Bash覆盖$0 -脚本的名称

Question

有一个类似busybox的shell脚本，它根据调用它的符号链接来决定该做什么。

我想直接调用它，传递符号链接的名称，而不实际创建符号链接。

Answer 1

如果它是一个脚本，如果它可以被引用：

bash -c '. script.sh' overridden-name param1 param2

为我的案子工作。

Answer 2

虽然这个想法永远不会出现在我的脑海中，但我尝试了一下，只是为了好玩。我的结论是:你不能(至少，我不能)。

这里是如何和为什么，假装你会用$1作为潜在的符号链接。

#!/bin/bash

# not surprisingly, you cannot do this:
# 0 = $1
# '0' is not a valid variable name, so bash try to use it as a command, hence:
# 0: command not found

# you CAN do this, as meaningless as it seem:
# $0 = $1
# but say your script is called as ./foobar.sh, this will actually result in:
# ./foobar.sh = whatever_$1_is
# ...which is a nice fork bomb, since you'll call your script again,
# which will call your script again,
# which will call your script again,
# which will call ...etc etc etc
# Try it! :-)

# I've just learned about $BASH_SOURCE from here:
# http://www.tldp.org/LDP/abs/html/debugging.html#BASHSOURCEREF
echo BASH_SOURCE $BASH_SOURCE
# you CAN do this:
BASH_SOURCE = $1
# ...but to no use, since the following will show that BASH_SOURCE
# rightfuly behave as read-only:
echo BASH_SOURCE $BASH_SOURCE #unmodified

因此，如果您的想法是在不影响创建符号链接的情况下建立"a la busybox“行为的话，我建议：

#!/bin/bash
called_as = $1 # to be replaced with called_as = $0 once you're done prototyping
# if called_as such name do this
# if called_as such other name do that