Quotes within Bash Shell variable do not work, why?

Question

#!/bin/sh
LOG='log --pretty=format:"%h - %an, %ar : %s"'
git $LOG

I expect this to output an actual git log in the specified format. However, all I get is a fatal argument error.

I have also tried the options below, and they also do not work:

LOG="log --pretty=format:\"%h - %an, %ar : %s\""
LOG='log --pretty=format:\"%h - %an, %ar : %s\"'

However, strangely the script below DOES work, and I don't understand why:

LOG='--pretty=format:"%h - %an, %ar : %s"'
git log "$LOG"

It has been argued that shell is considering the variable to be only one argument, however the following works fine:

LOG1LINE='log --pretty=oneline'
git $LOG1LINE

• glenn jackman: I'm trying to put a command in a variable, but the complex cases always fail

Answer 1

Here is one way to overcome this problem. Using bash arrays:

#!/bin/bash
#It's important to use bash instead of sh for this to work
LOG=(log --pretty=format:"%h - %an, %ar : %s")
git "${LOG[@]}"

• onlywei: I get "Syntax error: "(" unexpected" when I try to declare my array. I get it even if I use declare -a LOG=(b).
• anubhava: Are you sure you're using bash?
• onlywei: I'm using the default terminal that comes with Ubuntu. How can I check for sure?
• anubhava: Run this: echo $SHELL
• onlywei: echo $SHELL results in: /bin/bash

Answer 2

This is a product of how the shell handles arguments.

The first thing the shell does is substitute variables. Quotes that were used in the variable definition are not preserved here. Then it chunks them into arguments, based on either quotes or, if these are not present, word boundaries. Quotes that were quoted when they were first used are escaped at that time, and consequently cannot define word boundaries.

Consequently, the arguments passed to git in your first example are:

• log
• --pretty=format:"%h
• -
• %an,
• %ar
• :
• %s"

In the later example, the arguments passed to git are:

• log
• --pretty=format:"%h - %an, %ar : %s"

• onlywei: Is there any way to preserve the quotes?
• Gordon Davisson: @onlywei: Not really. The shell parses (& removes) quotes before replacing variable values, so by the time any quotes in the variable appear, it's too late for them to do any good. Solution: use an array instead (see anubhava's answer).