How to push Record producer name into a custom field on the form created by the very Record producer.

ravikiran0688
Tera Contributor

When a case is created using a record producer, we want the name of the producer to be updated in one of the custom field on the case form.

@Ankur Bawiskar @Allen Andreas @Sandeep Dutta 

1 ACCEPTED SOLUTION

Hi,

you mentioned your custom field is reference to sc_cat_item_producer table

So you just had to use this and use correct field name

cat_item.sys_id -> gives sysId

current.u_customField = cat_item.sys_id;

Regards
Ankur

Regards,
Ankur
Certified Technical Architect  ||  9x ServiceNow MVP  ||  ServiceNow Community Leader

View solution in original post

16 REPLIES 16

Ankur Bawiskar
Tera Patron
Tera Patron

Hi,

you can use this in record producer script

current.u_customField = cat_item.name;

OR

if that doesn't work then use this

var gr = new GlideRecord("sc_cat_item_producer");
gr.addQuery("sys_id", cat_item.sys_id);
gr.query();
if (gr.next()) {
	current.u_customField = gr.name;
}

Regards
Ankur

Regards,
Ankur
Certified Technical Architect  ||  9x ServiceNow MVP  ||  ServiceNow Community Leader

Hi @Ankur Bawiskar,

 

Both of the codes didnt work, May be i should have mentioned that this custom field is a reference field and is referring to "sc_cat_item_producer_list" table.

then just do this

current.u_customField = cat_item.sys_id;

Regards
Ankur

Regards,
Ankur
Certified Technical Architect  ||  9x ServiceNow MVP  ||  ServiceNow Community Leader

Hi Ankur,

We are avoiding using sys id in scripting.

Rgeards,

Ravikiran