Java json object call function by name

Question

I'm wondering if it is possible to call a function from a java json object?

Example

java:

JSONObject json = new JSONObject();
json.put("fnRowCallback", "test()");

jquery:

$ (function () {
    "use strict";
    function test() {
        alert('test');
    }
}(jQuery));

Ultimate accomplishment needed.

JSONObject json = new JSONObject();
json.put("fnRowCallback", function( nRow, aData, iDisplayIndex, iDisplayIndexFull ) {
                if (aData["rush"] == "Rush" ) {
                    $(nRow).addClass("gradeX");  
                }
            });

What are you trying to accomplish? What is your underlying goal? — Aaron Kurtzhals
– Aaron Kurtzhals, Commented Feb 4, 2013 at 20:12
Your JSON code is on the server. You're trying to call a jQuery function test() on the client. How is it likely to achieve? — Lion
– Lion, Commented Feb 4, 2013 at 20:14
I'm a tapestry5 user and I'm trying to use jquery datatable. The component is initialized with a java json object. The problem I'm facing is trying to add this configuration, json.put("fnRowCallback", "function( nRow, aData, iDisplayIndex, iDisplayIndexFull ) {alert('test'); "); However json objects do not accept functions, so I was hoping to be able to call the function. — Code Junkie
– Code Junkie, Commented Feb 4, 2013 at 20:15
JSON cannot force anything to happen. JSON is just a string - what happens is defined by the receiver - and a receiver shouldn't execute untrusted arbitrary code. — John Dvorak
– John Dvorak, Commented Feb 4, 2013 at 20:23

user400654 · Accepted Answer · 2013-02-04 20:41:42Z

1

You can accomplish this if you instead have your Java return javascript.

<script src="myjavapage.jsp"></script>

have your Java return

$(function(){
    alert('test');
});

however I can't think of a good reason to code it this way rather than instead having a js file that gets the content on demand using ajax and not including code in your json. It seems pretty pointless.

answered Feb 4, 2013 at 20:41

user400654

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5 Comments

Code Junkie Over a year ago

Perhaps I'm going about this all the wrong way. I'm working from the following example, tapestry5-jquery.com/docs/components/docsdatatables tab 2 line 60. I'm trying to incorporate this example, datatables.net/usage/callbacks "fnRowCallback"

user400654 Over a year ago

What does that("fnRowCallback") have to do with a json object?

Code Junkie Over a year ago

fnRowCallback would be my key in my json object. If you looked at the first link I posted, you'll notice the component is configured on the serverside with a json object, but in the example on datatables, they configure it directly in javascript which doesn't appear to be an option for me. My pair would be the function, I'm not sure how to pass that function in as a pair as json objects are nothing other than strings.

user400654 Over a year ago

How are you going to get it from the json string to the dataTable plugin?

Code Junkie Over a year ago

The actually tapestry component passes the json object into the datatable as an argument, but the problem is the function is being passed in as a string rather than an executable function. I thought maybe I could just pass a function name into the json object and use it to call the function. I honestly don't know how to handle this.

Code Junkie · Accepted Answer · 2013-02-06 19:02:43Z

1

I ended up answering my own question. The solution is to use a JSONLiteral like so,

Java

public JSONObject getOptions() {
    JSONObject json = new JSONObject();
    json.put("fnRowCallback", new JSONLiteral(String.format("fnRowCallback()")));
    return json;
}

JS.

function fnRowCallback(){
    var fnRowCallback = function( nRow, aData, iDisplayIndex, iDisplayIndexFull ) {
        if (aData["rush"] == "Rush" ) {
            nRow.className = nRow.className + " gradeX";  
        }
    };
    return fnRowCallback;
}

Thanks Everyone.

answered Feb 6, 2013 at 19:02

Code Junkie

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Comments

Ivan · Accepted Answer · 2013-02-05 06:57:00Z

0

See Function.prototype.call or Function.prototype.apply

var obj = {
   foo: function() {
    console.log('foo called');
   }
};

var json = {method: "foo"};

obj[json.method].call(this);

edited Feb 5, 2013 at 6:57

answered Feb 5, 2013 at 6:40

Ivan

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