‘Pretty print’ windows %PATH% variable – how to split on ‘;’ in CMD shell

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The simple way is to use

for %a in ("%path:;=";"%") do @echo %~a

This works for all without ; in the path and without " around a single element
Tested with path=C:\qt\4.6.3\bin;C:\Program Files;C:\documents & Settings

But a “always” solution is a bit complicated
EDIT: Now a working variant 

@echo off
setlocal DisableDelayedExpansion
set "var=foo & bar;baz<>gak;"semi;colons;^&embedded";foo again!;throw (in) some (parentheses);"unmatched ;-)";(too"

set "var=%var:"=""%"
set "var=%var:^=^^%"
set "var=%var:&=^&%"
set "var=%var:|=^|%"
set "var=%var:<=^<%"
set "var=%var:>=^>%"

set "var=%var:;=^;^;%"
rem ** This is the key line, the missing quote is intended
set var=%var:""="%
set "var=%var:"=""%"

set "var=%var:;;="";""%"
set "var=%var:^;^;=;%"
set "var=%var:""="%"
set "var=%var:"=""%"
set "var=%var:"";""=";"%"
set "var=%var:"""="%"

setlocal EnableDelayedExpansion
for %%a in ("!var!") do (
    endlocal
    echo %%~a
    setlocal EnableDelayedExpansion
)

What did I do there?
I tried to solve the main problem: that the semicolons inside of quotes should be ignored, and only the normal semicolons should be replaced with ";"

I used the batch interpreter itself to solve this for me.

  • First I have to make the string safe, escaping all special characters.
  • Then all ; are replaced with ^;^;
  • and then the trick begins with the line
    set var=%var:"=""%" (The missing quote is the key!).
    This expands in a way such that all escaped characters will lose their escape caret:
    var=foo & bar;;baz<>gak;;"semi^;^;colons^;^;^&embedded";;foo again!;;…
    But only outside of the quotes, so now there is a difference between semicolons outside of quotes ;; and inside ^;^;.
    Thats the key.

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