用于在bash printf内着色的ANSI转义码

Question

下面的行8.和9.让我感到困惑：

#!/bin/bash

a=foo
b=6
c=a
d="\e[33m"  # opening ansi color code for yellow text
e="\e[0m"   # ending ansi code
f=$d

printf "1. foo\n"
printf "2. $a\n"
printf "3. %s\n" "$a"
printf "4. %s\n" "${!c}"
printf "5. %${b}s\n" "$a"
printf "6. $d%s$e\n" "$a" # will be yellow
printf "7. $f%s$e\n" "$a" # will be yellow
printf '8. %s%s%s\n' "$d" "$a" "$e" # :(
printf "9. %s%s%s\n" "$f" "$a" "$e" # :(

是否可以使用%s扩展颜色变量并查看颜色开关？

输出：

1. foo
2. foo
3. foo
4. foo
5.    foo
6. foo
7. foo
8. \e[33mfoo\e[0m
9. \e[33mfoo\e[0m

注意：6.和7.确实黄色

---

修改

printf "10. %b%s%b\n" "$f" "$a" "$e" # :)

......终于来了！这是执行它的命令，感谢Josh！

Answer 1

您正在寻找一种格式说明符，它将在参数中展开转义字符。方便的是，bash支持（来自help printf）：

%b        expand backslash escape sequences in the corresponding argument

或者，bash还支持一种特殊的机制，通过它可以执行转义字符的扩展：

d=$'\e[33m'