How to call Alfresco from Share side

For code based on JavaScript (Surf) :

// set url to call
var url = "/url/to/some/webscript";

// get connector
var connector = remote.connect("alfresco");

// do HTTP get request
var result = connector.get(url);

// check success of call
if (result.status.code == status.STATUS_OK) {

    
    // parse response to object
    var obj = eval("(" + result.response + ")");
 
    // set some value to property of the model
    model.someProp = obj.someProp;
} else {

// do something
}

To call Alfresco from Share side in Java you need first write class:

import org.apache.commons.logging.Log;
import org.apache.commons.logging.LogFactory;
import org.json.JSONException;
import org.json.JSONObject;
import org.springframework.extensions.surf.WebFrameworkServiceRegistry;
import org.springframework.extensions.webscripts.ScriptRemote;
import org.springframework.extensions.webscripts.connector.Response;

/**
 * Created by s.palyukh on 07.09.15.
 */
public class CallAlfresco {

    private static Log logger = LogFactory.getLog(CallAlfresco.class);

    protected WebFrameworkServiceRegistry serviceRegistry = null;

    public JSONObject jsonGet(String uri)
    {
        ScriptRemote scriptRemote = serviceRegistry.getScriptRemote();
        Response response = scriptRemote.connect().get(uri);
        if (response.getStatus().getCode() == 200)
        {
            try
            {
                return new JSONObject(response.getResponse());
            }
            catch (JSONException e)
            {
                if (logger.isErrorEnabled())
                {
                    logger.error("An error occurred when parsing response to json from the uri '" + uri + "': " + e.getMessage());
                }
            }
        }
        return null;
    }

    public void setServiceRegistry(WebFrameworkServiceRegistry serviceRegistry) {
        this.serviceRegistry = serviceRegistry;
    }
}

And then setup Spring Bean for this class: