I started making this simple google chrome extension in javascript. And in the beginning of the code I have the following:

var req = new XMLHttpRequest();

req.open(
    "GET",
    "",
    true);

req.onreadystatechange(alert(req.readyState));

The value req.readyState es to be 1, which means the required page has not been properly fetched. I'm a newbie to Javascript. What's the problem in my code?

I started making this simple google chrome extension in javascript. And in the beginning of the code I have the following:

var req = new XMLHttpRequest();

req.open(
    "GET",
    "http://www.ldoceonline./dictionary/manga",
    true);

req.onreadystatechange(alert(req.readyState));

The value req.readyState es to be 1, which means the required page has not been properly fetched. I'm a newbie to Javascript. What's the problem in my code?

• javascript
• xmlhttprequest

2 Answers 2

how about something like this

var request = new XMLHttpRequest();

if (request == null){
        alert("Unable to create request");
    }else{

        var url = "http://www.ldoceonline./dictionary/manga";

        request.onreadystatechange = function()
            {
            if(request.readyState == 4)
            {
                LDResponse(request.responseText);
            }
        }

        request.open("GET", url, true);
        request.send(null);
    }

function LDResponse(response)
{
// do stuff with the response
}

Of course this is all assuming that they are giving you valid data back ie XML or json

On this line:

req.onreadystatechange(alert(req.readyState));

alert() is being called straight away, which I'm sure isn't your intention. It seems that you want to wait for the onreadystatechange event to fire and then alert the readyState. If that's the case then try this:

req.onreadystatechange = function() {
    alert(req.readyState);
};

And don't forget req.send(null)!