ANSI转义代码，用于在bash printf中着色 [英] ANSI escape codes for coloring inside bash printf

问题描述

下面的8.和9.行使我感到困惑:

Lines 8. and 9. below confound me:

#!/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?

Is it possible to use %s to expand a colour variable and see the colour switch?

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.确实是黄色

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

...终于！这是执行此命令的命令，这要感谢Josh！

... finally! That's the command that does it, thanks to Josh!

推荐答案

您正在寻找一种格式说明符，该说明符将扩展参数中的转义符.方便地，bash支持(来自help printf):

You're looking for a format specifier that will expand escape characters in the argument. Conveniently, bash supports (from help printf):

%b        expand backslash escape sequences in the corresponding argument

或者，bash还支持一种特殊的机制，通过该机制可以执行转义符的扩展:

Alternatively, bash also supports a special mechanism by which will perform expansion of escape characters:

d=$'\e[33m'

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