Bash 用子 shell 和子字符串进行错误替换

Question

一个人为的例子......给出

FOO="/foo/bar/baz"

这有效(在 bash 中)

BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1}       # result is BAZ="b"

这不是

BAZ=${$(basename $FOO):0:1} # result is bad substitution

我的问题是哪个规则导致这个 [subshel​​l substitution] 计算不正确？在 1 跳内完成此操作的正确方法是什么？

Answer 1

首先，请注意，当您这样说时:

BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1}       # result is BAZ="b"

BAZ 结构中的第一位是 BAR 而不是您要获取第一个字符的值。因此，即使 bash 允许变量名称包含任意字符，您在第二个表达式中的结果也不会是您想要的。

但是，关于阻止这种情况的规则，请允许我引用 bash 手册页中的内容:

DEFINITIONS
       The following definitions are used throughout the rest  of  this  docu‐
       ment.
       blank  A space or tab.
       word   A  sequence  of  characters  considered  as a single unit by the
              shell.  Also known as a token.
       name   A word consisting only of  alphanumeric  characters  and  under‐
              scores,  and beginning with an alphabetic character or an under‐
              score.  Also referred to as an identifier.

然后稍后:

PARAMETERS
       A parameter is an entity that stores values.  It can be a name, a  num‐
       ber, or one of the special characters listed below under Special Param‐
       eters.  A variable is a parameter denoted by a name.  A variable has  a
       value  and  zero or more attributes.  Attributes are assigned using the
       declare builtin command (see declare below in SHELL BUILTIN COMMANDS).

然后当它定义了你所询问的语法时:

   ${parameter:offset:length}
          Substring Expansion.  Expands to  up  to  length  characters  of
          parameter  starting  at  the  character specified by offset.

所以联机帮助页中阐明的规则说 ${foo:x:y} 构造必须有一个参数作为第一部分，并且参数只能是一个名称，一个数字，或少数特殊参数字符之一。 $(basename $FOO) 不是允许的参数之一。

至于在一次作业中执行此操作的方法，请使用其他响应中提到的其他命令的管道。