Tcl 正则表达式 [英] Tcl regular expressions

问题描述

set d(aa1) 1 
set d(aa2) 1                                                                                                                   
set d(aa3) 1
set d(aa4) 1
set d(aa5) 1
set d(aa6) 1
set d(aa7) 1
set d(aa8) 1
set d(aa9) 1
set d(aa10) 1
set d(aa11) 1
set regexp "a*\[1-9\]"
set res [array names d -glob $regexp]
puts "res = $res"

在这种情况下，结果是:

In this case, the result is:

res = aa11 aa6 aa2 aa7 aa3 aa8 aa4 aa9 aa5 aa1

但是当我将正则表达式从 a*\[1-9\] 更改为 a*\[1-10\] 时，结果变为:

But when I change the regexp from a*\[1-9\] to a*\[1-10\], the result becomes:

res = aa11 aa10 aa1

推荐答案

您的字符类有错误.

• [1-10] 不是 1 到 10 的数字
• 表示1-1，是一个从1到1的字符(即简单的一个1code>) 或 0.这解释了您的输出.
• 要表示从 1 到 10 的数字，请使用:(?:10?|[2-9])(作为几种方法之一.
• 因此你的正则表达式变成了 a*(?:10?|[2-9])
• 注意，如果您的引擎不允许非捕获组，则需要删除?:，为:a*(?:10?|[2-9])

• [1-10] does not mean a digit from 1 to 10
• It means 1-1, which is a character ranging from 1 to 1 (i.e., simply a 1), or a 0. This explains your output.
• to express a digit from 1 to 10, use this: (?:10?|[2-9]) (as one of several ways to do it.
• therefore your regex becomes a*(?:10?|[2-9])
• note that if your engine does not allow non-capturing group, you need to remove the ?:, for: a*(?:10?|[2-9])

这篇关于Tcl 正则表达式的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！