txt
Looking for suggestion to cat file | grep REGEX to get the lines with <version>anything</version>.
grep -F '<version>1.1.9-beta</version>' file
-F will match your pattern as literal text
you don't need that useless cat
if you really mean anything: try grep '<version>.*</version>' file or grep -P '<version>.*?</version>' file , however searching xml with regex is bad idea.
Use the -E option to match a regular expression:
grep -E "<version>.*</version>" file
Refer to these rules for the regular expression: https://www.gnu.org/savannah-checkouts/gnu/grep/manual/grep.html#Regular-Expressions
For example, to match the typical version format (3.14, or 13.14, or 0.1458) you can type:
grep -E "<version>[0-9]?\.[0-9]?</version>" file
You can do:
grep '<version>[^<]*</version>' file.xml
[^<]* will match zero or more characters upto next <.