当变量具有实际的颜色名称时，如何打印彩色字符串?

Question

我正在尝试根据错误代码打印具有特定颜色的字符串。颜色因每个错误代码而异，我将实际的颜色名称存储在变量中并在 printf 中使用它。

BLACK=$(tput setaf 0)
RED=$(tput setaf 1)
GREEN=$(tput setaf 2)
BLUE=$(tput setaf 4)
CYAN=$(tput setaf 5)
BOLD=$(tput bold)
NORMAL=$(tput sgr0)
# this color varies depending on error code
color=GREEN

printf "${$color} This is a String ${NORMAL} n"

但我得到的输出为

${GREEN} This is a String

预期输出(实际绿色(

This is a String

我可以使用

printf "${GREEN} This is a String ${NORMAL} n"

但是我希望使用颜色变量输出

Answer 1

Bash 不会像您尝试的那样在字符串中递归地处理${<variable>}。

你可以做color=$<colorVariable>，即color=$GREEN在printf之前，然后在printf字符串中做"$color This is a String ${NORMAL} n"

所以，最终结果：

BLACK=$(tput setaf 0)
RED=$(tput setaf 1)
GREEN=$(tput setaf 2)
BLUE=$(tput setaf 4)
CYAN=$(tput setaf 5)
BOLD=$(tput bold)
NORMAL=$(tput sgr0)
# Event-dependant color
color=$GREEN
# And finally,
printf "$color This is a String ${NORMAL} n"

还有另一种方法可以做到这一点。

# Define all colors, only GREEN and NORMAL here for brevity reasons
GREEN=$(tput setaf 2)
NORMAL=$(tput sgr0)
# Notice the missing $, as the original question
color=GREEN
# And then, use Bash's variable reference method (here notice both ! instead of $ and the missing  at the beginning
printf "${!color} This is a String ${NORMAL} n"

Answer 2

怎么样：

color=$GREEN

然后：

printf "$color This is a String ${NORMAL} n"

给我：

This is a String

绿色。