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- Posted on
May 20, 2009, 6:52 am
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script is receiving the information, I have checked using echo's and
also the results of the rest of the PHP script are shown in their own
window. In my naivity I expected the statement "echo "$callback
($json1)";" to make the $json1 available to the callback function
displays the JSON properties there. How can I make the JSON properties
pointers gratefully received.
<form id="postit" style="display:none" method="post" action="">
var phpScript = "http://mysite/postex.php",
callback = "loadJSON";
document.getElementById('postit').action = phpScript;
document.getElementById('spoke').value = points;
document.getElementById('callback').value = callback;
$latlngs = $_POST["spoke"];
$callback = $_POST["callback"];
$json1 = json_encode($heights);
Thanks in advance, Steve.
PHP can't open another window in the browser. All it can do is send
Remove the "x" from my email address
JDS Computer Training Corp.
Basically he needs to place an ajax request that has the callback
(in order to avoid reinventing the wheel) such as protype.js
Jerry Stuckle schreef:
And even if it was on topic it would be bad advise.
Using a framework to do something that basic is totally over the top.
"There are two ways of constructing a software design: One way is to
make it so simple that there are obviously no deficiencies, and the
other way is to make it so complicated that there are no obvious
deficiencies. The first method is far more difficult."
-- C.A.R. Hoare