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How to fine tune a DOS command to strip out the rows that represent line numbers within a file?

How would you modify the following DOS Command which so far strips out the rows with semi colons, to convert the following file's contents into plain text stripping out the rows with line number using DOS?

findstr /v /r "^$ --> ^[0-9]*$" TEST.txt > TEST_stripped.txt

I have attached a .SRT file with the following contents:

I was hoping to convert the following .SRT file's contents from

6
00:00:27,000 --> 00:00:30,000
and build some simple webpages.

7
00:00:30,000 --> 00:00:33,000
Part of what makes web design and web development so fun

8
00:00:33,000 --> 00:00:37,000
when compared to other forms of software development, is that you can

to revised values as follows:

and build some simple webpages.
Part of what makes web design and web development so fun
when compared to other forms of software development, is that you can
TEST.txt
0
zimmer9
Asked:
zimmer9
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1 Solution
 
Bill PrewCommented:
I'm confused about what you want.  I ran that command against you test file and seem to get what you said you are looking for?

and build some simple webpages.
Part of what makes web design and web development so fun
when compared to other forms of software development, is that you can

Open in new window

~bp
0
 
Gerwin Jansen, EE MVETopic Advisor Commented:
Look at my answer in your previous question, I suspect you did not copy/paste the findstr pattern correctly.
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Bill PrewCommented:
If you want a slightly different means to the same end, you could try:

findstr /v /r /c:"^$" /c:"^[0-9]*$" /c:"^[0-9:,]* --> [0-9:,]*$" TEST.txt > TEST_stripped.txt

It uses the /v option to select lines NOT matching any of the regex patterns.

The first pattern "^$" eliminates an blank lines.

The second pattern "^[0-9]*$" eliminates any lines with just a number on them.

The third pattern "^[0-9:,]* --> [0-9:,]*$" eliminates the "time stamp" lines with the " --> " on them.
0
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