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Date: Mon, 30 Nov 2009 02:18:48 -0800 (PST)
From: Jens Goldhammer <goldhammerdev@googlemail.com>
To: axis-user@ws.apache.org
Subject: Re: How to run .exe file located inside .aar
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Hello Kevin,

try to put the path into brackets! Windows does not like path-elements with
spaces in it :-(

path=3D"\"" + path +"\"";
Process p =3D Runtime.getRuntime().exec(path);

I hope, this helps...
Jens


Kevin Alonso wrote:
>=20
> I get the path of the file:
>=20
> URL url =3D=20
> getClass().getClassLoader().getResource("ColorClusteringFroga.exe");
>                      =09=09=09=09
> String path =3D url.getPath();
>=20
> (Path: C:/Program Files/Apache Software Foundation/Tomcat=20
> 6.0/temp/axis2-tmp-5629
> 138317308731761.tmp/axis2593833549232759380WebService.aar!/execution.ex
> e)
>=20
> But when I put it to execute:
>=20
> Process p =3D Runtime.getRuntime().exec(path);
>=20
>=20
>   I receive the following error:
>=20
> java.io.IOException: Cannot run program "C:/Program": CreateProcess=20
> error=3D2, The
>   system cannot find the file specified
>=20
> I think that it is becouse of !/
>=20
> Any idea to solve this?
>=20
> Thank you in advance.
>=20
> Kevin.
>=20
>=20
> Deepal jayasinghe escribi=C3=B3:
>> One way to do this is you can put the exe file somewhere else and you
>> can define the location from a parameter (in services.xml) and then use
>> the value of the parameter to load the exe file. One other way is you
>> can get the URI of the exe file from axisservice, and you may try to
>> execute it from that
>>=20
>> classloader.getResource("execution.exe")
>>=20
>> Thanks,
>> Deepal
>>> Hello,
>>>
>>> I have a web service that needs to call an .exe file and it is into
>>> the .aar file.
>>>
>>> How can I do this? Which is the path I have to use in the instruction..=
.
>>>
>>> Process p =3D Runtime.getRuntime().exec("execution.exe");
>>>
>>> Is there another way to do this?
>>>
>>> Thank you in advance.
>>>
>>> Kevin.
>>>
>>=20
>>=20
>=20
>=20

--=20
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