linux - Shell scripting input redirection oddities


Translate

Can anyone explain this behavior? Running:

#!/bin/sh
echo "hello world" | read var1 var2
echo $var1
echo $var2

results in nothing being ouput, while:

#!/bin/sh
echo "hello world" > test.file
read var1 var2 < test.file
echo $var1
echo $var2

produces the expected output:

hello
world

Shouldn't the pipe do in one step what the redirection to test.file did in the second example? I tried the same code with both the dash and bash shells and got the same behavior from both of them.


Toutes les réponses
  • Translate

    A recent addition to bash is the lastpipe option, which allows the last command in a pipeline to run in the current shell, not a subshell, when job control is deactivated.

    #!/bin/bash
    set +m      # Deactiveate job control
    shopt -s lastpipe
    echo "hello world" | read var1 var2
    echo $var1
    echo $var2
    

    will indeed output

    hello
    world
    

  • Translate
    #!/bin/sh
    echo "hello world" | read var1 var2
    echo $var1
    echo $var2
    

    produces no output because pipelines run each of their components inside a subshell. Subshells inherit copies of the parent shell's variables, rather than sharing them. Try this:

    #!/bin/sh
    foo="contents of shell variable foo"
    echo $foo
    (
        echo $foo
        foo="foo contents modified"
        echo $foo
    )
    echo $foo
    

    The parentheses define a region of code that gets run in a subshell, and $foo retains its original value after being modified inside them.

    Now try this:

    #!/bin/sh
    foo="contents of shell variable foo"
    echo $foo
    {
        echo $foo
        foo="foo contents modified"
        echo $foo
    }
    echo $foo
    

    The braces are purely for grouping, no subshell is created, and the $foo modified inside the braces is the same $foo modified outside them.

    Now try this:

    #!/bin/sh
    echo "hello world" | {
        read var1 var2
        echo $var1
        echo $var2
    }
    echo $var1
    echo $var2
    

    Inside the braces, the read builtin creates $var1 and $var2 properly and you can see that they get echoed. Outside the braces, they don't exist any more. All the code within the braces has been run in a subshell because it's one component of a pipeline.

    You can put arbitrary amounts of code between braces, so you can use this piping-into-a-block construction whenever you need to run a block of shell script that parses the output of something else.


  • Translate

    This has already been answered correctly, but the solution has not been stated yet. Use ksh, not bash. Compare:

    $ echo 'echo "hello world" | read var1 var2
    echo $var1
    echo $var2' | bash -s
    

    To:

    $ echo 'echo "hello world" | read var1 var2
    echo $var1
    echo $var2' | ksh -s
    hello
    world
    

    ksh is a superior programming shell because of little niceties like this. (bash is the better interactive shell, in my opinion.)


  • Translate
    read var1 var2 < <(echo "hello world")
    

  • Bob Lee
    Translate

    The post has been properly answered, but I would like to offer an alternative one liner that perhaps could be of some use.

    For assigning space separated values from echo (or stdout for that matter) to shell variables, you could consider using shell arrays:

    $ var=( $( echo 'hello world' ) )
    $ echo ${var[0]}
    hello
    $ echo ${var[1]}
    world
    

    In this example var is an array and the contents can be accessed using the construct ${var[index]}, where index is the array index (starts with 0).

    That way you can have as many parameters as you want assigned to the relevant array index.


  • Translate

    My take on this issue (using Bash):

    read var1 var2 <<< "hello world"
    echo $var1 $var2
    

  • Translate

    Allright, I figured it out!

    This is a hard bug to catch, but results from the way pipes are handled by the shell. Every element of a pipeline runs in a separate process. When the read command sets var1 and var2, is sets them it its own subshell, not the parent shell. So when the subshell exits, the values of var1 and var2 are lost. You can, however, try doing

    var1=$(echo "Hello")
    echo var1
    

    which returns the expected answer. Unfortunately this only works for single variables, you can't set many at a time. In order to set multiple variables at a time you must either read into one variable and chop it up into multiple variables or use something like this:

    set -- $(echo "Hello World")
    var1="$1" var2="$2"
    echo $var1
    echo $var2
    

    While I admit it's not as elegant as using a pipe, it works. Of course you should keep in mind that read was meant to read from files into variables, so making it read from standard input should be a little harder.


  • Betty Lee
    Translate

    It's because the pipe version is creating a subshell, which reads the variable into its local space which then is destroyed when the subshell exits.

    Execute this command

    $ echo $$;cat | read a
    10637
    

    and use pstree -p to look at the running processes, you will see an extra shell hanging off of your main shell.

        |                       |-bash(10637)-+-bash(10786)
        |                       |             `-cat(10785)
    

  • Translate

    Try:

    echo "hello world" | (read var1 var2 ; echo $var1 ; echo $var2 )
    

    The problem, as multiple people have stated, is that var1 and var2 are created in a subshell environment that is destroyed when that subshell exits. The above avoids destroying the subshell until the result has been echo'd. Another solution is:

    result=`echo "hello world"`
    read var1 var2 <<EOF
    $result
    EOF
    echo $var1
    echo $var2