Core
11/06/2023, 6:56 PMlet sublistFields = newRecord.getSublistFields({ sublistId: 'item'});
curFieldsInfo[sublistId] = [];
for (let sublistFieldId of sublistFields) {
curFieldsInfo[sublistId].push(sublistFieldId);
}
But, it returns so many fields and I want to get fields that appear on SO UI. Could anyone please help me?Anthony OConnor
11/06/2023, 7:13 PMlet sublistFields = newRecord.getSublistFields({ sublistId: 'item'});
curFieldsInfo.item = [];
for (let sublistFieldId of sublistFields) {
curFieldsInfo[sublistId].push(sublistFieldId);
}
Core
11/06/2023, 7:14 PMAnthony OConnor
11/06/2023, 7:15 PMCore
11/06/2023, 7:15 PMAnthony OConnor
11/06/2023, 7:18 PMerictgrubaugh
11/06/2023, 7:19 PMCore
11/06/2023, 7:19 PMCore
11/06/2023, 7:21 PMAnthony OConnor
11/06/2023, 7:22 PMCore
11/06/2023, 7:22 PMAnthony OConnor
11/06/2023, 7:23 PMCore
11/06/2023, 7:24 PMAnthony OConnor
11/06/2023, 7:25 PMCore
11/06/2023, 7:26 PMAnthony OConnor
11/06/2023, 7:29 PM