Batch:变量间接:通过动态构造的名称获取变量的值

问题描述

所以当我运行这段代码时:

So when I run this code:

@echo off
setlocal enabledelayedexpansion

set lstFolders= First Second

set intCounter=0
for %%i in (!lstFolders!) do (
    set /a intCounter += 1
    set strFlder=%%i
    set strFolder!intCounter!=!strFlder!
    echo %%i
    echo !strFlder!
    echo !strFolder%intCounter%!
    echo !strFolder1!
    echo !strFolder2!
)

:End
pause
endlocal

结果如下:

First
First
ECHO is off.
First
ECHO is off.
Second
Second
ECHO is off.
First
Second

为什么不允许我回显以以下格式创建的变量:！strFolder％intCounter％！?有没有其他方法可以引用此变量并获取其中的数据?

Why doesn't it allow me to echo the variable create with the format : !strFolder%intCounter%!? Is there another way to reference this variable and get the data that is inside of it?

推荐答案

注意:以下代码仅适用于列表值(如％lstFolders％的标记，例如 First ):

Caveat: The code below only works with list values (the tokens of %lstFolders% such as First) that:

• 都不包含空格
• 也不包含以下任何字符:&|<>"

需要一种不同的循环方法来处理这种情况.

A different looping approach would be needed to handle such cases.

@echo off
setlocal enabledelayedexpansion

set "lstFolders=First Second"

set intCounter=0
for %%i in (%lstFolders%) do (
      rem Increment the counter
    set /a intCounter += 1
      rem Echo the loop variable
    echo #!intCounter!=%%i
      rem Set variable strFolder<intCounter> to the loop variable's value
    set "strFolder!intCounter!=%%i"
      rem Echo the variable created using variable indirection with for /f ('...')
    for /f "delims=" %%v in ('echo "%%strFolder!intCounter!%%"') do set "thisFolder=%%~v"
    echo %%thisFolder%% ^(via %%strFolder!intCounter!%%^)=!thisFolder!
)

运行以上结果:

#1=First
%thisFolder% (via %strFolder1%)=First
#2=Second
%thisFolder% (via %strFolder2%)=Second

---

您要查找的是可变间接定向:

• 您可以间接设置一个变量(通过您根据另一个变量的值动态构造的名称)，例如，
  • 设置"strFolder！intCounter！= %% i" ，并且！intCounter！的值为 1 ，可以正确设置变量 strFolder1 设置为 %% i )的值，
  • While you can set a variable indirectly (by a name that you construct dynamically from the value of another variable), e.g.,
    • set "strFolder!intCounter!=%%i", with !intCounter! having a value of 1, correctly sets variable strFolder1 to the value of %%i),

    您无法以这种方式获取 的值；您需要一个额外的评估步骤，它可以为/f ...('echo ...')提供.:

    you cannot get a variable's value that way; you need an extra evaluation step, which for /f ... ('echo ...') can provide.:

    • 用于/f"delims =" %% v in('echo"%% strFolder！intCounter！%%"')... ...单引号( echo ... )并将结果整体( delims = )分配给变量 %% v
      ( %%〜v 删除了在 echo 参数周围添加的双引号，以使命令处理诸如& |<<>; 正确).

    • for /f "delims=" %%v in ('echo "%%strFolder!intCounter!%%"') do ... parses the output of the command in single quotes (echo ...) and assigns the result as a whole (delims=) to variable %%v
(%%~v removes the enclosing double quotes, which were added around the echo argument to make the command handle shell metacharacters such as & | < > correctly).

    %% strFolder！intCounter！%% 立即将 strFolder！intCounter！评估为 strFolder1 ，如果！intCounter！是 1 ，则由于包含了 doubled ％实例，该结果以 literal ％strFolder1％，这是 echo 命令在 for 命令运行时看到的内容，导致它评估变量引用并扩展为其值.

    %%strFolder!intCounter!%% immediately evaluates strFolder!intCounter! to strFolder1, if !intCounter! is 1, which, thanks to the enclosing doubled % instances, ends up as literal %strFolder1%, which is what the echo command sees when it is run by the for command, causing it to evaluate the variable reference and expand to its value.

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