When deserializing a JSON string to List<Object> using JSON.deserializeUntyped( resParams ) the list received is incomplete:

{Address=null, BillingAddress=2509, CompanyId=632, County=null, Fax=null, Id=2509, PostCode=null, ShortName=Test Site Ignore, SiteName=Test Site Ignore, Telephone=null, ...}

As you can see we get 10 elements and an ellipsis indicating missing elements. Is there a way to get all (12 )elements with deserialize()?

1 Answer 1


The problem isn't JSON.deserialize, it's Object.toString. Basically, when you debug a non-String, non-null item, System.debug calls Object.toString, which has the behavior of truncating big objects so they don't take up a ton of log space. There's no convenient way to debug the entire object. However, if you use checkpoints, you can open the checkpoint to observe the object's entire state by way of the checkpoint GUI, or you can convert your object back to a String and debug that via JSON.serialize(JSON.deserializeUntyped(resParams)).

  • Thanks for such a prompt answer, is there a way to use those missing elements that are in an ellipsis? Unfortunately, I can't pass them into say a SF field. E.g. Town = 'London' is hidden in the ellipsis and I cannot pass it into the SF field. Is there any other way to retrieve all the fields? Commented Dec 20, 2016 at 16:15
  • @Baraban Either re-serialize them, or you could process the results via something like Map<String, Object> values = (Map<String, Object>)JSON.deserializeUntyped(resParams); for(String key: values.keys()) { /* do something with values.get(key) */ }
    – sfdcfox
    Commented Dec 20, 2016 at 16:17
  • Thanks for your help! Basically, I serialized it and broken it into a map<String,Object> and got all the values in the end. All as you suggested! Commented Dec 20, 2016 at 16:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .