How to start an application without waiting in a batch file?

Is there any way to execute an 开发者_JAVA百科application without waiting in batch file? I have tried the start command but it just creates a new command window.

I'm making a guess here, but your start invocation probably looks like this:

start "\Foo\Bar\Path with spaces in it\program.exe"

This will open a new console window, using “\Foo\Bar\Path with spaces in it\program.exe” as its title.

If you use start with something that is (or needs to be) surrounded by quotes, you need to put empty quotes as the first argument:

start "" "\Foo\Bar\Path with spaces in it\program.exe"

This is because start interprets the first quoted argument it finds as the window title for a new console window.

I used start /b for this instead of just start and it ran without a window for each command, so there was no waiting.

If your exe takes arguments,

start MyApp.exe -arg1 -arg2

If start can't find what it's looking for, it does what you describe.

Since what you're doing should work, it's very likely you're leaving out some quotes (or putting extras in).

EDIT moved /B parameter/switch to before the command path, as per Nime Cloud's recommendation

I successfully used a combo of @joey's post, with added /B flag, as follows

START "" /B "Path\To\Application.exe"

Partial output from help command: HELP START...

  B  Start application without creating a new window. The
     application has ^C handling ignored. Unless the application
     enables ^C processing, ^Break is the only way to interrupt
     the application.

This command starts ClamAV in a seperate console window with custom icon of clamd.exe

start "c:\clamav\clamd.exe" "c:\clamav\clamd.exe"

So the generic format would be like:

start "Your new title" "c:\path\your.exe"