PHP returning JSON to JQUERY AJAX CALL

Question

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

<?php

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) {
    $('#msgid').html('<h1>Submitting Form (External Routine)</h1>');
    if ($('#formEnquiry').valid() ) {
        $("#msgid").append("<h1>(Outside Ready) VALIDATED send to PHP</h1>");
            $.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] + '<br />');
                }
            },
            error: function (data) {
                alert("ERROR: ");
                for(var key in data) {
                    $('#msgid').append(key);
                    $('#msgid').append('=' + data[key] + '<br />');
                }
            }
        });
    } else {
        $('#msgid').append('<h1>(Outside Ready) NOT VALIDATED</h1>');
    }
    evt.preventDefault();
};

Listing of supposed JSON data

readyState=4
setRequestHeader=function (a,b){if(!s){var c=a.toLowerCase();a=m[c]=m[c]||a,l[a]=b}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||(d.mimeType=a);return this}
etc etc

If anyone can advise as to what stupid mistake I have made, then I would be most grateful.

@Prisoner. I changed to your suggestion. I was obviously wrong. Copied code again, but still not working as I had hoped - same listing of spurious values — mcl
– mcl, Commented Aug 15, 2011 at 12:27
There should be an output status=... what is the value there? — marc
– marc, Commented Aug 15, 2011 at 12:29

Phil · Accepted Answer · 2018-09-06 23:22:30Z

116

You can return json in PHP this way:

header('Content-Type: application/json');
echo json_encode(array('foo' => 'bar'));
exit;

edited Sep 6, 2018 at 23:22

Phil

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answered Aug 15, 2011 at 11:39

Arnaud Le Blanc

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3 Comments

mcl Over a year ago

Thanks for the tip. I tried and I am getting the same result. Why is error: always being executed ?

mcl Over a year ago

Many thanks. I now understand why and you were right. Now fixed and working

Phil Over a year ago

I added exit as there should be nothing after the JSON response, hope you don't mind.

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