date - print string between two patterns but the first pattern is a range -

date - print string between two patterns but the first pattern is a range -

i have file containing dates in format mmddyyyy.

the file cotents below:

05192014 10212014 10222014 11232014 12242014

now wanted print dates have month 10 or 11 , year 2014, or can dates between months ranging 01 11.

for single month , year command working fine:

cat filename | sed -n '/10/,/2014/p'

the above command prints:

10212014 10222014

all good!

but if wanted dates having month either 10 or 11 did below:

cat filename | sed -n '/1[0-1]/,/2014/p'

but above command prints dates. there way out. can set range in first pattern.

also, should able dates when specify months in range example:

print dates month greater equal 01 , less equal 11 , year 2014.

then output should be:

05192014 10212014 10222014 11232014

i guess want this:

$ sed -n '/^1[01].*2014$/p' file 10212014 10222014 11232014

^1[01].*2014$ means: strings starting 1 followed either 0 or 1, bunch of strings end 2014.

if want check dates on format mmddyyyy, instead of ^1[01].*2014$/ can utilize ^1[01]..2014$. using ^ , $ avoid matching lines hello11112014bye , on.

also, note there no need cat ... | sed. sed ... file enough.

note current command not doing expect: sed -n '/10/,/2014/p' file prints files 1 containing 10 1 containing 2014, in between appear. see:

$ cat 05192014 10212014 asdfad 10222014 11232014 12242014 $ sed -n '/10/,/2014/p' 10212014 asdfad 10222014

also, should able dates when specify months in range example: print dates month greater equal 01 , less equal 11 , year 2014.

$ sed -rn '/^(0[1-9]|1[01]).*2014$/p' file 05192014 10212014 10222014 11232014

date sed