我想将 Bash 脚本的所有命令行参数存储到一个变量中
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I would like to store all command-line arguments to a Bash script into a single variable
提问by Mark Biek
Let's say I have a Bash script called foo.sh.
假设我有一个名为foo.sh的 Bash 脚本。
I'd like to call it like this:
我想这样称呼它：
foo.sh Here is a bunch of stuff on the command-line
And I'd like it to store all of that text into a single variable and print it out.
我希望它将所有文本存储到一个变量中并打印出来。
So my output would be:
所以我的输出是：
Here is a bunch of stuff on the command-line
How would I do this?
我该怎么做？
回答by David Z
echo "$*"
would do what you want, namely printing out the entire command-line arguments, separated by a space (or, technically, whatever the value of $IFSis). If you wanted to store it into a variable, you could do
会做你想做的，即打印出整个命令行参数，用空格分隔（或者，从技术上讲，无论值$IFS是多少）。如果你想把它存储到一个变量中，你可以这样做
thevar="$*"
If that doesn't answer your question well enough, I'm not sure what else to say...
如果这还不能很好地回答你的问题，我不知道还能说什么......
回答by Paused until further notice.
If you want to avoid having $IFS involved, use $@ (or don't enclose $* in quotes)
如果您想避免涉及 $IFS，请使用 $@（或不要将 $* 括在引号中）
$ cat atsplat
IFS="_"
echo " at: $@"
echo " splat: $*"
echo "noquote: "$*
$ ./atsplat this is a test
at: this is a test
splat: this_is_a_test
noquote: this is a test
The IFS behavior follows variable assignments, too.
IFS 行为也遵循变量赋值。
$ cat atsplat2
IFS="_"
atvar=$@
splatvar=$*
echo " at: $atvar"
echo " splat: $splatvar"
echo "noquote: "$splatvar
$ ./atsplat2 this is a test
at: this is a test
splat: this_is_a_test
noquote: this is a test
Note that if the assignment to $IFS were made after the assignment of $splatvar, then all the outputs would be the same ($IFS would have no effect in the "atsplat2" example).
请注意，如果在分配 $splatvar 之后对 $IFS 进行分配，则所有输出都将相同（$IFS 在“atsplat2”示例中无效）。
回答by Ayman Hourieh
Have a look at the $*variable. It combines all command line arguments into one.
看看$*变量。它将所有命令行参数合二为一。
echo "$*"
This should do what you want.
这应该做你想做的。
