on 09-18-2014 11:08 PM
I wonder how is the CRMD_ORDERADM_H table linked to IBIB via CRMD_LINK ?
or is there any other table via which both CRMD_ORDERADM_H and IBIB are linked?
Thanks
Karthik
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
Definitely CRMD_ORDERADM_H with COMM_PRODUCT and then IBIB did not help me.
Here is the situation I need to fetch the I base number based on the order number.
So I ve got the order number and I need to establish a link between the order header table and IB header table.
@Frederic Girod
Well I tried to use the FM CRM*_ORDER*_READ*. but I am not really sure on how to use that FM.
Let me know if you ve got any steps for this FM using which I could solve this problem?
Thanks,
Karthik.
Also, here is a code snippet from IBIB to CRMD_LINK you can do the inverse way:
"---------------------------------------------------------------------- *"
"Local Interface: *" IMPORTING *" REFERENCE(IV_IBASE) TYPE IB_IBASE
" EXPORTING *" REFERENCE(ET_GUID_LIST) TYPE CRMT_OBJECT_GUID_TAB *
"----------------------------------------------------------------------
TYPES: BEGIN OF ty_ibase,
guid TYPE crmt_object_guid,
END OF ty_ibase.
DATA: lv_ibguid TYPE guid_16,
lt_guid_ref TYPE STANDARD TABLE OF ty_ibase,
lt_guid_set TYPE STANDARD TABLE OF ty_ibase,
lt_guid TYPE crmt_object_guid_tab.
CHECK iv_ibase IS NOT INITIAL.
SELECT SINGLE ib_guid_16
FROM ibib INTO lv_ibguid
WHERE ibase = iv_ibase.
CHECK sy-subrc EQ 0.
SELECT guid_ref
FROM crmd_srv_refobj
INTO TABLE lt_guid_ref
WHERE ib_comp_ref_guid = lv_ibguid.
CHECK sy-subrc EQ 0.
SELECT guid_set
FROM crmd_srv_osset
INTO TABLE lt_guid_set
FOR ALL ENTRIES IN lt_guid_ref
WHERE guid = lt_guid_ref-guid.
CHECK sy-subrc EQ 0.
SELECT guid_hi FROM crmd_link
INTO TABLE lt_guid
FOR ALL ENTRIES IN lt_guid_set
WHERE guid_set = lt_guid_set-guid.
et_guid_list = lt_guid.
Hi,
I didn't have my notes here, but there are some functions module for CRM starting with CRM*_ORDER*_READ* .. you could test it with a program that have the exact same name.
Maybe in this programs/functions you will find all your answers
regards
Fred
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
User | Count |
---|---|
5 | |
1 | |
1 | |
1 | |
1 | |
1 | |
1 | |
1 | |
1 | |
1 |
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.