I am making an AJAX request in my Phonegap application. The code is:
function remoteCall()
{
alert("Remote call func called");
try
{
$.ajax({
url: 'http://192.168.1.200/testing/testConn.php',
contentType: "application/json; charset=utf-8",
dataType:"json",
success: function(data)
{
if(data == '')
alert("No data received from server");
else
alert("Data received from server = "+data.postcode);
},
error: function(jqXHR, textStatus, errorThrown)
{
console.log(textStatus);
alert('FAIL !!!');
},
});
}//end of try().
catch(e)
{
alert("error = "+e.message);
}
}//END OF FUNC remoteCall.
I have also added the following line in <head>
to avoid the jQuery conflict.
<script>jQuery.noConflict();</script>
My server side file is:
<?php
header("Access-Control-Allow-Origin: *");
header("Access-Control-Allow-Headers: X-Requested-With");
$retArray = array();
$retArray['postcode']= 'm14';
$retArray['brand']= '1';
$retArray['product_type']= '1';
$jsondata = json_encode($retArray);
return $jsondata;
?>
I am getting first alert message but later I am getting the error, "$ is not undefined". I have no idea how to fix this; can anyone help?
I am able to overcome that error of "$ is undefined", now i am not getting any errors and also I am not getting any output...
确保在添加包含代码的js文件之前添加jquery文件。
<script>jQuery.noConflict();</script>
means that you must refer to the jQuery object as jQuery
and not $
. Change your references from $
to jQuery
and it should work.
I solved my problem... :)
Following lines solved my $ and jQuery problem
<script src="http://ajax.aspnetcdn.com/ajax/jquery/jquery-1.7.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.ui/1.8.10/jquery-ui.js"></script>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
(Thanks to Ali and cfs :) )
And in server file i changed return to echo
<?php
header("Access-Control-Allow-Origin: *");
header("Access-Control-Allow-Headers: X-Requested-With");
$retArray = array();
$retArray["postcode"]= "m14";
$retArray["brand"]= "1";
$retArray["product_type"]= "1";
$jsondata = json_encode($retArray);
echo $jsondata;
?>
Now I am able to get the server details.
Thanks to all for the replies.... :)
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