Following is the JSON structure. This is the JSON structure

I want to deserialize this data structure in APEX to Update the Records, sobject is dynamic not every time 'Contact', I am confused how deserialize this data structure to Update the record Just for more info here is the Js function.

    let arrofRecs = [],Ids = cmp.get('v.selectedIds'),objName = cmp.get('v.objectName'),
    relatedApiName = cmp.get('v.RelatedListFieldAPI'),recID = cmp.get('v.recordId');
    Ids.map(val =>{
        let obj = {};
        obj['SobjectType'] = objName;
        obj[relatedApiName] = recID;
        obj['Id'] = val;
    let methodName = 'c.updateRecord',
        params = {
        callbackRess = (response) => {
            if (response) {
                let data = JSON.parse(response)[0];
                console.log('data', data);
    help.callApexMethod(cmp, methodName, params, callbackRess);
  • What could be other SObjectType here? You can pass this information along with in your apex method and utilize switch/if-else to de-serialize the JSON based on that information. – Jayant Das Apr 14 '19 at 16:10

This is easy. Just pass attributes as list of maps, and in Apex just do

(List<Contact>) JSON.deserialize(JSON.serialize(records), List<Contact>.class)


(List<SObject>) JSON.deserialize(JSON.serialize(records), List<SObject>.class)

The method signature would be the following.

public static void method(List<Map<String, Object>> records) {
  • getting error :Method does not exist or incorrect signature: void serialize(List<Map<String,Object>>) from the type List<Map<String,Object>> – Sarvesh Apr 14 '19 at 12:42
  • You missed JSON name of the class, you shouldn't call serialize(records) but JSON.serialize(records) – Patlatus Apr 14 '19 at 12:43
  • Here is the method i am try to using.... @AuraEnabled public static string updateRecord(List<Map<String, Object>> json){ System.debug('json'+json); List<SObject> lst = (List<SObject>) JSON.deserialize(JSON.serialize(json), List<SObject>.class); System.debug('lst'+lst); return 'Done'; } – Sarvesh Apr 14 '19 at 12:45
  • rename json to records or to something else, don't use standard class names like JSON, Test etc as the name for variable or class. – Patlatus Apr 14 '19 at 12:53
  • Sarvesh kumar, please avoid extended discussions, either update your post with the problem or create a new one – glls Apr 14 '19 at 13:47

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