bash 将第二个参数从 shell 脚本传递给 Java
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/3995067/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Passing second argument onwards from a shell script to Java
提问by cm_c
If I pass any number of arguments to a shell script that invokes a Java program internally, how can I pass second argument onwards to the Java program except the first?
如果我将任意数量的参数传递给在内部调用 Java 程序的 shell 脚本，如何将第二个参数继续传递给 Java 程序，但第一个参数除外？
./my_script.sh a b c d ....
./my_script.sh abcd ....
#my_script.sh
...
java MyApp b c d ...
回答by Bolo
First use shiftto "consume" the first argument, then pass "$@", i.e., the list of remaining arguments:
首先使用shift“消耗”第一个参数，然后传递"$@"，即剩余参数的列表：
#my_script.sh
...
shift
java MyApp "$@"
回答by bashfu
You can pass second argument onwards without using "shift" as well.
您也可以在不使用“shift”的情况下继续传递第二个参数。
set -- 1 2 3 4 5
echo "${@:0}"
echo "${@:1}"
echo "${@:2}" # here
