Java - returning JSON from servlet with Gson library

Question

I am using GSON library. I have a programm which returns JSON. JSON constructs and returns in this way:

Gson gson = new Gson();
//findNewComments returns List<Comment> comments
return gson.toJson(service.findNewComments(id,lastId));

So the result is:

[
    {
        "id": 43,
        "entryId": 19,
        "author": " m8w46",
        "body": "mw86",
        "date": "WED 9, 2011"
    },
    {
        "id": 44,
        "entryId": 19,
        "author": " n7w4",
        "body": "nw77w4",
        "date": "WED 9, 2011"
    }
]

But this array must be named as "comments"!

"comments": [
    {
        "id": 43,
        "entryId": 19,
        "author": " m8w46",
        "body": "mw86",
        "date": "WED 9, 2011"
    },
    {
        "id": 44,
        "entryId": 19,
        "author": " n7w4",
        "body": "nw77w4",
        "date": "WED 9, 2011"
    }
]

How can i do that?

return "\"comments\":" + gson.toJson(service.findNewComments(id,lastId)); ??? — SJuan76
– SJuan76, Commented Aug 10, 2011 at 15:38
Why does it have to have such a wrapper? That adds no real information in there, why not just return JSON array as is? — StaxMan
– StaxMan, Commented Aug 10, 2011 at 18:03
I only do it if the client code expects a particular format. — Kevin
– Kevin, Commented Aug 10, 2011 at 18:40

Kevin · Accepted Answer · 2011-08-10 15:42:20Z

3

Not sure if this is acceptable to you but:

public class CommentWrapper {
    List<Comments> comments;
    public CommentWrapper(List<Comment> comments) {
       this.comments = comments;
    }
}

Then you can do:

return new Gson().toJson(new CommentWrapper(service.findNewComments(id,lastId)));

Which results in:

{
    "comments": [
        ....your data here...
    ]
}

Not sure if the object syntax is acceptable to you or not.

answered Aug 10, 2011 at 15:42

Kevin

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4 Comments

Tw1sty Over a year ago

Ye, thank you. I thought that there is some easier way, but it seems that it is the only one.

Kevin Over a year ago

I usually make the wrapper class an inner class of whatever controller/servlet I'm working in, because typically it has no use outside that.

Tw1sty Over a year ago

Can I ask one more question? I am trying to show elements of this array with javascript, but it says, that they are undefined, but the cycle works fine!. What am I doing: var json = eval("("+xmlhttp.responseText+")"); for(var comment in json.comments){alert(comment.author);}

Kevin Over a year ago

@Lomer: I'm not a javascript expert you're probably better served by posting a new question.

StaxMan · Accepted Answer · 2011-08-13 20:04:33Z

0

Aside from wrapper object, you can also use a simple java.util.Map with single entry, possibly bit less code.

answered Aug 13, 2011 at 20:04

StaxMan

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