PHP returning JSON to JQUERY AJAX CALL

I am still struggling to get my head around the ins and out of JQUERY, AJAX and PHP.I can now call the PHP OK, process the form elements and send an email, but I am not handling the return to the AJAX. I am always getting the error: selector activated and when I try to list the supposed JSON …

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PHP returning JSON to JQUERY AJAX CALL

I am still struggling to get my head around the ins and out of JQUERY, AJAX and PHP.
I can now call the PHP OK, process the form elements and send an email, but I am not handling the return to the AJAX. I am always getting the error: selector activated and when I try to list the supposed JSON returned, I get info, that is obviously wrong.
PHP with supposed JSON return

touch(‘phpTouch.txt’);
// process email
$email=1;
if ($email)
$value = array(‘return’ => 1, ‘msg1′ => ‘Message sent OK, we will be in touch ASAP’);
else
$value = array(‘return’ => 0, ‘msg1′ => ‘Message Failed, please try later’);

$output = $json->encode($value);
echo $output;

?>Javascript and AJAX
function submitForm(evt)
$(‘#msgid’).html(‘

Submitting Form (External Routine)

‘);
if ($(‘#formEnquiry’).valid() )
$(“#msgid”).append(“

(Outside Ready) VALIDATED send to PHP

“);
$.ajax(
url: “ContactFormProcess3.php”,
type: “POST”,
data: $(‘#formEnquiry’).serialize(),
dataType: “json”,
success: function (data)
alert(“SUCCESS:”);
for(var key in data)
$(‘#msgid’).append(key);
$(‘#msgid’).append(‘=’ + data[key] + ‘
‘);

},
error: function (data)
alert(“ERROR: “);
for(var key in data)
$(‘#msgid’).append(key);
$(‘#msgid’).append(‘=’ + data[key] + ‘
‘);

}
});
} else
$(‘#msgid’).append(‘

(Outside Ready) NOT VALIDATED

‘);

evt.preventDefault();
};Listing of supposed JSON data
readyState=4
setRequestHeader=function (a,b)if(!s)var c=a.toLowerCase();a=m[c]=m[c]return this}
getAllResponseHeaders=function ()return s===2?n:null
getResponseHeader=function (a)var c;if(s===2)if(!o)o=;while(c=bF.exec(n))o[c[1].toLowerCase()]=c[2]}c=o[a.toLowerCase()]}return c===b?null:c}
overrideMimeType=function (a)s
etc etc If anyone can advise as to what stupid mistake I have made, then I would be most grateful.
…………………………………..

You can return json in PHP this way:
header(‘Content-Type: application/json’);
echo json_encode(array(‘foo’ => ‘bar’));

For more info: PHP returning JSON to JQUERY AJAX CALL

PHP Website Development » Search Results » ajax

PHP returning JSON to JQUERY AJAX CALL

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