3

I have this code

for /l %%a in (0, 1, 25) do (
if /i !TextAlphabet!==!Alphabet[%%a]! (
set AlphabetNumber=%%a
)
)

!Alphabet[]! is an array containing alphabets.

I tried echoing !TextAlphabet! and !Alphabet[%%a]! and when they are the same, %AlphabetNumber% is still the value I set before the loop.

I tried checking for white spaces in my variables but found none.

I also tried adding the variables in a "" tag.

  • This question appears to be off-topic because it is about programming and belongs on Stack Overflow. Try stackoverflow.com/help/how-to-ask – Kevin Panko Feb 22 '14 at 16:14
  • 5
    @KevinPanko The question is fine here. We allow scripting questions that pertain to power user environments and system administration. – slhck Feb 24 '14 at 11:54
4

If you want to check if a variable contains only letters you could use findstr, like this:

set var=Abc
echo %var%|findstr "^[A-Za-z]*$" >nul
if %errorlevel% == 0 (echo Variable is alphabetical)

How it works

The variable is echoed and then piped to the findstr command, which will use a regular expression to match alphabetical characters:

  • ^ matches beginning of line;
  • [A-Za-z] defines a character class which matches any character from A to Z, both upper case and lower case;
  • * repeats zero or more occurrences of the previous class;
  • $ matches end of line.

The %errorlevel% variable will be set to 0 if there's a match, or 1 otherwise. The findstr output will be then redirected to nul, thus ignored.

References

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