Ajax - Beginning JavaScript - page 439

while the XMLHttpRequest object awaits a response from the server. Asynchronous mode is the
default behavior of XMLHttpRequest , so you can usually omit the third argument to open() .
Note In the past, it was considered best practice to pass true as the third
argument.
The next step is to send the request; do this with the send() method. This method accepts one
argument, which is a string that contains the request body to send along with the request. GET
requests do not contain any information, so pass null as the argument:
var request = new XMLHttpRequest();
request.open("GET", " http://localhost/myTextFile.txt", f false);
request.send(null);
This code makes a GET request to retrieve a file called myTextFile.txt in synchronous mode.
Calling the send() method sends the request to the server.
WarNiNg The send() method requires an argument to be passed, even if
it is null .
Each XMLHttpRequest object has a status property. This property contains the HTTP status code
sent with the server's response. The server returns a status of 200 for a successful request, and one of
404 if it cannot find the requested file. With this in mind, consider the following example:
var request = new XMLHttpRequest();
request.open("GET", " http://localhost/myTextFile.txt", f false);
request.send(null);
var status = request.status;
if (status == 200) {
alert("The text file was found!");
} else if (status == 404) {
alert("The text file could not be found!");
} else {
alert("The server returned a status code of " + status);
}
This code checks the status property to determine what message to display to the user. If successful
(a status of 200 ), an alert box tells the user the request file exists. If the file doesn't exist (status
404 ), the user sees a message stating that the server cannot find the file. Finally, an alert box tells
the user the status code if it equals something other than 200 or 404 .
Many different HTTP status codes exist, and checking for every code is not feasible. Most of the
time, you should only be concerned with whether your request is successful. Therefore, you can cut
the previous code down to this:
var request = new XMLHttpRequest();
request.open("GET", " http://localhost/myTextFile.txt", f false);