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'));
if you only need the first email value,you don't need any loop.
Just use the code below;
if(a.Temporary_Email_address_coming_from_OTT__c != null && (a.Temporary_Email_address_coming_from_OTT__c).contains(',')){
a.Temporary_Email_address_coming_from_OTT__c = a.Temporary_Email_address_coming_from_OTT__c.split(',')[0];
}
Best Answer
Though billing address appears as one field on the UI it is a compound of multiple text fields
You can do
Refer: https://www.salesforce.com/developer/docs/api/Content/compound_fields_address.htm