正则表达式 - 没有时间的新线

Question

I need to preform a preg_replace for every empty line that is not followed by this:

00:00:02.800 --> 00:00:04.800

Its format is:

any 2 digits:any 2 digits:any 2 digits.any 3 digits --> any 2 digits:any 2 digits:any 2 digits.any 3 digits

I know how to search for an empty line:

"/(^[
]*|[
]+)[\s\t]*[
]+/"

And for the time row :

[0-9]{1,2}[:.,-]?[:][0-9]{1,2}[:.,-]?[:][0-9]{1,2}[:.,-]?[.][0-9]{1,3}[:.,-]?[\s][-][-][>][\s][0-9]{1,2}[:.,-]?[:][0-9]{1,2}[:.,-]?[:][0-9]{1,2}[:.,-]?[.][0-9]{1,3}[:.,-]?

But I wasn't able to create a regex that finds only the lines that are not followed by the time row.

EDIT: OPTION 1 File Input:

WEBVTT

00:00:00.300 --> 00:00:01.000
line1

line2
line3

00:00:01.000 --> 00:00:02.800
line1

00:00:02.800 --> 00:00:04.800
line1
line2


line3

File desired output:

WEBVTT

00:00:00.300 --> 00:00:01.000
line1  
line2
line3

00:00:01.000 --> 00:00:02.800
line1

00:00:02.800 --> 00:00:04.800
line1
line2
line3

My function:

 $content = preg_replace("/regex expresion/", "", $file_content);

EDIT 2:

Just found out I need to find another format: OPTION 2 File Input:

1
00:00:00,300 --> 00:00:01,000
line 1 line 1

line 2

2
00:00:01,000 --> 00:00:02,800
line 1 line 1 line 1

line2


line 3 line 3

3
00:00:02,800 --> 00:00:04,800
line 1

File desired output:

1
00:00:00,300 --> 00:00:01,000
line 1 line 1
line 2

2
00:00:01,000 --> 00:00:02,800
line 1 line 1 line 1
line2
line 3 line 3

3
00:00:02,800 --> 00:00:04,800
line 1

Totos answer worked great. I tried to modify it to my need and was unsuccessful. I tried:

/(\R){1,}(?!(\d\R\d\d:\d\d:\d\d\.\d{3}) --> (?2))

---

Solved

Solution:

option 1:

$regex = "/(\R){1,}(?=(\d\d:\d\d:\d\d\.\d{3}) --> (?2))/";

option 2:

$regex = "/(\R)(?!(\d\R\d\d:\d\d:\d\d\,\d{3}))/";

Answer 1

$str = <<<EOD
1
00:00:00,300 --> 00:00:01,000
line 1 line 1

line 2

2
00:00:01,000 --> 00:00:02,800
line 1 line 1 line 1

line2


line 3 line 3

3
00:00:02,800 --> 00:00:04,800
line 1
EOD;

$str =preg_replace('/(\R)+(?!\d)/', '$1', $str);
echo $str,"
";

Output for given example:

00:00:00,300 --> 00:00:01,000
line 1 line 1
line 2

2
00:00:01,000 --> 00:00:02,800
line 1 line 1 line 1
line2
line 3 line 3

3
00:00:02,800 --> 00:00:04,800
line 1

Explanation:

(\R)+       : group 1, any kind of linebreak, 2 or more times
(?!\d)      : negative lookahead, make sure we don't have digit after

Or, if lineX could begin with digit:

$str =preg_replace('/(\R){2,}(?!(\d\d:\d\d:\d\d\.\d{3}) --> (?2)|\d+)\R/', '$1', $str);