How to write a general command logging function in Bash

Question

Here's the general command logging function I wrote:

function exec_cmd() {
  {
    echo "$(date '+%Y-%m-%d %H:%M:%S')" "$1"
    $1
  } >>./cmd.log
}

When I call this function with exec_cmd "modprobe -r toa || true", I got an error "modprobe: FATAL: Module || not found."

Running the same command without the wrapper exec_cmd works as expected.

The whole script is running in strict mode. The || were included because only the error from this command can be ignored.

Although in this case the error value is discarded anyway, right? Or are you running it in strict mode? — user202729, Feb 22 '21 at 10:10
@RamanSailopal to make sure that even if toa mod is not installed, this error will be discarded — Blaise Wang, Feb 22 '21 at 10:13
@user202729 yes, the whole script is running in strict mode which means only this command can be failed — Blaise Wang, Feb 22 '21 at 10:14
Possible duplicate of [eval - Bash function calling command given as argument - Stack Overflow](https://stackoverflow.com/questions/3178604/bash-function-calling-command-given-as-argument) — user202729, Feb 22 '21 at 10:15
@user202729 Thanks. This question is indeed very useful. But back to the question, I would say using `$@` instead of `$1` is the key to answer the question. Anyway, thanks. — Blaise Wang, Feb 22 '21 at 10:33

score 1 · Answer 1 · answered Feb 22 '21 at 10:46

1

Bash has builtin support for printing commands as they are executed, namely the -x flag. From man bash:

-x Print commands and their arguments as they are executed.

So you can revise your script to call set -x as the first command, then every command after that is printed in the sequence they are called.

#!/bin/bin/env bash
set -x

modprobe -r toa || true

answered Feb 22 '21 at 10:46

Esa Lindqvist

311
2
6

Not a bad idea for logging, but this doesn't include the date. – user202729 Feb 22 '21 at 10:52
Of course the upside is that you don't have to remember to pass all commands to `exec_cmd`. – Esa Lindqvist Feb 22 '21 at 13:37
You can get the the date and time in the output after running `set -x` by first setting `PS4='\D{%Y-%m-%d %H:%M:%S} + '`. – pjh Feb 22 '21 at 19:01

score 0 · Accepted Answer · answered Feb 22 '21 at 10:31

0

I would write the function like this :

#!/usr/bin/env bash
  
function exec_cmd() {
  {
    printf "$(date '+%Y-%m-%d %H:%M:%S') %s\n" "$*"
    "$@"
  } >> ./cmd.log
}

exec_cmd modprobe -r toa || true

answered Feb 22 '21 at 10:31

Philippe

20,025
2
23
32

Thanks. I am also writing an almost identical answer after reading @user202729's comment. I will accept your answer since you posted the first answer. – Blaise Wang Feb 22 '21 at 10:36

How to write a general command logging function in Bash

2 Answers2