how to process a php code (in ajax) without returning any value

Question

I hope these lines would explain the question:


//in javascript
function foo(sessionName){
    var u = "test.php?q="+sessionName+"&r="+parseInt(Math.random()*9999999);
    xmlHttpObj.open("get", u, true);
    xmlHttpObj.send();
}       

//in php

$q = $_GET['q'];
unset($_SESSION[$q]);

Like you see, I don't want any value returned, I just want to unset the specific session.

This doesn't make any sense...I don't even know how to suggest an improvement. — rockerest
– rockerest, Commented Apr 18, 2011 at 5:53
you can't, ajax will erase the original value if you don't return any value, it's like returning echo""; — Bagong21
– Bagong21, Commented Apr 18, 2011 at 5:56
Sorry, "ajax will erase the original value" doesn't make sense. What's the concrete problem you're experiencing? — deceze
– deceze ♦, Commented Apr 18, 2011 at 6:11

No Results Found · Accepted Answer · 2011-04-18 05:58:28Z

2

You don't ever have to return a value, if you access and run the script it will do it's job. The problem will be that you will have no way of knowing if it succeeded, so it's best to return a success/failure value in general. However, since unset() doesn't actually return a value there is no need in this case.

answered Apr 18, 2011 at 5:58

No Results Found

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1 Comment

No Results Found Over a year ago

This is assuming typos rather than actual errors in your example :)

deceze · Accepted Answer · 2011-04-18 05:53:57Z

1

Yes, this works. I suppose your problem is a typo in the variable name?

var u = "test.php?q="+sessionName+"&r="+parseInt(Math.random()*9999999);
                  ^
$n = $_GET['n'];
            ^

answered Apr 18, 2011 at 5:53

deceze♦

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2 Comments

Bagong21 Over a year ago

just ignore if there's a typo, how to return nothing?

deceze Over a year ago

@Bagong Just... don't return anything. Looks like you're already doing just that.

gnur · Accepted Answer · 2011-04-18 06:08:14Z

0

It is probably better to still return a value indicating wether it worked or not. Even though this code should work just fine.

Adding this to the end of your script will give you an indicator wether it worked or not.

if (isset($_SESSION[$n])) {
    echo 'success';
}
else {
    echo 'fail';
}

answered Apr 18, 2011 at 6:08

gnur

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