源运算符在 bash 中的括号内不起作用
我想在 bash 脚本中包含一个配置文件，有两个条件：
- 配置文件名是即时构造的并存储在变量中，
- 如果配置文件不存在，脚本应该失败：
config .cfg:
CONFIGURED=yes
test.sh:
#!/bin/sh
$CFG=config.cfg
echo Source command doesn't work here:
[ -f $CFG ] && ( source $CFG ) || (echo $CFG doesnt exist; exit 127)
echo $CONFIGURED
echo ... but works here:
source $CFG
echo $CONFIGURED
[...] 语句有什么问题？
I'd like to include a config file in my bash script with 2 conditions:
- the config file name is constructed on-the-fly and stored in variable, and
- in case if config file doesn't exist, the script should fail:
config.cfg:
CONFIGURED=yes
test.sh:
#!/bin/sh
$CFG=config.cfg
echo Source command doesn't work here:
[ -f $CFG ] && ( source $CFG ) || (echo $CFG doesnt exist; exit 127)
echo $CONFIGURED
echo ... but works here:
source $CFG
echo $CONFIGURED
What's wrong in [...] statement?
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( ... )在单独的子 shell 中运行命令。如果您想在同一个 shell 中运行命令（如果可能），请使用{ ... ; }相反。( ... )runs the commands in a separate subshell. If you want to run the commands in the same shell if possible then use{ ... ; }instead.