如何在不删除分隔符的情况下拆分字符串？

Question

我的 AutoIt 脚本按句子解析文本。因为它们很可能以句号、问号或感叹号结尾，所以我用它来逐句拆分文本：

$LineArray = StringSplit($displayed_file, "!?.", 2)

问题；它会删除定界符（句号、问号和句末的感叹号）。例如，字符串 One. Two. Three. 被拆分为 One、Two 和 Three。

如何在保留句子结尾的句号、问号和感叹号的同时拆分成句子？

Answer 1

试试这个：

#include<Array.au3>
Global $str = "One. Two. Three. This is a test! Does it work? Yes, man! "
$re = StringRegExp($str, '(.*?[.!?])', 3)
_ArrayDisplay($re)

此模式在句子开头没有 space 的情况下有效

#include<Array.au3>
Global $str = "One. Two. Three.This is a test! Does it work? Yes, man! "
$re = StringRegExp($str, '(\S.*?[.!?])', 3)
_ArrayDisplay($re)

Answer 2

使用StringSplit() the delimiters are consumed in the process (and so are lost for the result). Using StringRegExp()：

#include <array.au3>
$string="This is a text. It has several sentences. Really? Of Course!"
$a = stringregexp($string,"(?U)(.*[.?!])",3)
_ArrayDisplay($a)

要删除前导 space(s)，请将模式更改为 "(?U)[ ]*?(.*[.?!])"。或 "(?U) *?(.*[.?!] )" 在 [.!?] 处拆分加上 <space> （在最后一句中添加一个 space）：

#include <array.au3>
$string = "Do you know Pi?   Yes!   What's it?    It's 3.14159!   That's correct."
$a = StringRegExp($string & " ", "(?U)[ ]*?(.*[.?!] )", 3)
_ArrayDisplay($a)

要在句子中保留 @CRLF (\r\n):

#include <array.au3>
$string = "Do you " & @CRLF & "know Pi?   Yes!  What's it?    It's" & @CRLF & "3.14159!   That's correct."
$a = StringRegExp($string & "  ", "(?s)(?U)[ ]*?(.*[.?!][ \R] )", 3)
_ArrayDisplay($a,"Sentences")   ;_ArrayDisplay doesn't show @CRLF

For $i In $a
    ;MsgBox(0,"",$i)
    ConsoleWrite(StringStripWS($i, 3) & @CRLF & "---------" & @CRLF)
Next

当行尾与句尾相同时，这不会保留 @CRLF：...line end!" & @CRLF & "Next line....

如何在不删除分隔符的情况下拆分字符串？

How do I split a string without deleting delimiters?

string

split

autoit