echo test.a.wav|sed 's/[^(.wav)]*//g'
.a.wav
What I want is to remove every character until it reaches the whole group .wav(that is, I want the result to be .wav), but it seems that sed would remove every character until it reaches any of the four characters. How to do the trick?
Groups do not work inside [], so the dot is part of the class as is the parens.
How about:
echo test.a.wav|sed 's/.*\(\.wav\)/\1/g'
Note, there may be other valid solutions, but you provide no context on what you are trying to do to determine what may be the best solution.
The feature you're requesting wouldn't be supported by sed (negative lookahead) but Perl does the trick.
$ echo 'test.a.wav' | perl -pe 's/^(?:(?!\.wav).)*//g'
.wav
Instead of regex, you can use awk like this:
echo test.a.wav.more | awk -F".wav" '{print FS$2}'
.wav.more
It splits the data with your pattern, then print pattern and the rest of the data.
This might work for you (GNU sed):
sed ':a;/^\.wav/!s/.//;ta;/./!d' file
or:
sed 's/\.wav/\n&/;s/^[^\n]*\n//;/./!d' file
N.B. This deletes the line if it is empty. If this is not wanted just remove /./!d from the above commands.
Related
I have a text file with the following pattern written to it:
TIME[32.468ms] -(3)-............."TEXT I WANT TO KEEP"
I would like to discard the first part of each line containing
TIME[32.468ms] -(3)-.............
To test the regular expression I've tried the following:
cat myfile.txt | egrep "^TIME\[.*\]\s\s\-\(3\)\-\.+"
This identifies correctly the lines I want. Now, to delete the pattern I've tried:
cat myfile.txt | sed s/"^TIME\[.*\]\s\s\-\(3\)\-\.+"//
but it just seems to be doing the cat, since it shows the content of the complete file and no substitution happens.
What am I doing wrong?
OS: CentOS 7
With your shown samples, please try following grep command. Written and tested with GNU grep.
grep -oP '^TIME\[\d+\.\d+ms\]\s+-\(\d+\)-\.+\K.*' Input_file
Explanation: Adding detailed explanation for above code.
^TIME\[ ##Matching string TIME from starting of value here.
\d+\.\d+ms\] ##Matching digits(1 or more occurrences) followed by dot digits(1 or more occurrences) followed by ms ] here.
\s+-\(\d+\)-\.+ ##Matching spaces91 or more occurrences) followed by - digits(1 or more occurrences) - and 1 or more dots.
\K ##Using \K option of GNU grep to make sure previous match is found in line but don't consider it in printing, print next matched regex part only.
.* ##to match till end of the value.
2nd solution: Adding awk program here.
awk 'match($0,/^TIME\[[0-9]+\.[0-9]+ms\][[:space:]]+-\([0-9]+\)-\.+/){print substr($0,RSTART+RLENGTH)}' Input_file
Explanation: using match function of awk, to match regex ^TIME\[[0-9]+\.[0-9]+ms\][[:space:]]+-\([0-9]+\)-\.+ which will catch text which we actually want to remove from lines. Then printing rest of the text apart from matched one which is actually required by OP.
This awk using its sub() function:
awk 'sub(/^TIME[[][^]]*].*\.+/,"")' file
"TEXT I WANT TO KEEP"
If there is replacement, sub() returns true.
$ cut -d'"' -f2 file
TEXT I WANT TO KEEP
You may use:
s='TIME[32.468ms] -(3)-............."TEXT I WANT TO KEEP"'
sed -E 's/^TIME\[[^]]*].*\.+//'
"TEXT I WANT TO KEEP"
The \s regex extension may not be supported by your sed.
In BRE syntax (which is what sed speaks out of the box) you do not backslash round parentheses - doing that turns them into regex metacharacters which do not match themselves, somewhat unintuitively. Also, + is just a regular character in BRE, not a repetition operator (though you can turn it into one by similarly backslashing it: \+).
You can try adding an -E option to switch from BRE syntax to the perhaps more familiar ERE syntax, but that still won't enable Perl regex extensions, which are not part of ERE syntax, either.
sed 's/^TIME\[[^][]*\][[:space:]][[:space:]]-(3)-\.*//' myfile.txt
should work on any reasonably POSIX sed. (Notice also how the minus character does not need to be backslash-escaped, though doing so is harmless per se. Furthermore, I tightened up the regex for the square brackets, to prevent the "match anything" regex you had .* from "escaping" past the closing square bracket. In some more detail, [^][] is a negated character class which matches any character which isn't (a newline or) ] or [; they have to be specified exactly in this order to avoid ambiguity in the character class definition. Finally, notice also how the entire sed script should normally be quoted in single quotes, unless you have specific reasons to use different quoting.)
If you have sed -E or sed -r you can use + instead of * but then this complicates the overall regex, so I won't suggest that here.
A simpler one for sed:
sed 's/^[^"]*//' myfile.txt
If the "text you want to keep" always surrounded by the quote like this and only them having the quote in the line starting with "TIME...", then:
sed -n '/^TIME/p' file | awk -F'"' '{print $2}'
should get the line starting with "TIME..." and print the text within the quotes.
Thanks all, for your help.
By the end, I've found a way to make it work:
echo 'TIME[32.468ms] -(3)-.............TEXT I WANT TO KEEP' | grep TIME | sed -r 's/^TIME\[[0-9]+\.[0-9]+ms\]\s\s-\(3\)-\.+//'
More generally,
grep TIME myfile.txt | sed -r ‘s/^TIME\[[0-9]+\.[0-9]+ms\]\s\s-\(3\)-\.+//’
Cheers,
Pedro
I'm trying to remove everything but 3 separate lines with specific matching pattern and leave just the 3 lines I want
Here is my code;
sed -n '/matching pattern/matching pattern/matching pattern/p' > file.txt
If you have multiple commands on the same line, you need to separate the commands by a ;:
sed -n '/matching pattern/p;/matching pattern2/p;/matching pattern3/p' file
Alternatively you can put them onto separate lines:
sed -n '/matching pattern/p
/matching pattern2/p
/matching pattern3/p' file
Beside that, you can also use regex alternation:
sed -rn '/(pattern|pattern2|pattern3)/p' file
or (better) use grep:
grep -E '(pattern|pattern2|pattern3)' file
However, this might get messy if the patterns getting longer and more complicated.
awk to the rescue!
awk '/pattern1/ || /pattern2/ || /pattern3/' filename
I think it's cleaner than alternatives.
Sed with Deletion
There's always more than one way to do this sort of thing, but one useful sed programming pattern is using alternation with deletion. For example:
# BSD sed
sed -E '/root|daemon|nobody/!d' /etc/passwd
# GNU sed
sed -r '/root|daemon|nobody/!d' /etc/passwd
This makes it possible to express ideas like "delete everything except for the listed terms." Even when expressions are functionally equivalent, it can be helpful to use a construct that most closely matches the idea you're trying to convey.
This might work for you (GNU sed):
sed '/pattern1/b;/pattern2/b;/pattern3/b;d' file
The normal flow of sed is to print what remains in the pattern space after processing. Therefore if the required pattern is in the pattern space let sed do its thing otherwise delete the line.
N.B. the b command is like a goto and if it has no following identifier, it means break out of any further sed commands and print (or not print if the -n option is in action) the contents of the pattern space.
If I understood you correctly:
sed -n '/\(pattern1\|pattern2\|pattern3\)/p' file > newfile
I'd like to grab the digits in a string like so :
"sample_2341-43-11.txt" to 2341-43-11
And so I tried the following command:
echo "sample_2341-43-11.txt" | sed -n -r 's|[0-9]{4}\-[0-9]{2}\-[0-9]{2}|\1|p'
I saw this answer, which is where I got the idea.
Use sed to grab a string, but it doesn't work on my machine:
it gives an error "illegal option -r".
it doesn't like the \1, either.
I'm using sed on MacOSX yosemite.
Is this the easiest way to extract that information from the file name?
You need to set your grouping and match the rest of the line to remove it with the group. Also the - does not need to be escaped. And the -n will inhibit the output (It just returns exit level for script conditionals).
echo "sample_2341-43-11.txt" | sed -r 's/^.*([0-9]{4}-[0-9]{2}-[0-9]{2}).*$/\1/'
Enhanced regular expressions are not supported in the Mac version of sed.
You can use grep instead:
echo "sample_2341-43-11.txt" | grep -Eo "((\d+|-)+)"
OUTPUT
2341-43-11
echo "one1sample_2341-43-11.txt" \
| sed 's/[^[:digit:]-]\{1,\}/ /g;s/ \{1,\}/ /g;s/^ //;s/ $//'
1 2341-43-11
Extract all numbers(digit) completed with - (thus allow here --12 but can be easily treated)
posix compliant
all number of the line are on same line (if several) separate by a space character (could be changed to new line if wanted)
You can try this ways also
sed 's/[^_]\+_\([^.]\+\).*/\1/' <<< sample_2341-43-11.txt
OutPut:
2341-43-11
Explanation:
[^_]\+ - Match the content untile _ ( sample_)
\([^.]\+\) - Match the content until . and capture the pattern (2341-43-11)
.* - Discard remaining character (.txt)
You can go with what the poster above said. Well, making use of this
pattern "\d+-\d+-\d+" would match what you are looking for. See demo here
https://regex101.com/r/kO2cZ1/3
I need to make a substitution using Sed or other program. I have these patterns <ehh> <mmm> <mhh> repeated at the beginning of a sentences and I need to substitute for nothing.
I am trying this:
echo "$line" | sed 's/<[a-zA-z]+>//g'
But I get the same result, nothing changes. Anyone can help?
Thank you!
For me, for the test file
<ahh> test
<mmm>test 1
the following
sed 's/^<[a-zA-Z]\+>//g' testfile
produces
test
test 1
which seems to be what you want. Note that for basic regular expressions, you use \+ whereas for extended regular expressions, you use + (and need to use the -r switch for sed).
NB: I added a ^to the check since you said: at the beginning of the line.
echo '<ehh> <mmm> <mhh>blabla bla' | \
sed '^Js/^\([[:space:]]*\<[a-zA-Z]\{3\}\>\)\{1,\}//'
remove all starting occurence of your pattern (including heading space)
I escape & to be sure due to sed meaning of this character in pattern (work without on my AIX)
I don't use g because it remove several occurence of full pattern and there is only 1 begin (^) and use a multi occurence counter with group instead \(\)\{1,\}
If the goal is to get the last parameter from lines like this:
<ahh> test
<mmm>test 1
You can do:
awk -F\; '/^<[[:alpha:]]+>/ {print $NF}' <<< "$line"
test
test 1
It will search for pattern <[[:alpha:]]+> and print last field on line, separated by ;
I am trying to remove As at the end of line.
alice$ cat pokusni
SALALAA
alice$ sed -n 's/\(.*\)A$/\1/p' pokusni
SALALA
one A is removed just fine
alice$ sed -n 's/\(.*\)A+$/\1/p' pokusni
alice$ sed -n 's/\(.*\)AA*$/\1/p' pokusni
SALALA
multiple occurrences not:(
I am probably doing just some very stupid mistake, any help? Thanks.
Try this one 's/\(.*[^A]\)AA*$/\1/p'
Why + does not work:
Because it is just a normal character here.
Why 's/\(.*\)AA*$/\1/p' does not work:
Because the reg-ex engine is eager, so .* would consume as many as As except the final A specified in AA*. And A* will just match nothing.
This might work for you:
sed -n 's/AA*$//p' file
This replaces an A and zero or more A's at the end of line with nothing.
N.B.
sed -n 's/A*$//p file'
would produce the correct string however it would operate on every line and so produce false positives.
Using awk
awk '{sub(/AA$/,"A")}1' pokusni
SALALA
EDIT
Correct version, removing all A from end of line.
awk '{sub(/A*$/,x)}1' pokusni
You can use perl:
> echo "SALALAA" | perl -lne 'if(/(.*?)[A]+$/){print $1}else{print}'
SALAL