Shulou(Shulou.com)06/03 Report--

This article will explain in detail how to retain blank line replacement strings in batches. The editor thinks it is very practical, so I share it for you as a reference. I hope you can get something after reading this article.

The code is as follows:

@ echo off

:: keep the blank line replacement string

When there is a colon at the beginning of the line, the first consecutive colons will be discarded (this problem can be solved)

:: lines starting with a semicolon are ignored

:: cannot handle exclamation points correctly (because variable delay is enabled)

::

::

Set file=

Set / p file= Please pull the file you want to process to this window:

Set "file=%file:" =% "

For I in ("file%") do set "file=%%~fi"

Echo.

Set replaced=

Set / p replaced= Please enter what is about to be replaced:

Echo.

Set all=

Set / p all= Please enter a replacement string:

Cd. > "% file%_tmp.txt"

Setlocal enabledelayedexpansion

For / f "tokens=1* delims=:"% I in ('findstr / n. * "% file%"') do (

Set str=%%j

If not "! str!" = "" set "stranded stranded% relocated% installed% all%!"

> > "file%_tmp.txt" echo.progresstr!

)

Copy "file%"file%_bak.txt" > nul 2 > nul

Move "file%_tmp.txt"file%"

Start "" file% "

This is the end of this article on "how to keep blank line replacement strings in batches". I hope the above content can be of some help to you, so that you can learn more knowledge. If you think the article is good, please share it for more people to see.

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