批处理文件来编辑ini文件行 [英] Batch file to edit line in ini file

问题描述

我有被自动生成ini文件。

I have an ini file that gets autogenerated.

它的第二行总是：
版本= W.XX.Y.ZZ

其中，是W 是主版本号， XX 是次版本，ÿ 是生成和 ZZ 是修订

Where W is the major version number, XX is the minor version, Y is the Build and ZZ is the Revision.

我需要打开ini文件和使用批处理文件编辑行，以便在该版本的内部版本号和修订号移除。因此，该行最终应该是这样的：
版本= W.XX

I need to open that ini file and edit that line using a batch file so that the build and revision numbers in that version get removed. Therefore, the line should end up like this: Version = W.XX

主要数字将永远是一个字符，而辅号永远是两个，因此整个字符串为14个字符（含空格）长。

The major number will always be one character and the minor number will always be two, therefore the entire string is 14 characters (inc spaces) long.

我希望我能得到的是字符串左该行14个字符，并替换该字符串该行。

I was hoping that I could get the string that is LEFT 14 characters of that line and replace that line with that string.

推荐答案

左的语法你要求的是使用一个变量子扩展：％VAR：〜14％

The "LEFT" syntax you're asking for is to use a variable substring expansion: %var:~,14%

以下code将包含字符串的每一行执行LEFT 14，版本

The following code will perform a "LEFT 14" on every line that contains the string "Version"

setlocal enabledelayedexpansion
del output.ini
for /f "tokens=*" %%a in (input.ini) do (
  set var=%%a
  if not {!var!}=={!var:Version=!} set var=!var:~,14!
  echo.!var! >> output.ini
)
endlocal

如果有其他的线，版本的话，你也可以修改循环使用一个计数器。

If there are other lines with the word "Version", you can also modify the loop to use a counter.

setlocal enabledelayedexpansion
del output.ini
set counter=0
for /f "tokens=*" %%a in (input.ini) do (
  set var=%%a
  set /a counter=!counter!+1
  if !counter! EQU 2 set var=!var:~,14!
  echo.!var! >> output.ini
)
endlocal

请注意，在这两种情况下，你可能需要做更多的工作，如果你的文件包含特殊符号像|，＆LT，或>

Note that in both cases, you might have to do more work if your file contains special symbols like |, <, or >

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