批处理文件计算字符串中字符的所有出现次数 [英] Batch File Count all occurrences of a character within a string

问题描述

我有一个版本号 17.06.01.01，我想知道有多少条目被一个句点分割.

I have a version number 17.06.01.01 and I would like to know how many entries there are split by a period.

我的最后一段代码是;

setlocal ENABLEDELAYEDEXPANSION
for /F "tokens=1-10 delims=." %%a in ("17.09.01.04.03") do (
set /a "iCount+=1"

echo %%a, !iCount!
)
endlocal

我尝试了许多For"命令，但似乎每次都越来越远.

I've tried a host of 'For' commands but seem to be getting further away each time.

推荐答案

用空格(作为分隔符)替换每个点.然后计算令牌的数量:

replace every dot with a space (as a delimiter). Then count number of tokens:

@echo off
set "string=17.09.01.04.03"
set count=0
for %%a in (%string:.= %) do set /a count+=1
echo %count%

(如果字符串中有其他空格、逗号或制表符，可能会给出错误的结果，但应该适用于仅包含数字和点的版本字符串示例)

(May give false results, if there are other spaces, commas or tabs in the string, but should work nice for your example of version strings containing only numbers and dots)

@treintje:

echo off
set "string=1>7.0 9.0&1.04.0!3"
set count=0
:again
set "oldstring=%string%"
set "string=%string:*.=%"
set /a count+=1
if not "%string%" == "%oldstring%" goto :again
echo %count%

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