Input file:
$ cat file.txt
sl909678342-slp10
sl919678362-slp11
sl929678322-slp12
sl929678382-slp12
Required output:
Replace first 5 characters in each line of the above file with 'XXXXX'. i.e. the required output:
XXXXX678342-slp10
XXXXX678362-slp11
XXXXX678322-slp12
XXXXX678382-slp12
Awk solution:
$ awk '{
first=substr($0,1,5)
gsub(/./,"X",first)
end=substr($0,6)
print first end
}' file.txt
Read about awk substr function here
Sed solution:
Using 'extended regular expressions' with sed (-r option)
$ sed -r "s/^(.{0})(.{5})/\1XXXXX/" file.txt
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3 comments:
Why is the sed version like this?
sed -r "s/^(.{0})(.{5})/\1XXXXX/" file.txt
The first match in regex is "anything 0 times" which amounts to nothing. This should suffice:
sed -r "s/^(.{5})/XXXXX/" file.txt
Some of the methods to replace first 5 characters with Xs:
1. partition the string into 2 halves (A5 and remaining, aka, A*),
then drop the first half & in it's place put 5 Xs
perl -ple '$_ = "X" x 5 . (unpack "A5A*")[1]' file.txt
2. here we make use of the lvalue form of 'substr'
perl -ple 'substr($_, 0, 5) = "X" x 5' file.txt
3. Sprinkle the "l" (list pattern space) command at various locations to see what's going on...
sed -e '
s/./&\n/5
h;s/.*\n//;x
s/\n.*//
s/./X/g
G
s/\n//
' file.txt
sed -e '
l
s/^/\n/
:loop
l
s/^\(X\{0,5\}\)\n./\1X\n/
tloop
l
s/\n//
l
' file.txt
Hi Jadu,
If i have line like below :
1234
First i want to create something like below :
1234XXXX
1234XXX
1234XX
How to solve this challenges jadu ?
Thanks & Regards,
Taufik
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