Hello Kevin,
try to put the path into brackets! Windows does not like path-elements with
spaces in it :-(
path="\"" + path +"\"";
Process p = Runtime.getRuntime().exec(path);
I hope, this helps...
Jens
Kevin Alonso wrote:
>
> I get the path of the file:
>
> URL url =
> getClass().getClassLoader().getResource("ColorClusteringFroga.exe");
>
> String path = url.getPath();
>
> (Path: C:/Program Files/Apache Software Foundation/Tomcat
> 6.0/temp/axis2-tmp-5629
> 138317308731761.tmp/axis2593833549232759380WebService.aar!/execution.ex
> e)
>
> But when I put it to execute:
>
> Process p = Runtime.getRuntime().exec(path);
>
>
> I receive the following error:
>
> java.io.IOException: Cannot run program "C:/Program": CreateProcess
> error=2, The
> system cannot find the file specified
>
> I think that it is becouse of !/
>
> Any idea to solve this?
>
> Thank you in advance.
>
> Kevin.
>
>
> Deepal jayasinghe escribió:
>> 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
>>
>> classloader.getResource("execution.exe")
>>
>> 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 = Runtime.getRuntime().exec("execution.exe");
>>>
>>> Is there another way to do this?
>>>
>>> Thank you in advance.
>>>
>>> Kevin.
>>>
>>
>>
>
>
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