Convert grails domain object to JSON and manipulate it

Question

I have a grails object that I am converting using def json = object as JSON. After I have it converted I want to add one more property called pin to the JSON which looks like the following.

[location:[lat:23.03, lon:72.58]]

Only way to do this so far seems like following

Serialize the DomainClass to JSON using grails.converters.json
Convert the JSON to string
Create JSONBoject using the string from Step 2
Add the property
Convert it back to String

Any other way to do this using grails.converters.json? I have tried using Gson but I do not want to go that route because I am getting many Circular Reference Errors

Alexander Tokarev · Answer 1 · 2014-01-21T12:46:22.183

2

Try this:

domainInstance.properties + [pin: pinInstance] as JSON

edited Jan 21 '14 at 12:46

answered Jan 21 '14 at 11:20

Alexander Tokarev

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Would this add pin as part of domain instance? – Sap Jan 21 '14 at 11:39
No, it wouldn't. But do you really need to merge your domain with additional fields? If so, maybe it will be better to add this field to your domain class? Or to make a surrogate field in it, providing a getter for `pin`? – Alexander Tokarev Jan 21 '14 at 11:46
That was my first attempt, I added lat and lon in the domain class and created `public getPin() { ['location':[lat:lat,lon:lon]] }` method but when I do `object as JSON` pin does not come in the string representation of the JSON – Sap Jan 21 '14 at 11:52
1

Had looked at the source code of `org.codehaus.groovy.grails.web.converters.marshaller.json.DomainClassMarshaller`, and found that only persistent properties of domain class are marshalled to JSON. So, the most straightforward way to get a JSON with custom structure is via creation of a Map and marshalling it. – Alexander Tokarev Jan 21 '14 at 12:21
Is there a way to get corresponding map of a domainobject? – Sap Jan 21 '14 at 12:29
Try `domainInstance.properties`. I've edited my answer. – Alexander Tokarev Jan 21 '14 at 12:44
There must be some kind of member object that is not letting this happne. When I do toString on the result json i get `Error converting Bean with class org.h2.jdbc.JdbcResultSet` – Sap Jan 23 '14 at 09:59
An issue I had with this solution, is that it added a lot of extraneous properties that I did not want in my response, such as includes, excludes, etc – adeady Apr 05 '16 at 15:07

score 0 · Answer 2 · edited Feb 19 '18 at 05:33

I recently needed to do a similar thing. Some caveats:

This is using Grails 2.4.5
I use MongoDB as a backend. As such, I created an object marshaller for MongoDB domain classes. It is printed below, and you can wrap a similar marshaller for your domain class(es):

Marshaller:

class MongodbObjectMarshaller implements ObjectMarshaller<JSON> {
    @Override
    boolean supports(Object o) { return o?.properties?.dbo }

    @Override
    void marshalObject(Object obj, JSON converter) throws    
    ConverterException {

        Map propertiesToOutput = obj.properties.dbo

        propertiesToOutput.remove("_id")  //don't print org.bson.types.ObjectId
        propertiesToOutput.remove("version")  //don't print gorm verson column

        converter.build {
            _id obj.id.toString()
            propertiesToOutput.each{ entry ->
                "$entry.key" entry.value
            }
        }

    }
}

What that marshaller does, it allow in JSON output any of the domain class's properties. obj.properties.dbo is special to MongoDB, but for a regular domain class, you can just grab the properties and exclude the ones you don't need.

Then in my controller, this works:

domainInstance.pin = [location:[lat:23.03, lon:72.58]]
def content = tacticalCard as JSON

because my marshaller now picks up the pin property and prints it.

Convert grails domain object to JSON and manipulate it

2 Answers2