CODEWITHSUNDEEP
javascriptjavaservletsxmlhttprequest
Cailean
Cailean

Reputation: 47

XMLHttpRequest with multiple parameters

On the web app I've been working, I've been using xmlhttprequests to pass single parameters to Java servlets as follows:

var xhr = new XMLHttpRequest();
xhr.open('GET', 'DCC?command=' + encodeURIComponent(command), true);
xhr.send(null);

The problem is, I still don't really understand the syntax and now I need to do something similar with multiple parameters. How is this done?

Upvotes: 2

Views: 6429

Answers (4)

Pedro S Cord
Pedro S Cord

Reputation: 1359

You could send the parameters in the send call.

var xhr = new XMLHttpRequest();
xhr.open('GET', 'DCC', true);
xhr.send(JSON.stringify(parameters));

in the java end use the InputStreamReader to read the request and deserialize the object.

Upvotes: 0

LinuxDisciple
LinuxDisciple

Reputation: 2379

The URL you're passing in the example above is:

'DCC?command=' + encodeURIComponent(command)

The DCC part is actually part of the path to the webpage. It's short because it's a relative path. The fully qualified path would look something like www.sitename.com/DCC
The part after that (after the ? character) is called the query string. That's the part of the URL that contains data that you're passing to the server (in a GET transaction), and it follows this pattern:

a=somevalue&b=anothervalue&c=yetanother

So add "&varnameA=valueA" to that string to pass both command and varnameA:

xhr.open('GET', 'DCC?command=' + encodeURIComponent(command)+"&varnameA=valueA",true);

You can keep tacking on &varname=value strings until your query is around 2000 characters, because that's where browsers commonly start crapping out because the URL is too long.
Remember to encode any special characters in the values (that's what encodeURIComponent() is being used for) or you'll get some weird behavior. That means that you're appending something like +"&varnameA="+encodeURIComponent("valueA") for each additional variable/value pair you want to pass to the server.

Upvotes: 1

Mario Mitterbauer
Mario Mitterbauer

Reputation: 433

Simple answer:

var url="index.php"+"?command="+cmd;
xhr.open("GET",url,true);

You can add more parameters, you only have to add a ? infront of each variable name.

Upvotes: 0

Jay
Jay

Reputation: 42

Syntax is : xhr.open("GET", url, true); in Url you can pass multitple parameter by appending "&"

Upvotes: 0

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