Question

当我从同一页面上的表单输入发布一个数字时，我试图让JustGage刷新而不重新加载整个页面。

$(document).ready(function(){
    //JSON REQUEST
    $.getJSON("test_server.php",function(result){
        var g = new JustGage({
            id: "gauge",
            value: result.total,
            min: 0,
            max: 100,
            title: "Data from Form"
        });
    });
    //Post Form
    $("#submitButtonId").on("click",function(e){
        e.preventDefault();
        var formdata = $(this.form).serialize();
        $.post('test_server.php', formdata, function(data){
            //Reset Form
            $('#myform')[0].reset(); 
        });
        return false;
    });
});
</script>

jQuery的：

<?php
$total= mysqli_real_escape_string($db_conx,$_POST['total']);
$my_data=array(total=>"$total");

// sending output
header('Content-Type: text/json');
echo json_encode($my_data,true);
?>

test_server.php：

let storyboard = UIStoryboard(name: "MyStoryboardName", bundle: nil)
let vc = storyboard.instantiateViewControllerWithIdentifier("someViewController") as! ViewB

/* Here you have the reference to the view, so you can set the parameters*/

vc.parameterInViewB = banana

/* At this point you can present the view to the user.*/

self.presentViewController(vc, animated: true, completion: nil)

刷新JustGage而不重新加载页面

0 个答案: