Linux的：我如何委派使用脚本异国情调的命令行参数？ [英] Linux: How do I delegate exotic commandline arguments using a script?

问题描述

我想编写一个包装bash脚本，和所有参数传递给调用的程序。我很肯定，这正常工作：

 ＃！/ bin / sh的
someProgam $ @
 

但经过异国参数时（在引号空，转义，...）失败。

例如：没有包装脚本， someProgram在参数1 23 结果，点击
[1〜2] 和 [3] 。结果
但是从脚本调用，我得到 [1] ， [2] ， [3 ] 。

的括号只是用于可视化。的

的注：这是一个Java程序，它调用。但我认为这并不重要。的

解决方案

 ＃！/ bin / sh的
someProgram$ @
 

又见特殊参数的bash文档

BTW1，$ @不是特定打坏。你也许可以依靠跨平台$ @ SH 来在任何地方运行的脚本。

BTW2，万一出现这种情况是在脚本中的最后一行，可以节省您的操作系统的几个字节，并通过更改行像

在进程表中的条目

  EXEC someProgram$ @
 

I want to write a wrapper bash script, and to pass all arguments to a called program. I was very sure, that this works correctly:

#!/bin/sh
someProgam $@

But when passing exotic arguments (empty, unescaped, in quotes, ...) this fails.

For example: without the wrapper script, someProgram "1 2" 3 results in the arguments
[1 2] and [3].
But called from the script, I get [1], [2], [3].

Braces are just for visualization.

NOTE: It's a Java program, which is called. But I think it doesn't matter.

解决方案

#!/bin/sh
someProgram "$@"

See also the bash docs on special parameters.

BTW1, "$@" is not specific to bash. You can probably rely on "$@" in cross-platform sh scripts to be run just about anywhere.

BTW2, in case this happens to be the last line in that script, you can save your operating system a few bytes and an entry in the process table by changing the line to something like

exec someProgram "$@"

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