I'm trying to make a replace with sed:
cat myfile | grep router | sed -e 's/Custom devices \(DiY\) \[CC2530 router\]\(http:\/\/ptvo\.info\/cc2530-based-zigbee-coordinator-and-router-112\/\) \(CC2530\.ROUTER\)/CC2530 router/g'
The output of the piped grep is:
<text text-anchor="middle" x="455.5" y="-31.3" font-family="Times,serif" font-size="14.00" fill="#ffffff">Custom devices (DiY) [CC2530 router](http://ptvo.info/cc2530-based-zigbee-coordinator-and-router-112/) (CC2530.ROUTER)</text>
This works well:
cat myfile | grep Xiaomi | sed -e 's/Xiaomi Aqara temperature, humidity and pressure sensor/AqaraTHP/g'
What am I missing here?
The man page says that
-e command
Append the editing commands specified by the command argument to the
list of commands.
You need to use -E or -r option (whichever is supported by your sed):
cat myfile | grep router | sed -E 's/Custom devices \(DiY\) \[CC2530 router\]\(http:\/\/ptvo\.info\/cc2530-based-zigbee-coordinator-and-router-112\/\) \(CC2530\.ROUTER\)/CC2530 router/g'
With -E, \( denotes a plain ( symbol and ( is the indicator of capturing group.
Related
This is mostly by curiosity, I am trying to have the same behavior as:
echo -e "test1:test2:test3"| sed 's/:/\n/g' | grep 1
in a single sed command.
I already tried
echo -e "test1:test2:test3"| sed -e "s/:/\n/g" -n "/1/p"
But I get the following error:
sed: can't read /1/p: No such file or directory
Any idea on how to fix this and combine different types of commands into a single sed call?
Of course this is overly simplified compared to the real usecase, and I know I can get around by using multiple calls, again this is just out of curiosity.
EDIT: I am mostly interested in the sed tool, I already know how to do it using other tools, or even combinations of those.
EDIT2: Here is a more realistic script, closer to what I am trying to achieve:
arch=linux64
base=https://chromedriver.storage.googleapis.com
split="<Contents>"
curl $base \
| sed -e 's/<Contents>/<Contents>\n/g' \
| grep $arch \
| sed -e 's/^<Key>\(.*\)\/chromedriver.*/\1/' \
| sort -V > out
What I would like to simplify is the curl line, turning it into something like:
curl $base \
| sed 's/<Contents>/<Contents>\n/g' -n '/1/p' -e 's/^<Key>\(.*\)\/chromedriver.*/\1/' \
| sort -V > out
Here are some alternatives, awk and sed based:
sed -E "s/(.*:)?([^:]*1[^:]*).*/\2/" <<< "test1:test2:test3"
awk -v RS=":" '/1/' <<< "test1:test2:test3"
# or also
awk 'BEGIN{RS=":"} /1/' <<< "test1:test2:test3"
Or, using your logic, you would need to pipe a second sed command:
sed "s/:/\n/g" <<< "test1:test2:test3" | sed -n "/1/p"
See this online demo. The awk solution looks cleanest.
Details
In sed solution, (.*:)?([^:]*1[^:]*).* pattern matches an optional sequence of any 0+ chars and a :, then captures into Group 2 any 0 or more chars other than :, 1, again 0 or more chars other than :, and then just matches the rest of the line. The replacement just keeps Group 2 contents.
In awk solution, the record separator is set to : and then /1/ regex is used to only return the record having 1 in it.
This might work for you (GNU sed):
sed 's/:/\n/;/^[^\n]*1/P;D' file
Replace each : and if the first line in the pattern space contains 1 print it.
Repeat.
An alternative:
sed -Ez 's/:/\n/g;s/^[^1]*$//mg;s/\n+/\n/;s/^\n//' file
This slurps the whole file into memory and replaces all colons by newlines. All lines that do not contain 1 are removed and surplus newlines deleted.
An alternative to the really ugly sed is: grep -o '\w*2\w*'
$ printf "test1:test2:test3\nbob3:bob2:fred2\n" | grep -o '\w*2\w*'
test2
bob2
fred2
grep -o: only matching
Or: grep -o '[^:]*2[^:]*'
echo -e "test1:test2:test3" | sed -En 's/:/\n/g;/^[^\n]*2[^\n]*(\n|$)/P;//!D'
sed -n doesn't print unless told to
sed -E allows using parens to match (\n|$) which is newline or the end of the pattern space
P prints the pattern buffer up to the first newline.
D trims the pattern buffer up to the first newline
[^\n] is a character class that matches anything except a newline
// is sed shorthand for repeating a match
//! is then matching everything that didn't match previously
So, after you split into newlines, you want to make sure the 2 character is between the start of the pattern buffer ^ and the first newline.
And, if there is not the character you are looking for, you want to D delete up to the first newline.
At that point, it works for one line of input, with one string containing the character you're looking for.
To expand to several matches within a line, you have to ta, conditionally branch back to label :a:
$ printf "test1:test2:test3\nbob3:bob2:fred2\n" | \
sed -En ':a s/:/\n/g;/^[^\n]*2[^\n]*(\n|$)/P;D;ta'
test2
bob2
fred2
This is simply NOT a job for sed. With GNU awk for multi-char RS:
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' '/1/'
test1
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' 'NR%2'
test1
test3
test5
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' '!(NR%2)'
test2
test4
test6
$ echo "foo1:bar1:foo2:bar2:foo3:bar3" | awk -v RS='[:\n]' '/foo/ || /2/'
foo1
foo2
bar2
foo3
With any awk you'd just have to strip the \n from the final record before operating on it:
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS=':' '{sub(/\n$/,"")} /1/'
test1
I have this list of files:
$ more files
one_this_2017_1_abc.txt
two_that_2018_1_abc.txt
three_another_2017_10.abc.txt
four_again_2018_10.abc.txt
five_back_2018_1a.abc.txt
I would like to get this output:
one_this_XXXX_YY_abc.txt
two_that_XXXX_YY_abc.txt
three_another_XXXX_YY.abc.txt
four_again_XXXX_YY.abc.txt
five_back_XXXX_YY.abc.txt
I am trying to remove the year and the bit after the year and replace them with another string--this is to generate test cases.
I can get the year just fine, but it's that one or two character piece after it I can't seem to match.
This should work, right?
~/test_cases
$ cat files | sed -e 's/_[[:digit:]]\{4\}_/_XXXX_/' -e 's/_[[:alnum:]]\{1,2\}_/_YY_/'
one_this_XXXX_YY_abc.txt
two_that_XXXX_YY_abc.txt
three_another_XXXX_10.abc.txt
four_again_XXXX_10.abc.txt
five_back_XXXX_1a.abc.txt
Except it doesn't for the 2 character cases.
$ cat files | sed -e 's/_[[:digit:]]\{4\}_/_XXXX_/' -e 's/_[[:alnum:]]\
{2\}_/_YY_/'
one_this_XXXX_1_abc.txt
two_that_XXXX_1_abc.txt
three_another_XXXX_10.abc.txt
four_again_XXXX_10.abc.txt
five_back_XXXX_1a.abc.txt
Doesn't work for the two character cases either, and this works not at all (but according to the docs it should):
$ cat files | sed -e 's/_[[:digit:]]\{4\}_/_XXXX_/' -e 's/_[[:alnum:]]\+_/_YY_/'
one_YY_XXXX_1_abc.txt
two_YY_XXXX_1_abc.txt
three_YY_XXXX_10.abc.txt
four_YY_XXXX_10.abc.txt
five_YY_XXXX_1a.abc.txt
Other random experiments that don't work:
$ cat files | sed -e 's/_[[:digit:]]\{4\}_/_XXXX_/' -e 's/_[a-zA-Z0-9]\+_/_YY_/'
one_YY_XXXX_1_abc.txt
two_YY_XXXX_1_abc.txt
three_YY_XXXX_10.abc.txt
four_YY_XXXX_10.abc.txt
five_YY_XXXX_1a.abc.txt
$ cat files | sed -e 's/_[[:digit:]]\{4\}_/_XXXX_/' -e 's/_[a-zA-Z0-9]\{1\}_/_YY_/'
one_this_XXXX_YY_abc.txt
two_that_XXXX_YY_abc.txt
three_another_XXXX_10.abc.txt
four_again_XXXX_10.abc.txt
five_back_XXXX_1a.abc.txt
$ cat files | sed -e 's/_[[:digit:]]\{4\}_/_XXXX_/' -e 's/_[a-zA-Z0-9]\{2\}_/_YY_/'
one_this_XXXX_1_abc.txt
two_that_XXXX_1_abc.txt
three_another_XXXX_10.abc.txt
four_again_XXXX_10.abc.txt
five_back_XXXX_1a.abc.txt
Tried with both GNU sed version 4.2.1 under Linux and sed (GNU sed) 4.4 under Cygwin.
And yes, I realize I can pipe this through multiple sed calls to get it to work, but that regex SHOULD work, right?
if your Input_file is same as shown sample then following may help you in same.
sed 's/\([^_]*\)_\([^_]*\)_\(.*_\)\(.*\)/\1_\2_XXXX_YY_\4/g' Input_file
Output will be as follows.
one_this_XXXX_YY_abc.txt
two_that_XXXX_YY_abc.txt
three_another_XXXX_YY_10.abc.txt
four_again_XXXX_YY_10.abc.txt
five_back_XXXX_YY_1a.abc.txt
With sed, I can replace the first match in a line using
sed 's/pattern/replacement/'
And all matches using
sed 's/pattern/replacement/g'
How do I replace only the last match, regardless of how many matches there are before it?
Copy pasting from something I've posted elsewhere:
$ # replacing last occurrence
$ # can also use sed -E 's/:([^:]*)$/-\1/'
$ echo 'foo:123:bar:baz' | sed -E 's/(.*):/\1-/'
foo:123:bar-baz
$ echo '456:foo:123:bar:789:baz' | sed -E 's/(.*):/\1-/'
456:foo:123:bar:789-baz
$ echo 'foo and bar and baz land good' | sed -E 's/(.*)and/\1XYZ/'
foo and bar and baz lXYZ good
$ # use word boundaries as necessary - GNU sed
$ echo 'foo and bar and baz land good' | sed -E 's/(.*)\band\b/\1XYZ/'
foo and bar XYZ baz land good
$ # replacing last but one
$ echo 'foo:123:bar:baz' | sed -E 's/(.*):(.*:)/\1-\2/'
foo:123-bar:baz
$ echo '456:foo:123:bar:789:baz' | sed -E 's/(.*):(.*:)/\1-\2/'
456:foo:123:bar-789:baz
$ # replacing last but two
$ echo '456:foo:123:bar:789:baz' | sed -E 's/(.*):((.*:){2})/\1-\2/'
456:foo:123-bar:789:baz
$ # replacing last but three
$ echo '456:foo:123:bar:789:baz' | sed -E 's/(.*):((.*:){3})/\1-\2/'
456:foo-123:bar:789:baz
Further Reading:
Buggy behavior if word boundaries is used inside a group with quanitifiers - for example: echo 'it line with it here sit too' | sed -E 's/with(.*\bit\b){2}/XYZ/' fails
Greedy vs. Reluctant vs. Possessive Quantifiers
Reference - What does this regex mean?
sed manual: Back-references and Subexpressions
This might work for you (GNU sed):
sed 's/\(.*\)pattern/\1replacement/' file
Use greed to swallow up the pattern space and then regexp engine will step back through the line and find the first match i.e. the last match.
A fun way to do this, is to use rev to reverse the characters of each line and write your sed replacement backwards.
rev input_file | sed 's/nrettap/tnemecalper/' | rev
thanks in advance for the help.
I have the following line that does work on linux.
myfile (extract)
active_instance_count=
aq_tm_processes=1
archive_lag_target=0
audit_file_dest=?/rdbms/audit
audit_sys_operations=FALSE
audit_trail=NONE
background_core_dump=partial
background_dump_dest=/home1/oracle/app/oracle/admin/iopecom/bdump
...
cat myfile |sed -r 's/ {1,}//g'|sed -r 's/\t*//g' |grep -v "^#"|sed -s "/^$/d" |sed =|sed 'N;s/\n/\t/'|sed -r "s/#.*//g" | sed "s/\t/;/g"|sed "s/\t/;/g"|sed -e "s,',\o042,g"
The result will be:
1;O7_DICTIONARY_ACCESSIBILITY=TRUE
2;active_instance_count=
3;aq_tm_processes=1
4;archive_lag_target=0
5;audit_file_dest=?/rdbms/audit
6;audit_sys_operations=FALSE
7;audit_trail=NONE
8;background_core_dump=partial
9;background_dump_dest=/home1/oracle/app/oracle/admin/iopecom/bdump
But, I can't figure out, how to perform the same command on AIX server.
Help is very welcome.
Regards.
Antonio.
Unless you have a compelling reason to use sed, you could use alternate tools:
awk -v OFS=';' '{print NR,$0}' filename
would produce the desired output.
You could also use perl:
perl -ne 'print "$.;$_"' filename
It appears that your sed expression would skip lines beginning with a #. As such, you could say:
perl -ne '$,=";"; !/^#/ && print ++$i,$_' filename
or something like:
grep -v '^#' filename | awk ...
reformatting your pipeline:
cat myfile |
sed -r 's/ {1,}//g' | # strip all spaces (1)
sed -r 's/\t*//g' | # strip all tabs (2)
grep -v "^#" | # delete all lines beginning `#` (3)
sed -s "/^$/d" | # delete all empty lines (4)
sed = | # interleave with line numbers (5)
sed 'N;s/\n/\t/' | # join line number and line with `\t` (6)
sed -r "s/#.*//g" | # strip all `#` comments (7)
sed "s/\t/;/g" | # replace all tabs with `;` (8)
sed "s/\t/;/g" | # do it again (9)
sed -e "s,',\o042,g" # replace all ' with " (10)
Boiling that down and using cat -n to provide the line numbers up front gets:
cat -n myfile |
sed "$(print 's/\t/;/')
$(print 's/[ \t]*//g')
s/#.*//g
/^$/d
s/'/\"/g"
which behaves identically unless I'm misreading the aix docs. The $(...) construction is command substitution, it runs that command and substitutes its output. print would be printf on linux.
I'm running a command line like this:
filename_listing_command | xargs -0 action_command
Where filename_listing_command uses null bytes to separate the files -- this is what xargs -0 wants to consume.
Problem is that I want to filter out some of the files. Something like this:
filename_listing_command | sed -e '/\.py/!d' | xargs ac
but I need to use xargs -0.
How do I change the line separator that sed wants from newline to NUL?
If you've hit this SO looking for an answer and are using GNU sed 4.2.2 or later, it now has a -z option which does what the OP is asking for.
Pipe it through grep:
filename_listing_command | grep -vzZ '\.py$' | filename_listing_command
The -z accepts null terminators on input and the -Z produces null terminators on output and the -v inverts the match (excludes).
Edit:
Try this if you prefer to use sed:
filename_listing_command | sed 's/[^\x0]*\.py\x0//g' | filename_listing_command
If none of your file names contain newline, then it may be easier to read a solution using GNU Parallel:
filename_listing_command | grep -v '\.py$' | parallel ac
Learn more about GNU Parallel http://www.youtube.com/watch?v=OpaiGYxkSuQ
With help of Tom Hale and that answer we have:
sed -nzE "s/^$PREFIX(.*)/\1/p"