php+html5使用FormData对象提交表单及上传图片的方法_php技巧

var formdata = new FormData(); 
formdata.append('name','fdipzone'); 
formdata.append('gender','male');

var form = document.getElementById('form1'); 
var formdata = new FormData(form);

 
 
  
   
   
   
   
  
  
   

 
       

name:

 
       

gender:male female

 
       

photo:

 
       

 
   

 
     
  

$name = isset($_POST['name'])? $_POST['name'] : ''; 
$gender = isset($_POST['gender'])? $_POST['gender'] : ''; 
$filename = time().substr($_FILES['photo']['name'], strrpos($_FILES['photo']['name'],'.')); 
$response = array(); 
if(move_uploaded_file($_FILES['photo']['tmp_name'], $filename)){ 
    $response['isSuccess'] = true; 
    $response['name'] = $name; 
    $response['gender'] = $gender; 
    $response['photo'] = $filename; 
}else{ 
    $response['isSuccess'] = false; 
} 
echo json_encode($response); 
?>