Shell Programming and Scripting

Hi All,

I have the below input and i only want to filter out some un-wanted info from here. Expected output is below. Can somebody help ?

The catch is that i want to grep those lines with term "k=" and lines with term "**" as the 1st column and "07" as the last column. And the number of unwanted lines in between the wanted lines are random. On top of that, the number of field of the wanted lines are also random.
Expected output shown below.


Input:
aaa k= aaaa
dddd
dddd
bbb k= cccc
zzz
zzz
zzzzz
ggg k= pppp
vvvvvv
vvv
v
vvvvv
vvvvvv
** XXXXX ZZZ VVV 07
** ccc kkk lll
** ggg lll ppp
** hhh iii kkk
xxxxxx
mmm k= qqqq
xxxxx
ddd
eee k= yyyy
ffff
ff
rrr k= vvvv
ttt
gggggg
zzzzz

Expected Output:
aaa k= aaaa
bbb k= cccc
ggg k= pppp
** XXXXX ZZZ VVV 07
mmm k= qqqq
eee k= yyyy
rrr k= vvvv

Code:

sed -n -e "/k=/p" -e "/^\*\*.*07$/p" input.txt

Hi Vino,

I encounter some error. Can you help?

Code:

$ sed -n -e "/k=/p" -e "/^\*\*.*07$/p" inputfile
Variable syntax
$

I guess you are using csh.

Try this.

Code:

sed -n -e "/k=/p" -e '/^\*\*.*07$/p' inputfile

See this thread as well - https://www.unix.com/shell-programmin...d-c-shell.html

HI Vino,

Yeah i am using csh.
Thanks it really works!!
You are such a pro at this!!

But i am not familiar with sed. Can you explain you code so that i can understand better ?

sed -n -e "/k=/p" -e '/^\*\*.*07$/p' inputfile

sed's default behaviour is to print out the pattern space it reads. Turn it off with the -n flag.

-e "/k=/p" Look for all lines containing the pattern k= and then print those lines.

-e '/^\*\*.*07$/p' Look for all lines starting with ** and ending with 07 and the print those lines. The ^ is a placeholder for the start of a line and $ is likewise for the end-of-line.

We need to check for actual ** and so escape it as \*\*.

Hi Vino,

Thanks for the explanation.
But what does the red dot *.*07$/ signify ?