CODEWITHSUNDEEP
stringbatch-filecmdfindstr
Angie
Angie

Reputation: 45

Batch - find a string in first 10 characters on line and print whole line

I am having trouble with findstr command. I am using batch file to execute a proccess that will search all .txt in folder and print out lines that have a string in first 10 characters that is previously defined. So far I have been using this batch: 

for /f "delims=" %%I in ('dir/B *.txt') do (
    for /f "delims=" %%J in (%%I) do (
        set "=%%J"
        call echo %%:~0,10%%|findstr "R0621 32411"&&call echo %%_%% >> search.txt
    ) 
)
endlocal
Somehow this batch however does not print out lines that consists strings like R0621 or 32411. Is this a bug? When I try the typical findstr batch it works and prints out lines just fine. For example this: 
findstr "R0621 32411" *.txt >> search.txt
.txt files that this batch searches through looks like this: 
AA32411   AAA RANDOMTEXTANDNUMBERS 13121313212153
BBR0621   BBB RANDOMTEXTANDNUMBERS 78975487798797
CCY4488   CCC RANDOMTEXTANDNUMBERS 44455577799998
I cant use the findstr because it finds string after 10 characters and those lines i dont need (i need only those that have strings i define in first 10 chars per line).

Is there any alternative? I tried to search internet but couldnt find any help anywhere. Also for better understanding you can check my previous thread Batch to find a string of number&letter combination only in first 10 characters (in a txt file) per row and print the whole row

Upvotes: 2

Views: 3410

Answers (2)

MC ND
MC ND

Reputation: 70923

@echo off
    setlocal enableextensions

    set "tempFile=%temp%\%~nx0.tmp"
    set "outputFile=search.out"

    (for %%a in ( 
        R0621 .R0621 ..R0621 ...R0621 ....R0621 .....R0621
        32411 .32411 ..32411 ...32411 ....32411 .....32411
    ) do @echo %%a) > "%tempFile%"

    type nul > "%outputFile%"
    for %%a in (*.txt) do findstr /r /b /g:"%tempFile%" "%%~a" >> "%outputFile%"

    del /q "%tempFile%" >nul 2>nul

    endlocal

It simply construct a temporary file with the adecuated regular expressions to search for. As the strings should be located in the first 10 characters and they are five character long (all this is hardcoded, but may be adapted), the search string can start at the first, second, ... up to the sixth position from line start (a dot in a regular expressions means that any character can be at that position)

Once the search file has been constructed, a for loop will iterate over the .txt files and for each of them we search with findstr, searching at the begin of line (/b) for a regular expression (/r) that is read from the generated file (/g:)

Upvotes: 2

Aacini
Aacini

Reputation: 67196

@echo off
setlocal EnableDelayedExpansion

rem Seek the target strings in all .txt files
(for /F "tokens=1* delims=:" %%a in ('findstr "R0621 32411" *.txt') do (
   rem Check that the target string(s) exists in first 10 characters of found lines
   set "line=%%b"
   rem Get just the first 10 characters
   set "tenChars=!line:~0,10!"
   rem If R0621 was in first 10 characters
   if "!tenChars:R0621=!" neq "!tenChars!" (
      echo !line!
   rem If 32411 was in first 10 characters
   ) else if "!tenChars:32411=!" neq "!tenChars!" (
      echo !line!
   )
)) > search.out

Please note that if the output file have also .txt extension it would be included in the original findstr! You may use a ren search.out search.txt command at end to rename the output file with the right extension, or create the output file with .txt extension in a different directory.

Upvotes: 3

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