问题
I am using grep
to produce output that will be parsed by another program.
However, that program expects output only to be numeric or zero-bytes.
Now grep
outputs a newline character after its output. I've checked the -Z
option but it doesn't seem to work as I'm using grep for counting (-c
).
I am executing in sh
, not bash
. So nesting it into echo -n "$(grep -c pattern)"
doesn't work either.
How can I get rid off the trailing newline?
回答1:
You can pipe it through tr
and translate the \n
to a \0
character:
tr '\n' '\0'
回答2:
Use tr -d to delete characters in a string:
$ grep -c ' ' /etc/passwd | tr -d '\n'
69$ grep -c ' ' /etc/passwd | tr -d '\n' | xxd
0000000: 3639 69
$
回答3:
I know this is old, and tr works just as well, but I happened across this question and noticed OP stated: I am executing in sh, not bash. So nesting it into echo -n "$(grep -c pattern)" doesn't work either.
This isn't grep or sh so much as how echo is being used. For future visitors, the only reason this didn't work is due to the double quotes around the substituted command. The following does, in fact, work even using sh.
echo -n $(grep -c pattern)
Examples:
$ ls /dev/sd? #example of formatted output
/dev/sda /dev/sdc /dev/sde /dev/sdg /dev/sdi /dev/sdk
/dev/sdb /dev/sdd /dev/sdf /dev/sdh /dev/sdj
$ echo $(ls /dev/sd?) #without -n, appends \n only at the end
/dev/sda /dev/sdb /dev/sdc /dev/sdd /dev/sde /dev/sdf /dev/sdg /dev/sdh /dev/sdi /dev/sdj /dev/sdk
$ echo -n $(ls /dev/sd?) #with -n, does not append the \n, but still strips the line breaks from the string
/dev/sda /dev/sdb /dev/sdc /dev/sdd /dev/sde /dev/sdf /dev/sdg /dev/sdh /dev/sdi /dev/sdj /dev/sdk
$ echo -n "$(ls /dev/sd?)" #output when double quotes are used
/dev/sda
/dev/sdb
/dev/sdc
/dev/sdd
/dev/sde
/dev/sdf
/dev/sdg
/dev/sdh
/dev/sdi
/dev/sdj
/dev/sdk
来源:https://stackoverflow.com/questions/7249755/how-to-prevent-grep-from-printing-a-trailing-newline