Aha,
That explains the ! . Thanks Mohammad.
On 5/26/07, Mohammad Nour El-Din <[EMAIL PROTECTED]> wrote:
Hi Karan...
On 5/27/07, Karan Malhi <[EMAIL PROTECTED]> wrote:
>
> I looked at the documentation on the openejb website, wanted to start
the
> snapshot version of the server. The docs say that there should be a bin
> folder within snapshot jar, I did not have one. So, i thought maybe
maven
> did not build the assembly properly, so i clean installed again, i did
not
> get it this time either. looked into the source code and found
> org.apache.openejb.cli.Bootstrap. I guess this is the starting point
(this
> was the main class in openejb-core-3.0-snapshot.jar) for starting from
the
> command-line. Is that correct? Also, should the Snapshot jar have a bin
> folder (mine doesnt have one)?
I will try to build you a new assembly, cauz I am not sure why the
assembly
is not built properly.
I am a little puzzled as to why this line of code is in the Bootstrap
>
> propsString = propsString.substring(0, propsString.indexOf("!"));
>
> When i look at openejb-core/src/main/resources/openejb-
version.properties,
> I
> don't find a ! in there. So the above code should always throw an
> exception.
> Why dont i see the exception when i run Bootstrap.java from SNAPSHOT,
> whereas when i run it from my IDE i.e. the checked out version from SVN,
i
> get the exception.
>
> Could anybody clarify this for me?
For the line of code you are asking about, this line is there cauze the
properties file is searched for within the jar file and AFAIK looking for
resources this way the path of the resource is returened as the path of
the
jar file itself, the "!", and then the fully qualified name of the
resource,
this y the code looks for the "!" and this is y it troughs an exception
while you are running the code from within ur IDE.
--
> Karan Malhi
>
--
Thanks
- Mohammad Nour
--
Karan Malhi
