在shell脚本中，$@是什么意思？

Question

在shell脚本中，美元符号后跟at符号(@)是什么意思？

例如：

umbrella_corp_options $@

Answer 1

$@是传递给脚本的所有参数的。

例如，如果调用./someScript.sh foo bar，则$@将等于foo bar。

如果您这样做了：

./someScript.sh foo bar

然后在someScript.sh引用内部：

umbrella_corp_options "$@"

这将传递给umbrella_corp_options，每个参数都用双引号括起来，允许从调用者那里获取带有空格的参数并将它们传递出去。

Answer 2

$@与$*几乎相同，都意味着“所有命令行参数”。它们通常被用来简单地将所有参数传递给另一个程序(从而在另一个程序周围形成一个包装器)。

当你有一个包含空格的参数时，这两种语法之间的差异就会显现出来(例如)并将$@放在双引号中：

wrappedProgram "$@"
# ^^^ this is correct and will hand over all arguments in the way
#     we received them, i. e. as several arguments, each of them
#     containing all the spaces and other uglinesses they have.
wrappedProgram "$*"
# ^^^ this will hand over exactly one argument, containing all
#     original arguments, separated by single spaces.
wrappedProgram $*
# ^^^ this will join all arguments by single spaces as well and
#     will then split the string as the shell does on the command
#     line, thus it will split an argument containing spaces into
#     several arguments.

示例:调用

wrapper "one two    three" four five "six seven"

将导致：

"$@": wrappedProgram "one two    three" four five "six seven"
"$*": wrappedProgram "one two    three four five six seven"
                             ^^^^ These spaces are part of the first
                                  argument and are not changed.
$*:   wrappedProgram one two three four five six seven

Answer 3

以下是命令行参数，其中：

$@ =将所有参数存储在字符串列表中

$* =将所有参数存储为单个字符串

$# =存储参数的数量