Question

  # The **variable slicing** notation in the following conditional
  # needs some explanation: `${var#expr}` returns everything after
  # the match for 'expr' in the variable value (if any), and
  # `${var%expr}` returns everything that doesn't match (in this
  # case, just the very first character. You can also do this in
  # Bash with `${var:0:1}`, and you could use cut too: `cut -c1`.

这实际上是什么意思？我能得到一个例子

Answer 1

你引用的解释并不准确。此机制允许您从变量的值中删除前缀或后缀。

vnix$ v=foobar
vnix$ echo "${v#foo}"
bar
vnix$ echo "${v%bar}"
foo

表达式可以是glob，因此您不限于静态字符串。

vnix$ echo "${v%b*}"
foo

还有##和%%来修剪最长的匹配而不是最短的匹配。

Answer 2

这是一个简单的例子：

#!/bin/bash

message="hello world!"
var1="hello"
var2="world!"
echo "${message#$var1}"
echo "${message%$var2}"
echo "${message%???}"
echo "${message}"

输出：

 world!
hello 
hello wor
hello world!

Answer 3

问题中引用的文字是一个可怕的解释。以下是the sh language standard的文字：

${parameter%word}
    Remove Smallest Suffix Pattern. The word shall be expanded to produce a pattern.
    The parameter expansion shall then result in parameter, with the smallest portion
    of the suffix matched by the pattern deleted.
${parameter%%word}
    Remove Largest Suffix Pattern. The word shall be expanded to produce a pattern. 
    The parameter expansion shall then result in parameter, with the largest portion
    of the suffix matched by the pattern deleted.
${parameter#word}
    Remove Smallest Prefix Pattern. The word shall be expanded to produce a pattern.
    The parameter expansion shall then result in parameter, with the smallest portion
    of the prefix matched by the pattern deleted.
${parameter##word}
    Remove Largest Prefix Pattern. The word shall be expanded to produce a pattern.
    The parameter expansion shall then result in parameter, with the largest portion
    of the prefix matched by the pattern deleted.

这些表达式在unix中代表什么？

3 个答案: