如何返回批量数组的元素? [英] How to return an element of an array in Batch?
问题描述
我在数组列表中的两个元素在我的计划。我该如何分配变量等于要素之一?
I have two elements in my array list in my program. How can I assign a variable to equal one of the elements?
这里的code:
@echo off
setlocal enabledelayedexpansion
set /p string=
for /l %%a in (0,1,1000) do if not "!String:~%%a,1!"=="" set /a length=%%a+1
set i=0
:input
set str=%string:~0,1%
if "%str%"=="M" set array[i]=1000
if "%str%"=="D" set array[i]=500
if "%str%"=="C" set array[i]=100
if "%str%"=="L" set array[i]=50
if "%str%"=="X" set array[i]=10
if "%str%"=="I" set array[i]=1
set string=%string:~1%
set /a i=i+1
if %i%==%length% goto logic
goto input
:logic
我真的,虽然有这样的标准方法。
I really though there was a standard way of doing this.
推荐答案
主要的问题是,你的code不会产生任何的批量阵列。结果
您code只能创建一个名为数组[我]
一个变量,但我想你想创建一个数组:
The main problem is that your code doesn't create any batch-array.
Your code create only one variable named array[i]
, but I suppose you want to create an array with:
array[0]=1000
array[1]=500
然后,你需要像
setlocal EnableDelayedExpansion
set i=0
:inputLoop
set "str=%string:~0,1%"
if "%str%"=="M" set array[%i%]=1000
if "%str%"=="D" set array[%i%]=500
if "%str%"=="C" set array[%i%]=100
if "%str%"=="L" set array[%i%]=50
if "%str%"=="X" set array[%i%]=10
if "%str%"=="I" set array[%i%]=1
set "string=%string:~1%"
set /a i+=1
if NOT %i%==%length% goto :inputLoop
:logic
rem ** logic begins
for /L %%n in (1 1 %i%) do (
echo !array[%%n]!
set /a value=array[%%n]
)
和在逻辑的你可以看到如何访问数组元素的一部分。
And in the logic part you can see how to access an array element.
顺便说一句。你strlen函数有点慢,它可以用一个二进制搜索会更快。结果
如何在批量字符串算上字符?
Btw. Your strlen function is a little bit slow, it could be faster with a binary search.
How to count the characters in a string with Batch?
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