Link Button in grid

[Former Member]

hello expert

i create one form which have grid .i added the LINK BUTTON in grid by following code

oCol = objGrid.Columns.Item("Prs No")

oCol.LinkedObjectType = "30"

it show link buton in grid in perticular cell but when i clicked it .it give error like

"yu are not permitted to perform action".i am also change -1 in place of 30 then it not show link button

sir pls help me for it .i want when i click button it show perticular form acording to cell data

[Former Member · Accepted solution]

You have to replace the value "30" fro the correspondant object ID.

Example:

For Items the value to enter would be "4"

For Warehouses the value to enter would be "64"

For Employees the value to enter would be "171"

For a list of object IDs check SAP BO's help, look for "Objects ID".

Hope this helps.

Regards.

[Former Member]

Hi,

Click administration module and click license and click license administration check box check all the three.

Regards,

Billa

[Former Member]

Hello Expert

i have allready cheked this check box.

[Former Member]

Dear vmaster,

Try with existing TransId of JE. I facing with this error message when I link a docnum for any document linked object. When I try with Docentry it's worked.

Hope it help you.

Nghia

[Former Member]

Hello,

The error message says: You are not permitted....

It means you have no authorization for object type 30 which is Journal Entry.

Did you checked your permission?

-1 is not defined object type. do not use it.

Regards,

J.

[Former Member]

hello expert

sir i want open user defiend object by this link button which is in grid then how it do it