如何将字符串作为 bash 脚本的命令行参数传递？
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How to pass a string as command line argument for bash script?
提问by sivakumarspnv
I want to pass a string as command line argument to a bash script; Simply my bash script is:
我想将字符串作为命令行参数传递给 bash 脚本；简单地说，我的 bash 脚本是：
>cat test_script.sh
for i in $*
do
echo $i
done
I typed
我打字
bash test_script.sh test1 test2 "test3 test4"
bash test_script.sh test1 test2 "test3 test4"
Output :
输出 ：
test1
test2
test3
test4
Output I am expecting:
我期待的输出：
test1
test2
test3 test4
I tried with backslashes (test1 test2 "test3\ test4") and single quotes but I haven't got the expected result.
我尝试使用反斜杠（test1 test2 "test3\ test4"）和单引号，但没有得到预期的结果。
How do I get the expected output?
我如何获得预期的输出？
回答by Jonathan Leffler
You need to use:
您需要使用：
for i in "$@"
do echo $i
done
or even:
甚至：
for i in "$@"
do echo "$i"
done
The first would lose multiple spaces within an argument (but would keep the words of an argument together). The second preserves the spaces in the arguments.
第一个会在参数中丢失多个空格（但会将参数的单词保持在一起）。第二个保留参数中的空格。
You can also omit the in "$@"clause; it is implied if you write for i(but personally, I never use the shorthand).
你也可以省略in "$@"子句；如果你写for i（但就我个人而言，我从不使用速记），这是暗示的。
回答by Danny Hong
Try use printf
尝试使用 printf
for i in $* do $(printf '%q' $i) done
