Shell Programming and Scripting

Hi,
Can anyone please give me the grep command to find all the lines in a file
that exceed 80 columns

Thanks,
gubbala

Code:

awk -F<delimiter> ' NF > 80 { print } ' filename

or just this

Code:

awk -F<delimiter> ' NF > 80  ' filename

sed -n '/\(.\)\{80\}/p' filename

This commands works out for 80 chars length of a line
and the OP had requested for lines greater than 80 columns !

Hi Matrixmadhan,

Have a look at this..It works well when I tested..

u142115@linux2alm:~/aps/aps4/product/den/dennis/test> awk '{print length($0)" "$0}' ne
36 gagkasdadsgfkdaskdgkagsdkgkasgdkgasd
8 assddasd
4 aaas
4 sddd
6 asasas
5 asddd
2 sa
9 saddasdas
2 sd
u142115@linux2alm:~/aps/aps4/product/den/dennis/test> sed -n '/\(.\)\{5\}/p' ne
gagkasdadsgfkdaskdgkagsdkgkasgdkgasd
assddasd
asasas
asddd
saddasdas
u142115@linux2alm:~/aps/aps4/product/den/dennis/test>


advise me if something wrong here...as I am a newbie in Unix..

with GNU grep

Code:

grep  '.\{80,\}' file

Well possibly I could be wrong!

Meaning of columns is ambiguous here

What i meant is,

Code:

c1 <delimiter> c2 <delimiter> c3 ... <cn>

literal meaning of column as such being delimited by some delimiter ( including the default delimiter as well )

Its OP who should correct us, giving the meaning of columns here,

whether is 80 chars in a line or
columns as I had said .

Till OP clears it, I had to get back my statement that your command is wrong.

Am sorry about that !