Publishing automatically WSDL to WSRR

Jose Miguel Ordax Cassa (2.4k●4●126●100) | asked Nov 13 '10, 3:08 a.m.

Hi, is there any way to setup RAM to be able to publish automatically to
WSRR a WSDL within an Asset when the Asset arrives to Approved state. I
would like to avoid to go into RAM and manually press the publish button.

Thanks in advance,

Chemi.

2 answers

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Gili Mendel (1.8k●5●6) | answered Nov 22 '10, 1:30 p.m.
JAZZ DEVELOPER

RAM 7.2.x has a special ALE Policy. When you create an SOA life cycle, RAM automatically places this policy on every state.

This policy (when executed on a state), reads an XML configuration file that tells it what to do (WSRR related thing).

When you configure a WSRR connection it uses a
ALEConfiguration.xml ... check it out ... you can override this configuration file to do a publish on the state you want.

For instance:


&lt;Trigger source=&quot;RAM&quot; type=&quot;Specification&quot; state=&quot;Identified&quot;&gt;

     &lt;Action target=&quot;WSRR&quot; action=&quot;create&quot;/&gt; 

&lt;/Trigger&gt;

Tells the policy, that Asset of type Specification in the Identified state should be published to WSRR.

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Jose Miguel Ordax Cassa (2.4k●4●126●100) | answered Nov 23 '10, 2:23 p.m.

On 11/22/2010 7:38 PM, gmendel wrote:

RAM 7.2.x has a special ALE Policy. When you create an SOA life
cycle, RAM automatically places this policy on every state.

This policy (when executed on a state), reads an XML configuration
file that tells it what to do (WSRR related thing).

When you configure a WSRR connection it uses a
ALEConfiguration.xml ... check it out ... you can override this
configuration file to do a publish on the state you want.

For instance:

Trigger
Action target="WSRR" action="create"/

/Trigger

Tells the policy, that Asset of type Specification in the Identified
state should be published to WSRR.

Understood. Thanks for the answer,

Chemi.

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