unix - How can I take the output of a shell script and place it in a file on the command line?

06
2014-04
  • Questioner

    How can I take the output of a shell script and place it in a file on the command line?

    • Answers
  • Am. 
    # write to file
sh myscript > out.txt
# append to file
sh myscript >> out.txt
# write both output and stderr to file
sh myscript 2&1>> out.txt
  • Oren Mazor 
    $ ./foo >> myoutputfile.txt

    • Related Question

    unix - bash shell script which adds output of commands
  • John Kube

    Let's say I have a command called foo which prints a number to the screen when called:

    $ foo
3

    Let's also say I have another command called bar which prints another number to the screen when called:

    $ bar
5

    I'm looking to write a shell script which will arithmetically add together the output of foo and bar (3+5=8). How would I do that? (The outputs from the commands are not known ahead of time. They just so happen to have been 3 and 5 the last time they were run. They could have been something else.)


    • Related Answers
  • Benjamin Bannier

    Use bash's let to evalutate arithmetric expressions.

    #!/bin/bash
a=`echo 3`
b=`echo 5`

let c=$a+$b
echo $c

    Just substitute the calls to echo with your program calls.

  • Dennis Williamson

    An alternative to let is to use double-parenthesis syntax:

    (( c = $(foo) + $(bar) ))

    or

    echo $(( $(foo) + $(bar) ))

    or using variables, you can omit the dollar sign on the right hand side of the equals:

    (( c += $(foo) + num ))

    (which also illustrates an incremental assignment)

    If you're using non-integers you can use bc:

    echo "$(foo) + $(bar)" | bc

    or

    c=$(echo "$(foo) + $(bar)" | bc)

    One advantage of using double parentheses is that you can put spaces around the operands and operators to make things more readable:

    (( c = ( a + b ) * ( i - j ) ))
  • Ignacio Vazquez-Abrams

    bash:

    bc < <({ foo ; echo + ; bar ; } | tr '\n' ' ' ; echo)

    If the output is integers only:

    $(( $(foo) + $(bar) ))