在 unix/linux 命令行中定义函数（例如 BASH）

Question

Sometimes I have a one-liner that I am repeating many times for a particular task, but will likely never use again in the exact same form. It includes a file name that I am pasting in from a directory listing. Somewhere in between and creating a bash script I thought maybe I could just create a one-liner function at the command line like:

numresults(){ ls "$1"/RealignerTargetCreator | wc -l }

I've tried a few things like using eval, using numresults=function... , but haven't stumbled on the right syntax, and haven't found anything on the web so far. (Everything coming up is just tutorials on bash functions).

Answer 1

Quoting my answer for a similar question on Ask Ubuntu:

Functions in bash are essentially named compound commands (or code blocks). From man bash :

 Compound Commands A compound command is one of the following: ... { list; } list is simply executed in the current shell environment. list must be terminated with a newline or semicolon. This is known as a group command. ... Shell Function Definitions A shell function is an object that is called like a simple command and executes a compound command with a new set of positional parameters. ... [C]ommand is usually a list of commands between { and }, but may be any command listed under Compound Commands above.

There's no reason given, it's just the syntax.

Try with a semicolon after wc -l :

numresults(){ ls "$1"/RealignerTargetCreator | wc -l; }

Answer 2

Don't use ls | wc -l ls | wc -l as it may give you wrong results if file names have newlines in it. You can use this function instead:

numresults() { find "$1" -mindepth 1 -printf '.' | wc -c; }

Answer 3

You can also count files without find . Using arrays,

numresults () { local files=( "$1"/* ); echo "${#files[@]}"; }

or using positional parameters

numresults () { set -- "$1"/*; echo "$#"; }

To match hidden files as well,

numresults () { local files=( "$1"/* "$1"/.* ); echo $(("${#files[@]}" - 2)); }
numresults () { set -- "$1"/* "$1"/.*; echo $(("$#" - 2)); }

(Subtracting 2 from the result compensates for . and .. .)

Answer 4

You can get a

bash: syntax error near unexpected token `('

error if you already have an alias with the same name as the function you're trying to define.

Answer 5

The easiest way maybe is echoing what you want to get back.

function myfunc()
{
    local  myresult='some value'
    echo "$myresult"
}

result=$(myfunc)   # or result=`myfunc`
echo $result

Anyway here you can find a good how-to for more advanced purposes