Question

我不知道什么＃？是的，我走了，什么也没找到

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Answer 1

${tag#?}扩展为$tag的值，其中第一个字符被删除。

引用POSIX shell specification：

${parameter#[word]}
删除最小前缀模式。 单词应为展开以产生图案。然后参数扩展结果为 parameter ，且前缀的最小部分匹配由图案删除。如果存在， word 不得以未引用的'#'。

在这种情况下，模式为?，它与单个字符匹配。

如果您使用的是bash，则Bash manual也涵盖了这一点（您可以使用info bash在系统上查看手册）

'${PARAMETER#WORD}'
'${PARAMETER##WORD}'
     The WORD is expanded to produce a pattern just as in filename
     expansion (*note Filename Expansion::).  If the pattern matches the
     beginning of the expanded value of PARAMETER, then the result of
     the expansion is the expanded value of PARAMETER with the shortest
     matching pattern (the '#' case) or the longest matching pattern
     (the '##' case) deleted.  If PARAMETER is '@' or '*', the pattern
     removal operation is applied to each positional parameter in turn,
     and the expansion is the resultant list.  If PARAMETER is an array
     variable subscripted with '@' or '*', the pattern removal operation
     is applied to each member of the array in turn, and the expansion
     is the resultant list.

Answer 2

#!/bin/bash
tag='123'
version=${tag#?}
echo ${version}  # output is: 23

=======
所以＃？是删除第一个字符？

Shell脚本中$ {var＃？}是什么意思

2 个答案: