Read values into a shell variable from a pipe

问题:

I am trying to get bash to process data from stdin that gets piped into, but no luck. What I mean is none of the following work:

echo "hello world" | test=($(< /dev/stdin)); echo test=$test
test=

echo "hello world" | read test; echo test=$test
test=

echo "hello world" | test=`cat`; echo test=$test
test=

where I want the output to be test=hello world. I've tried putting "" quotes around "$test" that doesn't work either.

回答1:

Use

IFS= read var << EOF
$(foo)
EOF

You can trick read into accepting from a pipe like this:

echo "hello world" | { read test; echo test=$test; }

or even write a function like this:

read_from_pipe() { read "$@" <&0; }

But there's no point - your variable assignments may not last! A pipeline may spawn a subshell, where the environment is inherited by value, not by reference. This is why read doesn't bother with input from a pipe - it's undefined.

FYI, http://www.etalabs.net/sh_tricks.html is a nifty collection of the cruft necessary to fight the oddities and incompatibilities of bourne shells, sh.

回答2:

if you want to read in lots of data and work on each line separately you could use something like this:

cat myFile | while read x ; do echo $x ; done

if you want to split the lines up into multiple words you can use multiple variables in place of x like this:

cat myFile | while read x y ; do echo $y $x ; done

alternatively:

while read x y ; do echo $y $x ; done < myFile

But as soon as you start to want to do anything really clever with this sort of thing you're better going for some scripting language like perl where you could try something like this:

perl -ane 'print "$F[0]\n"' < myFile

There's a fairly steep learning curve with perl (or I guess any of these languages) but you'll find it a lot easier in the long run if you want to do anything but the simplest of scripts. I'd recommend the Perl Cookbook and, of course, The Perl Programming Language by Larry Wall et al.

回答3:

This is another option

$ read test < <(echo hello world)

$ echo $test
hello world

回答4:

read won't read from a pipe (or possibly the result is lost because the pipe creates a subshell). You can, however, use a here string in Bash:

$ read a b c <<< $(echo 1 2 3)
$ echo $a $b $c
1 2 3

回答5:

I'm no expert in Bash, but I wonder why this hasn't been proposed:

stdin=$(cat)

echo "$stdin"

One-liner proof that it works for me:

$ fortune | eval 'stdin=$(cat); echo "$stdin"'

回答6:

bash 4.2 introduces the lastpipe option, which allows your code to work as written, by executing the last command in a pipeline in the current shell, rather than a subshell.

shopt -s lastpipe
echo "hello world" | read test; echo test=$test

回答7:

The syntax for an implicit pipe from a shell command into a bash variable is

var=$(command)

or

var=`command`

In your examples, you are piping data to an assignment statement, which does not expect any input.

回答8:

In my eyes the best way to read from stdin in bash is the following one, which also lets you work on the lines before the input ends:

while read LINE; do
    echo $LINE
done < /dev/stdin

回答9:

The first attempt was pretty close. This variation should work:

echo "hello world" | { test=$(< /dev/stdin); echo "test=$test"; };

and the output is:

test=hello world

You need braces after the pipe to enclose the assignment to test and the echo.

Without the braces, the assignment to test (after the pipe) is in one shell, and the echo "test=$test" is in a separate shell which doesn't know about that assignment. That's why you were getting "test=" in the output instead of "test=hello world".

回答10:

Piping something into an expression involving an assignment doesn't behave like that.

Instead, try:

test=$(echo "hello world"); echo test=$test

回答11:

Because I fall for it, I would like to drop a note. I found this thread, because I have to rewrite an old sh script to be POSIX compatible. This basically means to circumvent the pipe/subshell problem introduced by POSIX by rewriting code like this:

some_command | read a b c

into:

read a b c << EOF
$(some_command)
EOF

And code like this:

some_command |
while read a b c; do
    # something
done

into:

while read a b c; do
    # something
done << EOF
$(some_command)
EOF

But the latter does not behave the same on empty input. With the old notation the while loop is not entered on empty input, but in POSIX notation it is! I think it's due to the newline before EOF, which cannot be ommitted. The POSIX code which behaves more like the old notation looks like this:

while read a b c; do
    case $a in ("") break; esac
    # something
done << EOF
$(some_command)
EOF

In most cases this should be good enough. But unfortunately this still behaves not exactly like the old notation if some_command prints an empty line. In the old notation the while body is executed and in POSIX notation we break in front of the body.

An approach to fix this might look like this:

while read a b c; do
    case $a in ("something_guaranteed_not_to_be_printed_by_some_command") break; esac
    # something
done << EOF
$(some_command)
echo "something_guaranteed_not_to_be_printed_by_some_command"
EOF

回答12:

The following code:

echo "hello world" | ( test=($(< /dev/stdin)); echo test=$test )

will work too, but it will open another new sub-shell after the pipe, where

echo "hello world" | { test=($(< /dev/stdin)); echo test=$test; }

won't.

---

I had to disable job control to make use of chepnars' method (I was running this command from terminal):

set +m;shopt -s lastpipe
echo "hello world" | read test; echo test=$test
echo "hello world" | test="$(</dev/stdin)"; echo test=$test

Bash Manual says:

lastpipe

If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.

Note: job control is turned off by default in a non-interactive shell and thus you don't need the set +m inside a script.

回答13:

I think you were trying to write a shell script which could take input from stdin. but while you are trying it to do it inline, you got lost trying to create that test= variable. I think it does not make much sense to do it inline, and that's why it does not work the way you expect.

I was trying to reduce

$( ... | head -n $X | tail -n 1 )

to get a specific line from various input. so I could type...

cat program_file.c | line 34

so I need a small shell program able to read from stdin. like you do.

22:14 ~ $ cat ~/bin/line 
#!/bin/sh

if [ $# -ne 1 ]; then echo enter a line number to display; exit; fi
cat | head -n $1 | tail -n 1
22:16 ~ $ 

there you go.

回答14:

How about this:

echo "hello world" | echo test=$(cat)