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我有一些 SRT 数据以 \r 和 \n 标记作为每个句子中间的换行符返回。如何仅在文本/句子中间找到那些 \r 和 \n 标记，而不是表示其他换行符的其他标记。

示例来源：

18
00:00:50,040 --> 00:00:51,890
All the women gather
at the hair salon,

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters
and they dye their hair orange.

期望的输出：

18
00:00:50,040 --> 00:00:51,890
All the women gather at the hair salon,

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters and they dye their hair orange.

我对正则表达式绝对是垃圾，但我最好的猜测（无济于事）是这样的

var reg = /[\d\r][a-zA-z0-9\s+]+[\r]/

然后对其进行 split() 以删除其中一个值中间的任何 \r 。我敢肯定这甚至不接近正确的方式，所以...stackoverflow！:)

4

2 回答 2

1

This regex should do the trick:

/(\d+\r\d{2}:\d{2}:\d{2},\d{3} --> \d{2}:\d{2}:\d{2},\d{3}\r)([^\r]+)\r([^\r]+)(\r|$)/g

To make this work with more lines (has to be a set number) then just add more ([^\r]+)\r's. (Remember to also add $'s to the match replace as so (with 3 lines): '$1$2 $3 $4\r').

Usage

mystring = mystring.replace(/(\d+\r\d{2}:\d{2}:\d{2},\d{3} --> \d{2}:\d{2}:\d{2},\d{3}\r)([^\r]+)\r([^\r]+)(\r|$)/g, '$1$2 $3\r');

Limitations

  • If there is more than 2 lines of text this won't work.

Example 1

Works fine!

Input:

18
00:00:50,040 --> 00:00:51,890
All the women gather
at the hair salon,

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters
and they dye their hair orange.

Output:

18
00:00:50,040 --> 00:00:51,890
All the women gather at the hair salon,

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters and they dye their hair orange

Example 2

Doesn't work; more than 2 lines

Input:

18
00:00:50,040 --> 00:00:51,890
All the women gather
at the hair salon,
and they just talk

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters
and they dye their hair orange.
Except for Maria who dyes it pink.

Output:

18
00:00:50,040 --> 00:00:51,890
All the women gather at the hair salon,
and they just talk

19
00:00:52,080 --> 00:00:56,210
all the mothers and daughters and they dye their hair orange.
Except for Maria who dyes it pink.
于 2012-10-22T21:13:44.637 回答
1

这将匹配您想要摆脱的换行符，捕获它之前和之后的字符，将这两个放回空格周围：

var regex = /([a-z,.;:'"])(?:\r\n?|\n)([a-z])/gi;
str = str.replace(regex, '$1 $2');

关于正则表达式的一些事情。我使用了修饰符i并g使其不区分大小写并查找字符串中的所有换行符，而不是在第一个换行符之后停止。此外，它假定可移动换行符可以出现在字母、逗号、句点、分号、冒号或单引号或双引号之后以及另一个字母之前。正如@nnnnnn 在上面的评论中提到的那样，这不会涵盖所有可能的句子，但它至少不应该被大多数标点符号窒息。换行符必须是单个换行符，但它与平台无关（可以是\r,\n或\r\b）。我捕获了换行符之前的字符和换行符之后的字母（带括号），因此我可以在替换字符串中使用$1and访问它们$2。这基本上就是它的全部内容。

于 2012-10-22T20:57:05.397 回答