Remove characters from the file

Shell Programming and Scripting
1

i know , the below question has been repeated.
can you guys guide me .

I have the below input

999999 xxxxxxxxxxxxxx    123.45  2013-05-02 08:14 1 1 1 xxxx
999999 xxxxxxxxxxxxxx    123.45  2013-06-02 02:14 1 4 1 dddd

i need to remove from the column 54 to 70 , as like the below output.

999999 xxxxxxxxxxxxxx    123.45   1 1 1 xxxx
999999 xxxxxxxxxxxxxx    123.45   1 4 1 dddd

thanks in advance ! ! !

2 
sed 's#[1-9][0-9][0-9][0-9]-.*:[0-9][0-9] ##g' file
3

something like: cut -f1-3,6-9

$ echo 999999 xxxxxxxxxxxxxx    123.45  2013-05-02 08:14 1 1 1 xxxx | cut -d" " -f1-3,6-9
999999 xxxxxxxxxxxxxx 123.45 1 1 1 xxxx
4

It can also be done either of the below way:

awk '{print $1 " " $2 " " $3 " " $6 " " $7 " " $8 " " $9}' file

awk '{$4=" "; $5=" "; print}' file

Is this what you were looking for?

1 Like
5

Please elaborate. The strings you want to remove are not column 54 to 70.

6

Your example text did not appear to have 70+ columns. My count showed 56 characters across.

7

Since the default value of OFS is a space, you do not need to explicitly concatenate.

print $1, $2, $3, $6, $7, $8, $9

Regards,
Alister

8

Thanks alister :slight_smile:
Just started learning AWK !

9

i need the character to cut from 38 to 54.
and also i cannot use awk since it the second column"xxxxxxxxxxxxxx " will have special character or space..

---------- Post updated at 09:47 PM ---------- Previous update was at 09:41 PM ----------

thanks guys, ihave finally achived it using the substr command, which i have forgot earlier :slight_smile:

awk '{print substr($0,1,38),substr($0,55,90)}'
10

The simplest solution is right there in your post. Hint:

Regards,
Alister