如何在 Shell 脚本中将参数作为别名运行？

Question

我有一个别名alias ll='ls -lF'；

我创建了一个 bash 脚本temp.sh：

#!/bin/bash 

# Allow aliases to work in bash NON-interactive mode!
shopt -s expand_aliases

# .. and load them
source ~/.bash_aliases
$1

但是当我运行它时它给了我：

$ ./temp.sh ll
./temp.sh: line 10: ll: command not found

当我更改脚本并直接输入别名时：

#!/bin/bash 

# Allow aliases to work in bash NON-interactive mode!
shopt -s expand_aliases

# .. and load them
source ~/.bash_aliases
ll

...它正在工作：

$ ./temp.sh   
total 12
-rwxrwxr-x 1 sobi3ch sobi3ch 423 Apr 19 14:21 script.sh*
-rwxrwxr-x 1 sobi3ch sobi3ch 196 Apr 26 12:28 temp.sh*
-rwxrwxr-x 1 sobi3ch sobi3ch 173 Apr 26 12:02 script2.sh

alias...此外，当我在脚本中运行命令而不是ll（或$1）时，我可以看到所有别名之间的别名ll。

当我将别名作为参数传递时，为什么它不起作用？

Answer 1

文档中没有明确指出这一点（至少我没有注意到），但问题是别名扩展优先于变量扩展；这意味着$1检查标记以查看它是否对应于别名，将其作为潜在别名丢弃，然后才将其扩展为参数。由于没有名为的命令ll，Bash 会出错。

您可以使用eval使参数扩展两次，第一次作为参数，第二次作为别名：

#!/bin/bash 

# Allow aliases to work in bash NON-interactive mode!
shopt -s expand_aliases

# .. and load them
source ~/.bash_aliases
eval "$1"

~$ cat temp.sh 
#!/bin/bash 

# Allow aliases to work in bash NON-interactive mode!
shopt -s expand_aliases

# .. and load them
source .bash_aliases
eval "$1"  
~$ cat .bash_aliases 
alias ll='ls -l'
~$ ./temp.sh ll
drwxrwxr-x  3 user user 4096 apr 24 15:18 articles
drwxrwxr-x  2 user user 4096 apr 24 00:20 bin
drwxr-xr-x  2 user user 4096 apr 21 20:22 Documenti
-rw-r--r--  1 user user 8980 apr 21 20:18 examples.desktop
drwxr-xr-x  2 user user 4096 apr 21 21:59 Immagini
drwxr-xr-x  2 user user 4096 apr 21 20:22 Modelli
drwxrwxr-x  6 user user 4096 apr 23 20:45 MT7630E
drwxr-xr-x  2 user user 4096 apr 21 20:22 Musica
drwxr-xr-x  2 user user 4096 apr 21 20:22 Pubblici
drwxr-xr-x  2 user user 4096 apr 24 23:02 Scaricati
drwxr-xr-x  2 user user 4096 apr 24 13:44 Scrivania
-rwxrwxr-x  1 user user  149 apr 26 13:22 temp.sh
drwxrwxr-x  2 user user 4096 apr 26 13:22 tmp
drwxrwxr-x 24 user user 4096 apr 23 13:45 util-linux-2.28
drwxr-xr-x  2 user user 4096 apr 21 20:22 Video