How to get OCI Hyperlink

[Former Member]

Dear all,

In SRM 5.0 I customized the OCI settings for an external catalog provider. In my Shop transaction BBPSC02 I see the link and can open the external catalog.

When I check the HTML code of the page, I see that the link is generated via a javascript: function launchCTLG(nameCTLG, InFrame)

How can I get the link that is being generated. I would like to execute some tests using this link and therefore I would like to add it manually in the internet explorer..

Thanks and regards

AD

[masa_139]

Hi,

You can manually test the OCI call by typing the URL. You can also trace the post data using httpWatch tool.

http://catalog.com:1080?username=username&password=password&HOOK_URL=http://test.com:8123/sap(cz1TSU...?_OkCode=ADDI&Target=_top&_caller=CTLG&client=777&_language=JA&HTTP_CONTENT_CHARSET=shift_jis&_{client=777&sap-client=777&}language=J&sap-language=J&http_content_charset=shift_jis&OCI_VERSION=4.0&OPI_VERSION=1.0&returntarget=_top

Regards,

Masa

[Former Member]

Hi,

If you just need to form the URL to access the catalog website, you can do the following.

1. Find the Base URL from the URL parameter defined in the IMG for that catalog.

2. add ? at the end of the URL

3. For each parameter in the catalog definition, keep adding param=value&

For example: If your URL is http://www.google.com and the parameters are userid - testuser; password - welcome. Your final URL would look like http://www.google.com?userid=testuser&password=welcome&'.

Note that with this URL you can only launch your catalog website and test shopping on that site, you cannot transfer anything back to SRM.

SG

Edited by: SRMUSER123 on Sep 21, 2010 8:56 AM

[Former Member]

Hi all,

Thanks for your help. I know how to built the link by myself.

The problem is that I would like to see the exact link that is automatically generated by SRM. The reason is that the behaviour of the catalog call is different if I generate the link manually or if it is generated automatically by SRM.

Thanks again.

AD

[Former Member]

AD,

You can use tools like HTTPWatch to capture the URL that is being generated.

I am assuming you are trying to capture the URL to record tests. SRM, while generating the URL, will put unique sessionId information on URL. So, the generated URL will change from session to session. So, the generated URL will not be unique.

SG

[Former Member]

Hi all,

Thanks for your reply. This is not what I am looking for.

I want to the the full real link including all parameters that have been defined here:

SPRO->Supplier Relationship management-> SRM server-> Master data-> Define External Web Service.

Best regards

AD

[Former Member]

Hi,

please refer to below wiki page:

http://wiki.sdn.sap.com/wiki/display/SRM/SRM7.0catalog+call

You can place a break point in the code and check the URL . Also you can execute the FM /SAPSRM/CLL_WSI_CALL with webservice id and URL will be returned.

If the FM does not bring complete URL as you want then you can build the URL the URL yourself with parameter like &Username etc etc..

Thanks

Iftekhar Alam

[Former Member]

Hi,

Thanks. This is what I am looking for. Unfortunately we are in SRM 5.0 and not 7.0 so the class and FM names do not match. Do you know the same for SRM 5.0?

Best regards

AD

[Former Member]

Hi ,

goto SPRO->Supplier Relationship management-> SRM server-> Master data-> Define External Web Service.

there you select the catalog id or webservice id and click on standard call structure or integrated call structure , the first parameter should be the URL if the external catalog .

Thanks

Iftekhar Alam

[Former Member]

Hi Achim,

Please go to table - BBP_WS_CALL_V

It will promt you for the Catalog id - Enter the catalog id and execute - you can retrive the required URL and other login details.

Thanks,

-Devi Swain