Question

对不起，我的措辞首先不知道怎么说这个。我的问题是我有一个带有2个变量的循环，我想简单地检查是否存在多个字符的示例代码是：

if %time:~8,1% EQU "" echo Less than 8 characters

我头脑中的代码（这是错误的）看起来像这样

set "AN1=Little String"
set "AN2=Along String More Character"
set "AN3=Extra String With A Lot Character"
set "AN3=SmallSt"
for /l %%A IN (1,1,28) DO (
    If Not !AN%%!:~8,1 EQU "" (echo String has Less than 8 characters) ELSE (
    If Not !AN%%!:~16,1 EQU "" (echo String has Less than 16 characters) ELSE (
    If Not !AN%%!:~24,1 EQU "" (echo String has Less than 24 characters) ELSE (
    )
)

Answer 1

您希望根据长度阈值测试字符串的“数组”。你的逻辑是倒置的（如果第8个字符不为空，那么字符串短于8：错误）除了它已经接近。

关键是：set V=!AN%%A%!。这允许从另一个“生成”变量名称。 EnableDelayedExpansion部分对于能够使用除%之外的其他分隔符非常有用：!否则嵌套会失败。

我采用了5个字符串的例子来涵盖所有测试。

@echo off

setlocal EnableDelayedExpansion

set "AN1=Little String"
set "AN2=Along String More Character"
set "AN3=Extra String With A Lot Character"
set "AN4=SmallSt"
set "AN5=Along String More "

for /l %%A IN (1,1,5) DO (

    set V=!AN%%A%!

    If "!V:~8,1!"=="" (echo "!V!" String has Less than 8 characters) else (
    If "!V:~16,1!"=="" (echo "!V!" String has Less than 16 characters) else (
    If "!V:~24,1!"=="" (echo "!V!" String has Less than 24 characters)
  )
   )
)

结果：

"Little String" String has Less than 16 characters
"SmallSt" String has Less than 8 characters
"Along String More " String has Less than 24 characters

CMD使用带有拆分字符串的EnableDelayedExpansion

1 个答案: