douguazhi5966 2012-12-19 03:47
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JSON返回值

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.

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2条回答 默认 最新

  • douzhi9478 2012-12-19 03:51
    关注

    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.

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