Using awk or sed how can I select lines which are occurring between two different marker patterns? There may be multiple sections marked with these patterns.
For example: Suppose the file contains:
abc
def1
ghi1
jkl1
mno
abc
def2
ghi2
jkl2
mno
pqr
stu
And the starting pattern is abc and ending pattern is mno
So, I need the output as:
def1
ghi1
jkl1
def2
ghi2
jkl2
I am using sed to match the pattern once:
sed -e '1,/abc/d' -e '/mno/,$d' <FILE>
Is there any way in sed or awk to do it repeatedly until the end of file?
This question is related to
shell
unix
sed
awk
pattern-matching
perl -lne 'print if((/abc/../mno/) && !(/abc/||/mno/))' your_file
From the previous response's links, the one that did it for me, running ksh on Solaris, was this:
sed '1,/firstmatch/d;/secondmatch/,$d'
1,/firstmatch/d: from line 1 until the first time you find firstmatch, delete./secondmatch/,$d: from the first occurrance of secondmatch until the end of file, delete.Using sed:
sed -n -e '/^abc$/,/^mno$/{ /^abc$/d; /^mno$/d; p; }'
The -n option means do not print by default.
The pattern looks for lines containing just abc to just mno, and then executes the actions in the { ... }. The first action deletes the abc line; the second the mno line; and the p prints the remaining lines. You can relax the regexes as required. Any lines outside the range of abc..mno are simply not printed.
sed '/^abc$/,/^mno$/!d;//d' file
golfs two characters better than ppotong's {//!b};d
The empty forward slashes // mean: "reuse the last regular expression used". and the command does the same as the more understandable:
sed '/^abc$/,/^mno$/!d;/^abc$/d;/^mno$/d' file
This seems to be POSIX:
If an RE is empty (that is, no pattern is specified) sed shall behave as if the last RE used in the last command applied (either as an address or as part of a substitute command) was specified.
This might work for you (GNU sed):
sed '/^abc$/,/^mno$/{//!b};d' file
Delete all lines except for those between lines starting abc and mno
I tried to use awk to print lines between two patterns while pattern2 also match pattern1. And the pattern1 line should also be printed.
e.g. source
package AAA
aaa
bbb
ccc
package BBB
ddd
eee
package CCC
fff
ggg
hhh
iii
package DDD
jjj
should has an ouput of
package BBB
ddd
eee
Where pattern1 is package BBB, pattern2 is package \w*. Note that CCC isn't a known value so can't be literally matched.
In this case, neither @scai 's awk '/abc/{a=1}/mno/{print;a=0}a' file nor @fedorqui 's awk '/abc/{a=1} a; /mno/{a=0}' file works for me.
Finally, I managed to solve it by awk '/package BBB/{flag=1;print;next}/package \w*/{flag=0}flag' file, haha
A little more effort result in awk '/package BBB/{flag=1;print;next}flag;/package \w*/{flag=0}' file, to print pattern2 line also, that is,
package BBB
ddd
eee
package CCC
Don_crissti's answer from Show only text between 2 matching pattern?
firstmatch="abc"
secondmatch="cdf"
sed "/$firstmatch/,/$secondmatch/!d;//d" infile
which is much more efficient than AWK's application, see here.
something like this works for me:
file.awk:
BEGIN {
record=0
}
/^abc$/ {
record=1
}
/^mno$/ {
record=0;
print "s="s;
s=""
}
!/^abc|mno$/ {
if (record==1) {
s = s"\n"$0
}
}
using: awk -f file.awk data...
edit: O_o fedorqui solution is way better/prettier than mine.
Source: Stackoverflow.com