12

线条8.9.以下让我感到困惑:

#!/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:确实是黄色6.7.


编辑

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

... 最后!这就是执行此操作的命令,感谢 Josh!

4

1 回答 1

19

您正在寻找可以扩展参数中的转义字符的格式说明符。方便地,bash 支持(来自help printf):

%b        expand backslash escape sequences in the corresponding argument

或者,bash 还支持一种特殊机制,通过该机制将执行转义字符的扩展:

d=$'\e[33m'
于 2013-04-01T00:44:05.847 回答