How to pipe command output to other commands?
There is a distinction between command line arguments and standard input. A pipe will connect standard output of one process to standard input of another. So
ls | echo
Connects standard output of ls to standard input of echo. Fine right? Well, echo ignores standard input and will dump its command line arguments - which are none in this case to - its own stdout. The output: nothing at all.
There are a few solutions in this case. One is to use a command that reads stdin and dumps to stdout, such as cat.
ls | cat
Will 'work', depending on what your definition of work is.
But what about the general case. What you really want is to convert stdout of one command to command line args of another. As others have said, xargs
is the canonical helper tool in this case, reading its command line args for a command from its stdin, and constructing commands to run.
ls | xargs echo
You could also convert this some, using the substitution command $()
echo $(ls)
Would also do what you want.
Both of these tools are pretty core to shell scripting, you should learn both.
For completeness, as you indicate in the question, the other base way to convert stdin to command line args is the shell's builtin read
command. It converts "words" (words as defined by the IFS
variable) to a temp variable, which you can use in any command runs.
ls | echo
prints just a blank line because echo
reads no input; the last command of the pipeline is actually echo
that prints nothing but a blank line.
Generally:
a | b
makes sure that the output of a
become the input of b
. I suggest you to read the Pipelines
section of man bash
.
If you really want to use ls
and echo
together here's some (pretty useless) examples:
ls | xargs -L 1 echo
echo `ls`
for i in `ls` ; do echo $i ; done
If you want to echo
the ls
command try:
ls | xargs echo
This will call echo
with the output of ls
.