如何将字符串参数传递给 bash 脚本中的命令？
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How to pass a string argument to a command in a bash script?
提问by qiuxiafei
I concat all the arguments in a variable $ARGS and pass all of them to a command onece togather, just like what the ./test_arg.shdo.
我将所有参数连接到一个变量 $ARGS 中，并将它们全部传递给一个命令 onece togather，就像./test_arg.sh所做的那样。
But i found that the arguments with quoted string was splitted into multiple ones.
但我发现带引号字符串的参数被拆分为多个。
Here is my test scripts, test_arg.shperform the test case and arg.shjust print every argument on seperated line.
这是我的测试脚本，test_arg.sh执行测试用例，arg.sh只是在单独的行上打印每个参数。
$ cat ./test_arg.sh
#!/bin/sh
ARGS="$ARGS OPT_1"
ARGS="$ARGS OPT_2"
ARGS="$ARGS 'OPT_3 is a string with space'"
echo "./arg.sh $ARGS"
echo '----------------------------'
./arg.sh $ARGS
[ xiafei@xiafeitekiMacBook-Pro ~/tmp ]
$ cat ./arg.sh
#!/bin/sh
for var in "$@"
do
echo "arg -> $var"
done
Here coms the result:
结果如下：
$ ./test_arg.sh
./arg.sh OPT_1 OPT_2 'OPT_3 is a string with space'
----------------------------
arg -> OPT_1
arg -> OPT_2
arg -> 'OPT_3
arg -> is
arg -> a
arg -> string
arg -> with
arg -> space'
But if I put argument directly after the command it works correctly:
但是，如果我在命令之后直接放置参数，它可以正常工作：
$ cat test_arg.sh
#!/bin/sh
./arg.sh OPT_1 OPT_2 'OPT_3 is a string with space'
[ xiafei@xiafeitekiMacBook-Pro ~/tmp ]
$ sh test_arg.sh
arg -> OPT_1
arg -> OPT_2
arg -> OPT_3 is a string with space
I guess the problem is the way bash process quotes. Anyone knows about it?
我想问题是 bash 进程引用的方式。有人知道吗？
回答by Lars Brinkhoff
The way the unix shell works, your call to ./arg.sh $ARGSwill not evalutate the ' quotes.
unix shell 的工作方式，您的调用./arg.sh $ARGS不会评估 ' 引号。
One possible solution is to use eval "./arg.sh $ARGS".
一种可能的解决方案是使用eval "./arg.sh $ARGS".
