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
