Regular Expression with findstr (ms-dos) - regex

Question

I am trying to use ms-dos command findstr to find a string and eliminate it from the file.
At the moment I can find an explicit string but I am really struggling with regular expressions.
The file looks something like the below:
PLs - TULIP Report
Output_Format, PLS - TULIP REPORT
NUMLINES, 110907
VARIABLE_TYPES,T1,T8,I,T9,T2,N,N,N
[[data below]]
The file is an export from some system and annoyingly has that header in it - so I would like to clean it before using SQL Loader to bring it into an Oracle database.
There's more than just the one file and all would have the same type of header but ever so slightly different in every file.
Although I am happy to first remove the first 2 lines using hardcoded values, e.g.:
findstr /v "PLs - TULIP Report" "c:\myfiles\file1.PRO" > "c:\myfiles\file1.csv"</code><br>
findstr /v "Output_Format, PLS - TULIP REPORT" "c:\myfiles\file1.csv" > "c:\myfiles\file2.csv"
(note how I do that in 2 steps - any suggestions to make this happen in a single step, would be massivelly appreciated)
The third line is mnore complicated for me, it will always be in that format:
NUMLINES, 110907
except that the number at the end would be different for each file. So how do I get to find this entire line using a regular expression? I have tried:
findstr /v /b /r "\D+ \s+ \d+"
but without any luck.
FYI, the data in [[data below]] looks like
*,"00000161",456823,"017896532","FU",23.95,3.34,20.61
etc ..
Obviously, I do not want to modify the data area.
I hope the above makes sense,
Thanks

Answer 1

You must exclude single lines, findstr cannot match multiple lines. Just separate the different regexes with a space
findstr /r /b /v "NUMLINES PLs Output_Format" *.txt
^regex1 ^2 ^3
Specifying /b allows you to find matches only at the beginning of the lines and /v excludes those lines.
EDIT:
Of course the usage is
findstr /r /b /v "NUMLINES PLs Output_Format" yourfile > yourtarget
And in yourtarget you will find the data of yourfile except the lines excluded by the regex.
EDIT 2:
Based on your comments you need just to add VARIABLE_TYPES to your regex making it
findstr /r /b /v "NUMLINES PLs Output_Format VARIABLE_TYPES" yourfile > yourtarget
This is the way to complete the whole operation in one single instruction.

Answer 2

Here is a one liner using regex that will exclude all four lines. (I used line continuation so that the code looks better.) Each line must match exactly. I allow for each line to end in any number of spaces because I wasn't sure of your format. Note - FINDSTR regex support is very limited and non-standard. There are many other FINDSTR quirks and bugs. See What are the undocumented features and limitations of the Windows FINDSTR command? for more info.
findstr /vrx /c:"PLs - TULIP Report *"^
/c:"Output_Format, PLS - TULIP REPORT *"^
/c:"NUMLINES, *[0-9]* *"^
/c:"VARIABLE_TYPES,T1,T8,I,T9,T2,N,N,N *"^
"c:\myfiles\file1.PRO" >"c:\myfiles\file1.csv"
If all you need to do is skip the first 4 lines, then you normally should be able to use MORE. But there are some circumstances with large files where MORE can hang, but I can't remember the specifics. Also MORE will convert tabs into a series of spaces.
more +4 "c:\myfiles\file1.PRO" >"c:\myfiles\file1.csv"
Another option is to use a FOR /F loop. The FOR /F skips empty lines, but I don't think that is a concern for you.
>"c:\myfiles\file1.csv" (
for "usebackq skip=4 delims=" %%A in ("c:\myfiles\file1.PRO") do echo(%%A
)
If any of your data can begin with a ; then the code gets a bit uglier. You would then want to disable the EOL option by setting it to a line feed character.
set LF=^
::above 2 blank lines are critical - do not remove
>"c:\myfiles\file1.csv" (
for usebackq^ skip^=4^ eol^=^%LF%%LF%^ delims^= %%A in ("c:\myfiles\file1.PRO") do echo(%%A
)