在bash中解析命令行参数-奇怪的行为 [英] Parsing command line arguments in bash - strange behavior

问题描述

#!/bin/bash

for arg
do
        echo "$arg"
done

在名为compile的文件中.

In file called compile.

我尝试过:

compile ha -aa -bb -cc -dd -ee -ff -gg
ha
-aa
-bb
-cc
-dd

-ff
-gg

-ee为什么不显示?实际上，似乎[[e] +没有出现.

Why does -ee not show up? In fact, it seems that -[e]+ does not show up.

推荐答案

因为您使用的 echo 版本采用 -e 作为选项(意思是扩展转义字符").

Because the version of echo you use takes -e as an option (meaning "expand escape characters").

编辑后添加:

对此类型问题的标准答案是"use printf"，这是严格准确的，但有些不令人满意. printf 使用起来很烦人，因为它处理多个参数的方式.因此:

The standard response to this type of question is "use printf", which is strictly accurate but somewhat unsatisfactory. printf is a lot more annoying to use because of the way it handles multiple arguments. Thus:

$ echo -e a b c
a b c
$ printf "%s\n" -e a b c
-e
a
b
c
$ printf "%s" -e a b c
-eabc$ # That's not what I wanted either

当然，您只需要记住将整个参数序列都用引号引起来，但这会导致烦人的用引号转义的问题.

Of course, you just need to remember to quote the entire argument sequence, but that can lead to annoying quote-escaping issues.

因此，我为您提供回声:

Consequently, I offer for your echoing pleasure:

$ ech-o() { printf "%s\n" "$*"; }
$ ech-o -e a b c
-e a b c
$

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