批量计数器不增加

Question

代码不言自明。我已经在注释行中尝试了相同结果的命令。最后一行是增量分配和证据enabledelayedexpansion工作的测试。该故障必须位于for循环内。

@echo off 
setlocal EnableDelayedExpansion 
set count_k=5 

for /l %%a in (1,1,5) do (
rem set a/ count_k+=1 
rem set a/ "count_k+=1" 
set a/ count_k=count_k+1 
echo This is count_k per %count_k% 
echo This is count_k exc !count_k! 
) 

echo After loop this is count_k %count_k% 

set _var=first 
set _var=second & echo %_var% !_var! 

set count = 0 
(
set /a count+=1 
echo %count% fails 
echo !count! works 
) 

这是上面的批处理文件的输出：

This is count_k per 5 
This is count_k exc 5 
This is count_k per 5 
This is count_k exc 5 
This is count_k per 5 
This is count_k exc 5 
This is count_k per 5 
This is count_k exc 5 
This is count_k per 5 
This is count_k exc 5 
After loop this is count_k 5 
first second 
fails 
1 works 

Answer 1

我从来没有在“set”命令之前看过“a /”参数。你确定它不打算成为“/ a”，这可能会影响你的结果和你的代码吗？

我讨厌交出鱼来代替教鱼，但是这与你想要做的相似吗？

@ECHO OFF 
SET COUNT=0 
ECHO Before the loop count is: %COUNT% 

FOR /L %%A IN (1,1,5) DO (
    @ECHO Loop %%A 
    SET /A COUNT=%COUNT%+%%A 
) 

ECHO Outside the loop count is: %COUNT% 

输出看起来是这样的：

Before the loop count is: 0 
Loop 1 
Loop 2 
Loop 3 
Loop 4 
Loop 5 
Outside the loop count is: 5 

通过品评出@ECHO OFF，它看起来像这样：

C:\Users\loginID>REM @ECHO OFF 

C:\Users\loginID>SET COUNT=0 

C:\Users\loginID>ECHO Before the loop count is: 0 
Before the loop count is: 0 

C:\Users\loginID>FOR /L %A IN (1 1 5) DO (

SET /A COUNT=0+%A 
) 

C:\Users\loginID>(

SET /A COUNT=0+1 
) 
Loop 1 

C:\Users\loginID>(

SET /A COUNT=0+2 
) 
Loop 2 

C:\Users\loginID>(

SET /A COUNT=0+3 
) 
Loop 3 

C:\Users\loginID>(

SET /A COUNT=0+4 
) 
Loop 4 

C:\Users\loginID>(

SET /A COUNT=0+5 
) 
Loop 5 

C:\Users\loginID>ECHO Outside the loop count is: 5 
Outside the loop count is: 5 

注意内循环，引用％无论我们在循环内重置了多少次，COUNT％始终为0（从进入循环之前）。如果我让内环路的变化来

SET/A + COUNT = %%一个

我的最终结果呼应％COUNT％会导致15在此之外，我不知道你想达到什么样的。

Answer 2

这适用于Windows 7控制台窗口：

@echo off 
setlocal EnableDelayedExpansion 
set count_k=5 
for /l %%a in (1 1 5) do (
set /a count_k=!count_k!+1 
echo !count_k! 
) 

注意呼应％count_k％将输出5，因为％...％类似于C预处理程序，它在实际运行批处理文件之前执行字符串替换。