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Jack Daniels
Jack Daniels

Reputation: 31

Batch file - Variables in Variables

I've used a for loop to split a string by spaces. I've also used a loop to make each word in its own variable such as var1=this, var2=is, var3=a, var4=test. That looks like "set var!count! = %%A"

That works. I just need to recall it. How do I do that? Logically, I think it would look like this: %var%count%%

Can someone explain to me how to get that? If I have the 'count' 1, what do I do to get "var1"?

Upvotes: 3

Views: 232

Answers (4)

user10870307
user10870307

Reputation:

I don't see why you would need to do that. What you are looking for seems to be an array variable. You could solve the issue with:

set variable=this;is;a;test
for %%a in (%variable%) do (echo %%a)

For each value you want separate it with a ";" The output of this code will be:

this
is
a
test

Upvotes: 0

Monacraft
Monacraft

Reputation: 6630

When going through arrays in batch just use for /l:

@echo off
setlocal enabledelayedexpansion
var0 = A
var1 = B
var2 = C
var3 = D
var4 = E

for /l %%a in (0, 1, 4) do (
Echo var%%a = !var%%a!
)

Output:

var0 = A
var1 = B
var2 = C
var3 = D
var4 = E

Upvotes: 0

Rafael
Rafael

Reputation: 3112

There is an easy way, you'll need to enable the delayed expansion, so first of all place setlocal enabledelayedexpansion, then use exclamation marks to access these variables. Your script should look like this:

@echo off
setlocal enabledelayedexpansion
:: Here comes your loops to set the variables
echo/!var%count%!

Upvotes: 2

mordecai
mordecai

Reputation: 71

If you want to print out the contents of var1 outside the for statement use 

  Echo %var1%

To print out the contents of var1 inside the block first , delayed expansion needs to be enabled and instead of enclosing the variable references with percent signs , you use 

   Echo !var1!

Upvotes: -1

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