JSON return value

Question

Currently I am having issue with returning some values from PHP to jQuery - not sure how to do it

$(document).ready(function(){
  $('#testForm').submit(function(e){
    $.post('submit.php',$(this).serialize(),function(msg){

        $('#submit').val('Submit');

        if(msg.status){
            $('#testForm').html(msg);
        }
        else {
            $('#testForm').html("fail");
        }
    },'json');

});

});

<?php
$name = $_POST['name'];
$email = $_POST['email'];

//echo json_encode(array('status'=>1,'html'=>$name." : ".$email));
echo '{"status":1,'.$name.'}';
?>

I would like to return the name variable value from PHP to jQuery once status = 1 means success, but I'm still having no luck in doing it.

Why are you not using the commented-out line?

Niet the Dark Absol
– Niet the Dark Absol

2012-12-19 03:50:35 +00:00
Commented Dec 19, 2012 at 3:50 — Niet the Dark Absol
– Niet the Dark Absol, Commented Dec 19, 2012 at 3:50
it has special char on a name try htmlentities($name)

Kotzilla
– Kotzilla

2012-12-19 03:50:52 +00:00
Commented Dec 19, 2012 at 3:50 — Kotzilla
– Kotzilla, Commented Dec 19, 2012 at 3:50
echo '{"status":1,name:"'.$name.'"}'; Try this

Amit Garg
– Amit Garg

2012-12-19 03:54:30 +00:00
Commented Dec 19, 2012 at 3:54 — Amit Garg
– Amit Garg, Commented Dec 19, 2012 at 3:54

Niet the Dark Absol · Accepted Answer · 2012-12-19 03:51:23Z

5

JSON has a very strict syntax.

In your case, however, you're failing because you're not even specifying a property name, you just have a bare value with no quotes.

Just use json_encode, it will handle all edge cases for you.

answered Dec 19, 2012 at 3:51

Niet the Dark Absol

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1 Comment

user1897151 Over a year ago

how does the syntax looks like for the json_encode if i would like to set status 1/0 together with the data return from php to jquery?

Hernan Velasquez · Accepted Answer · 2012-12-19 03:53:14Z

1

Are you sure that you are returning a valid json string?

From this instruction:

echo '{"status":1,'.$name.'}';

Assuming that $name is a plain string, for example "hello", you will return this json string:

{"status":1, hello}

And this is invalid.

answered Dec 19, 2012 at 3:53

Hernan Velasquez

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