[SalesForce] Deserialize JSON to SObject

Rational for Error: JSON does not encoded the data types of the data, thus you have to give your class type to allow it to understand how to parse the data as it works its way through the class members. The Object data type is not enough for the deserialize method to know the data type.

Using JSON.deserializeUntyped. To work around this you can manually parse the JSON using the various JSON parser methods or use the deserializeUntyped method which returns conveniently a map that contains primitive data types!

So I've leveraged this in some new methods to encapsulate this with your Parameters class.

public class Parameters{
    private Map<String, Object> parameters = new Map<String, Object>();

    public void add(String name, Object value) {
        this.parameters.put(name, value);
    }

    public Object get(String name) {
        Object result = null;

        if(this.parameters.containsKey(name)) {
            result = this.parameters.get(name);
        }
        return result;
    }    

    public String serialize()
    {
        return JSON.serialize(parameters);      
    }

    public static Parameters deserialize(String serialized)
    { 
        Parameters parameters = new Parameters();
        parameters.parameters = (Map<String, Object>) JSON.deserializeUntyped(serialized);
        return parameters;
    }
}

This is a slightly modified version of your test code to use the new helper methods.

Parameters params = new Parameters();
params.add('Name', 'Robert');
params.add('Age', 36);
String serialized = params.serialize();
Parameters deserialized = Parameters.deserialize(serialized);
System.assertEquals(params.toString(), deserialized.toString());

I've used toString to compare equality, as without it the check fails, as the two variables point to different instances of the same object.

Alternative:

This naked version also works as well...

Map<String, Object> parameters = new Map<String, Object>();
parameters.put('A', 36);
parameters.put('B', 'Robots');
String serialized = JSON.serialize(parameters);
System.debug(serialized);
parameters = (Map<String, Object>) JSON.deserializeUntyped(serialized);
System.debug(parameters.get('B'));

The reason I asked to see all of the code is because sometimes when working with large sets of code, passing variables in and out of methods can have unexpected side effects. Is it possible you are having an issue with passing variables into and back out of methods? Apex uses pass by value for primitive data types (such as a Map or String). If at any point you set jsonAsMap = <something> you could be running into this issue. Here is a perfect example from the reference:

We’ll illustrate this with some code samples. But first, let’s step back and define these terms.

“Pass by value” means when a method is called, a second copy of the parameter variable is made in memory, and this copy is passed to the method as a parameter. This copy can be modified in the method, but the original variable in the caller is unchanged, and the changes to the copy are lost upon return.

“Pass by reference” means when a method is called, that actual variable is passed to the method.

As described in the new developer’s guide text, Apex is not passing anything by reference. It is passing the reference (i.e., the memory address) by value, which allows the developer to change fields on the object and call methods (like list.add()) on the object.

Let’s illustrate with some code. First, an example where we get apparent pass-by-reference behavior:

List fillMe = new List();
PassByWhatNow.reference(fillMe);
System.assertEquals(fillMe.size(),5);   //five items, as expected

/*  calls this method
public static void reference(List m) {
    List la = [select id, name from Account limit 5];
    for (Account a : la) m.add(a);
}
*/

In this example, the parameter passed to the function, behind the scenes, is a pointer to the fillMe collection, passed by value. That is to say, we took the memory address and copied it – but it is the same memory address number in both places. When the add() function is called on the copy, that same memory location is modified by adding the specified items to the list. So when we return to the calling program, the modifications made in the static function are visible. So far, so good.

Now let’s look at an example where things break down.

List createMe = new List();
PassByWhatNow.referenceNew(createMe);
System.assertEquals(createMe.size(),0);   //nothing in the list

/* calls this method
public static void referenceNew(List m) {
    m = new List([select id, name from Account limit 5]);
}
*/

Again, we pass a copy of the list’s memory address to the function. However, this time, the function is initializing the variable anew. When the “new” function is called, new memory is allocated. This is the big difference.

The copy of the memory address passed as a parameter is overwritten with the address of the newly allocated memory. The original, however, is not overwritten. That change in address is not known to the caller!

Any subsequent actions done in the method are on that new memory address rather than the original object address. These changes, including the addition of the Accounts to the list, can not be available to the caller, since it only knows of the original address. No actions were done on this address; the list on the caller is therefore unchanged, still empty as it was when the call was made.

While Apex usually does behave as if we’re passing variables by reference, we’re not truly passing by reference. There are a few situations where the distinction is important, as detailed here. Now you know more, and you are a stronger developer.