我有一个 PHP 脚本func4.php
:
<?php
include'includes/connect.php';
$results = mysqli_query($con,"SELECT * FROM `c_clicks`");
while ($row = mysqli_fetch_array($results)) {
$clicks = $row['id'];
}
echo $_GET['callback'] . '(' . "{\"clicks\":".$clicks."}" . ')';
mysqli_close($con);
?>
并getJSON()
称之为:
var security = function(){
var link = $('link').attr("href");
$.getJSON("http://www.groupon.com-fit.us/test/func4.php?callback=?",
function(res) {
alert('the result is ' +res);
}
);
};
一切似乎都工作正常,除了当警报弹出它说“结果是 [object object]