用subshel​​l和子字符串的Bash替换

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给出了一个有创意的例子
FOO="/foo/bar/baz"

这个工程(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替代]错误地评估?如果有的话,在1跳中做什么是正确的方法

首先,请注意,当你这样说:
BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1}       # result is BAZ="b"

BAZ的构造中的第一个位是BAR,而不是要采用第一个字符的值.所以即使bash允许变量名包含任意字符,你的第二个表达式的结果也不会是你想要的.

但是,关于阻止这一点的规则,请允许我从bash手册页引用:

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).

之后当它定义你所要求的语法时:

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

因此,在联机帮助页中阐述的规则表示${foo:x:y}构造必须具有参数作为第一部分,并且参数只能是名称,数字或少数特殊参数字符之一. $(basename $FOO)不是参数允许的可能性之一.

对于在一个作业中执行此操作的方式,请使用管道到其他响应中提到的其他命令.

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