I know I can pass arguments to a bash function like
function output () {
echo 'text' > text.txt;
echo $1 >> text.txt;
echo 'more text' >> text.txt;
}
output something;
But I actually need to pass the output of another script or program and have it in one variable, so I am able to call the function like
output $(ls);
... and have the output of ls in one single variable in the function. Is that possible? How?
Or can I at least echo every single input to the function?
Try quoting the arguments:
output "$(ls)";
or changing your function to
function output () {
echo 'text' > text.txt;
echo "$@" >> text.txt;
echo 'more text' >> text.txt;
}
#!/bin/bash
output=`ls -larht`
/bin/echo -e "$output"
It is possible.
Consider this decimal to hex function, which takes the output of printf, and stores it in a var called NUM.
d2h()
{
NUM=$(printf "%x\n" $1)
echo $NUM
}
printf has this wonderful -v switch so that a=$(printf ...) is usually better written as printf -v a ... (think of C's sprintf). So that your d2h function would be more elegantly and more efficiently (think no subshells) written as d2h() { printf -v NUM '%x\n' $1; echo $NUM; }
total seemed to be the output of ls.