How can I pass arguments to a batch file?

Question

I need to pass an ID and a password to a batch file at the time of running rather than hardcoding them into the file.

Here\'s what the command line looks like:

Answer 1

Accessing batch parameters can be simple with %1, %2, ... %9 or also %*,
but only if the content is simple.

There is no simple way for complex contents like "&"^&, as it's not possible to access %1 without producing an error.

set  var=%1
set "var=%1"
set  var=%~1
set "var=%~1"

The lines expand to

set  var="&"&
set "var="&"&"
set  var="&"&
set "var="&"&"

And each line fails, as one of the & is outside of the quotes.

It can be solved with reading from a temporary file a remarked version of the parameter.

@echo off
SETLOCAL DisableDelayedExpansion

SETLOCAL
for %%a in (1) do (
    set "prompt="
    echo on
    for %%b in (1) do rem * #%1#
    @echo off
) > param.txt
ENDLOCAL

for /F "delims=" %%L in (param.txt) do (
  set "param1=%%L"
)
SETLOCAL EnableDelayedExpansion
set "param1=!param1:*#=!"
set "param1=!param1:~0,-2!"
echo %%1 is '!param1!'

The trick is to enable echo on and expand the %1 after a rem statement (works also with %2 .. %*).
So even "&"& could be echoed without producing an error, as it is remarked.

But to be able to redirect the output of the echo on, you need the two for-loops.

The extra characters * # are used to be safe against contents like /? (would show the help for REM).
Or a caret ^ at the line end could work as a multiline character, even in after a rem.

Then reading the rem parameter output from the file, but carefully.
The FOR /F should work with delayed expansion off, else contents with "!" would be destroyed.
After removing the extra characters in param1, you got it.

And to use param1 in a safe way, enable the delayed expansion.