批量計數器不增加

Question

代碼不言自明。我已經在註釋行中嘗試了相同結果的命令。最後一行是增量分配和證據enabledelayedexpansion工作的測試。該故障必須位於for循環內。

@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 
) 

這是上面的批處理文件的輸出：

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

我從來沒有在「set」命令之前看過「a /」參數。你確定它不打算成爲「/ a」，這可能會影響你的結果和你的代碼嗎？

我討厭交出魚來代替教魚，但是這與你想要做的相似嗎？

@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% 

輸出看起來是這樣的：

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

通過品評出@ECHO OFF，它看起來像這樣：

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 

注意內循環，引用％無論我們在循環內重置了多少次，COUNT％始終爲0（從進入循環之前）。如果我讓內環路的變化來

SET/A + COUNT = %%一個

我的最終結果呼應％COUNT％會導致15在此之外，我不知道你想達到什麼樣的。

Answer 2

這適用於Windows 7控制檯窗口：

@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! 
) 

注意呼應％count_k％將輸出5，因爲％...％類似於C預處理程序，它在實際運行批處理文件之前執行字符串替換。