I use sed to get the content of file from a desire point but I have a problem.
I can not print $variable value into this sed command
count=$(sed -n '/$variable/,$p' file.log | grep '"KO"' -c)
I try with double quotes and close the single but not working
count=$(sed -n "/$variable/,$p" file.log | grep '"KO"' -c) ERROR unexpected `,'
count=$(sed -n '/'$variable'/,$p' file.log | grep '"KO"' -c) ERROR unterminated address regex
I know that the sed reseach is letteral "$variable" but I can not pass the value...
Thanks in advance.
It's a question of getting the quoting right.
Your first example:
count=$(sed -n '/$variable/,$p' file.log | grep '"KO"' -c)
doesn't expand $variable because it's in single quotes, the second:
count=$(sed -n "/$variable/,$p" file.log | grep '"KO"' -c)
expands $variable but has issues with its contents, as mentioned by choroba. It also has issue with the $p which will be interpreted as a shell variable. Your third example:
count=$(sed -n '/'$variable'/,$p' file.log | grep '"KO"' -c)
comes pretty close to what you need, but still suffers if $variable contains characters that sed treats specially, so these need to be escaped, e.g. the following works:
variable="\[17-09-12 00:01:03\]"
count=$(sed -n '/'$variable'/,$p' file.log
And as brackets are also special to the shell you can escape them automatically with the printf %q directive:
variable="[17-09-12 00:01:03]"
variable=$(printf "%q" "$variable")
count=$(sed -n '/'$variable'/,$p' file.log
[ has a special meaning in sed. I would use something more powerful than sed, i.e. Perl. It can escape the variable for you:
perl -ne '/\Q'"$variable"'\E/ and print'
Related
I am trying to separate out parts of a path as follows. My input path takes the following possible forms:
bucket
bucket/dir1
bucket/dir1/dir2
bucket/dir1/dir2/dir3
...
I want to separate the first part of the path (bucket) from the rest of the string if present (dir1/dir2/dir3/...), and store both in separate variables.
The following gives me something close to what I want:
❯ BUCKET=$(echo "bucket/dir1/dir2" | sed 's#\(^[^\/]*\)[\/]\(.*\)#\1#')
❯ EXTENS=$(echo "bucket/dir1/dir2" | sed 's#\(^[^\/]*\)[\/]\(.*\)#\2#')
echo $BUCKET $EXTENS
❯ bucket dir1/dir2
HOWEVER, it fails if I only have bucket as input (without a slash):
❯ BUCKET=$(echo "bucket" | sed 's#\(^[^\/]*\)[\/]\(.*\)#\1#')
❯ EXTENS=$(echo "bucket" | sed 's#\(^[^\/]*\)[\/]\(.*\)#\2#')
echo $BUCKET $EXTENS
❯ bucket bucket
... because, in the absence of the first '/', no capture happens, so no substitution takes place. When the input is just 'bucket' I would like $EXTENS to be set to the empty string "".
Thanks!
For something so simple you could use bash built-in instead of launching sed:
$ path="bucket/dir1/dir2"
$ bucket="${path%%/*}"
$ extens="${path#$bucket}"
$ printf '|%s|%s|\n' "$bucket" "$extens"
|bucket|/dir1/dir2|
$ path="bucket"
$ bucket="${path%%/*}"
$ extens="${path#$bucket}"
$ printf '|%s|%s|\n' "$bucket" "$extens"
|bucket||
But if you really want to use sed and capture groups:
$ declare -a bucket_extens
$ mapfile -td '' bucket_extens < <(printf '%s' "bucket/dir1/dir2" | sed -E 's!([^/]*)(.*)!\1\x00\2!')
$ printf '|%s|%s|\n' "${bucket_extens[#]}"
|bucket|/dir1/dir2|
$ mapfile -td '' bucket_extens < <(printf '%s' "bucket" | sed -E 's!([^/]*)(.*)!\1\x00\2!')
$ printf '|%s|%s|\n' "${bucket_extens[#]}"
|bucket||
We use the extended regex (-E) to simplify a bit, and ! as separator of the substitute command. The first capture group is simply anything not containing a slash and the second is everything else, including nothing if there's nothing else.
In the replacement string we separate the two capture groups with a NUL character (\x00). We then use mapfile to assign the result to bash array bucket_extens.
The NUL trick is a way to deal with file names containing spaces, newlines... NUL is the only character that cannot be part of a file name. The -d '' option of mapfile indicates that the lines to map are separated by NUL instead of the default newline.
Don't capture anything. Instead, just match what you don't want and replace it with nothing:
BUCKET=$(echo "bucket" | sed 's#/.*##'). # bucket
BUCKET=$(echo "bucket/dir1/dir2" | sed 's#/.*##') # bucket
EXTENS=$(echo "bucket" | sed 's#[^/]*##') # blank
EXTENS=$(echo "bucket/dir1/dir2" | sed 's#[^/]*##') # /dir1/dir2
As you are putting a slash in the regex. the string with no slashes will not
match. Let's make the slash optional as /\?. (A backslash before ?
is requires due to the sed BRE.) Then would you please try:
#!/bin/bash
#path="bucket/dir1/dir2"
path="bucket"
bucket=$(echo "$path" | sed 's#\(^[^/]*\)/\?\(.*\)#\1#')
extens=$(echo "$path" | sed 's#\(^[^/]*\)/\?\(.*\)#\2#')
echo "$bucket" "$extens"
You don't need to prepend a backslash to a slash.
By convention, it is recommended to use lower cases for user variables.
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$/,"")} /1/'
test1
thanks in advance for the help.
I have the following line that does work on linux.
myfile (extract)
active_instance_count=
aq_tm_processes=1
archive_lag_target=0
audit_file_dest=?/rdbms/audit
audit_sys_operations=FALSE
audit_trail=NONE
background_core_dump=partial
background_dump_dest=/home1/oracle/app/oracle/admin/iopecom/bdump
...
cat myfile |sed -r 's/ {1,}//g'|sed -r 's/\t*//g' |grep -v "^#"|sed -s "/^$/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.
Hi I have lots of logfiles with ^[[1m (as vim displays it) in them. I want to watch a logfile life via
tail -n 1000 -f logfile.log | grep <expression-for-escape-sequence>
and only get lines that have bold in them.
I am not sure which grep options I should use and have tried the following already:
tail -n 1000 -f logfile.log | grep "\033\0133\061\0155"
tail -n 1000 -f logfile.log | grep "\033\01331m"
tail -n 1000 -f logfile.log | grep "\033\[1m"
It does not work though... And yes there are bold lines in the last 1000 lines of logfile.log, testing with
echo -e "\033\01331mTest\033\01330m" | grep ...
same results... ;)
Appreciate any help!
Use single quotes with a dollar sign in front—as in $'...'—to have the shell convert the \033 escape sequence into an ESC character:
tail -n 1000 -f logfile.log | grep $'\033\[1m'
From man bash:
Words of the form $'string' are treated specially. The word expands to string, with backslash-escaped characters replaced as specified by the ANSI C standard.
This works (in a POSIX shell, not necessarily bash):
echo -e "\033\01331mTest\033\01330m" | grep "$(printf "\x1b\\[1m")"
more file
param1=" 1,deerfntjefnerjfntrjgntrjnvgrvgrtbvggfrjbntr*rfr4fv*frfftrjgtrignmtignmtyightygjn 2,3,4,5,6,7,8,
rfcmckmfdkckemdio8u548384omxc,mor0ckofcmineucfhcbdjcnedjcnywedpeodl40fcrcmkedmrikmckffmcrffmrfrifmtrifmrifvysdfn drfr4fdr4fmedmifmitfmifrtfrfrfrfnurfnurnfrunfrufnrufnrufnrufnruf"****
need to match the content of param1 as
sed -n "/$param1/p" file
but because the line length (very long line) I cant match the line
what’s the best way to match very long lines?
The problem you are facing is that param1 contains special characters which are being interpreted by sed. The asterisk ('*') is used to mean 'zero or more occurrences of the previous character', so when this character is interpreted by sed there is nothing left to match the literal asterisk you are looking for.
The following is a working bash script that should help:
#!/bin/bash
param1=' 1,deerfntjefnerjfntrjgntrjnvgrvgrtbvggfrjbntr\*rfr4fv\*frfftrjgtrignmtignmtyightygjn 2,3,4,5,6,7,8, rfcmckmfdkckemdio8u548384omxc,mor0ckofcmineucfhcbdjcnedjcnywedpeodl40fcrcmkedmrikmckffmcrffmrfrifmtrifmrifvysdfn'
cat <<EOF | sed "s/${param1}/Bubba/g"
1,deerfntjefnerjfntrjgntrjnvgrvgrtbvggfrjbntr*rfr4fv*frfftrjgtrignmtignmtyightygjn 2,3,4,5,6,7,8, rfcmckmfdkckemdio8u548384omxc,mor0ckofcmineucfhcbdjcnedjcnywedpeodl40fcrcmkedmrikmckffmcrffmrfrifmtrifmrifvysdfn
EOF
Maybe the problem is that your $param1 contains special characters? This works for me:
A="$(perl -e 'print "a" x 10000')"
echo $A | sed -n "/$A/p"
($A contains 10 000 a characters).
echo $A | grep -F $A
and
echo $A | grep -P $A
also works (second requires grep with built-in PCRE support. If you want pattern matching you should use either this or pcregrep. If you don't, use the fixed grep (grep -F)).
echo $A | grep $A
is too slow.