How to exclude strings that do not begin with (insert here)

[eax22]

From this I would get all strings that are eight numbers in length, but what if I want to exclude all of the results that do not begin with two specific numbers? For example out of 12345678 23456789 34567890 I want only those that begin with 34, so the last result would be the only one to show up.

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findstr /m /R "[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]" *.txt

I thought about printing them to a file and using "34.*" but is there another way?

[aGerman]

Your explanation is somehow confusing. You're talking about strings but you used option /m.
From what I understood:

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set "first2digits=34"
for /f "tokens=1* delims=:" %%i in ('findstr /rb "%first2digits%[0-9][0-9][0-9][0-9][0-9][0-9]" *.txt') do echo %%j

Steffen

[eax22]

Apologize for any noobie mistakes, but I thought /m would print the filename containing the string(s) that I want? It's basically meant to be a program that finds all phone numbers that I have saved on the computer. They length is always the same and the first two numbers are always the same, that is why I thought about applying these two to the search.

Your code is exactly what I wanted to do, it's more elegant than the approach I initially thought about, but then again, I naturally seem to take the harder routes (beginner habit?)

Thanks 4 the help.

[aGerman]

I thought /m would print the filename containing the string(s)

That's correct. But only the file names without the found strings.
If you also want to see the file names execute only the findstr command line without the FOR loop.

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findstr /rb "%first2digits%[0-9][0-9][0-9][0-9][0-9][0-9]" *.txt

The output will be in that format:
File Name:Found String

Steffen

[eax22]

I made a text file with the content 34111111 and another one with 34111111111111111111111. Both of them came up, instead of only the one with eight numbers. Is not the limit set by the amount of times I write [0-9]? F.ex., [0-9][0-9] = 2, [0-9][0-9][0-9] = 3. A bit confused here, any ideas as to why this happens?

[aGerman]

If the line only consists of the number then replace option b with x.
If something else is following in the line then append a word boundary to the search pattern

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findstr /rb "%first2digits%[0-9][0-9][0-9][0-9][0-9][0-9]\>" *.txt

Steffen

[ShadowThief]

Nope, you're saying "find a string that contains [0-9][0-9][0-9][0-9][0-9][0-9][0-9]", and 3411111111111111111111111 does that, it just contains a lot of other numbers.

If the phone numbers are one per line, you can say that there will be 8 digits and then the end of the line

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findstr /rb "%first2digits%[0-9][0-9][0-9][0-9][0-9][0-9]$" *.txt

If there are other things on the line, you can just throw a space at the end of the search string

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findstr /rb "%first2digits%[0-9][0-9][0-9][0-9][0-9][0-9] " *.txt

[aGerman]

ShadowThief I assume the latter won't work without option c.

[ShadowThief]

It would if /C and /R played nice together. Unfortunately, what I've suggested won't work at all and they should stick to your recommendation of using \> for word barriers.

I really there was only one flavor of regex; it would make things so much easier. Just a /^([0-9]{7}).*/ and you're done.

[eax22]

ShadowThief I assume the latter won't work without option c.

You mean

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findstr /rb /c:"[3][4][0-9][0-9][0-9][0-9][0-9][0-9] " *.txt

?

[aGerman]

Yes. Or simply

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findstr /rbc:"%first2digits%[0-9][0-9][0-9][0-9][0-9][0-9] " *.txt

But note that this will also match the space next to the number. If the line ends or there follows something else (like a tab or hyphen) it would fail.

Steffen