error `1"" was unexpected at this time.` when first command line argument is double quoted

Question

Running the following minimal example:

(
    echo if "%%1" == "" (
    echo   echo Success ^|^| goto :fail
    echo ^) else (
    echo   echo Failure ^|^| goto :fail
    echo ^)
) >"testresult.bat"

call "testresult.bat" "first 1" --flag="other options" --verbose

Results the following as contents of testresult.bat:

if "%1" == "" (
  echo Success || goto :fail
) else (
  echo Failure || goto :fail
)

And calling testresult.bat with the command line arguments "first 1" --flag="other options" --verbose gives me:

C:\User>test.bat

C:\User>(
echo if "%1" == "" (
 echo   echo Success || goto :fail
 echo ) else (
 echo   echo Failure || goto :fail
 echo )
) 1>"testresult.bat"

C:\User>call "testresult.bat" "first 1" --flag="other options" --verbose
1"" was unexpected at this time.

C:\User>if ""first 1"" == "" (

C:\User>

The error 1"" was unexpected at this time. is caused because my first arguments has a double quotes.

How can I properly write my if "%1" == "" to check whether the first argument is empty, regardless it has double quotes or not?

Have you tried `if "%~1"==""` (note the `~`)? See [this](https://ss64.com/nt/syntax-args.html)... — aschipfl, Nov 15 '19 at 22:12

score 3 · Accepted Answer · answered Nov 16 '19 at 11:04

The issue here is referencing argument 1 with "%1" instead of "%~1".

The help output on running in a Windows command prompt window call /? explains that %1 references the first argument as passed to the batch file without or with being enclosed in double quotes.

For that reason passing the string "first 1" as first argument to the created batch file results in an IF condition with ""first 1"" left to comparison operator == which is a serious syntax error on the IF condition.

The usually used solution is using "%~1" on IF command line. The modifier ~ results in removing surrounding " from first argument string if there are surrounding double quotes at all. So the first string on IF condition is now nearly always correct enclosed with one pair of double quotes.

The batch file can be called with if "%~1" == "" with first argument being:

argument1 ... not quoted argument string because of " not necessary.
"argument 1" ... quoted argument string because of space character.
"" ... empty string as argument string.
The condition is true although the batch file is run with an argument string. Most batch files requiring one or more argument strings do not allow that one of the argument strings is an empty string. But some batch scripts support also empty argument strings.
"argument 1 ... not correct quoted argument string with space character working nevertheless on using %~1 as the single " at beginning is removed in this case.
argument1" ... not correct quoted argument string.
Windows command processor does not remove " at end of the not correct specified argument string resulting in a syntax error on command line if "%~1" == "" because of resulting command line is if "argument1"" == "".

See my answer on What is the difference between "…" and x"…" in an IF condition in a Windows batch file? It explains in detail how to write a batch file which works even on last argument example.

error `1"" was unexpected at this time.` when first command line argument is double quoted

1 Answers1