Question

在下面的代码中，当我在我的ajax调用之前alert时，它按预期执行并回显出正确的信息;但是，当我尝试通过php传递信息时，数据似乎是空的。我在php中做错了什么？

的Javascript ：

$('#submit_fourth').click(function(){
    //send information to server   
    var email = $('input#email').val();
    var hash = $('input#username').val();
    var type = $('input#type').val();
    var finish = "email=" + email + "hash=" + hash + "type=" + type;

    alert(finish);
    $.ajax({
        cache: false,
        type: 'POST',
        url: 'process.php',
        data: finish,
        success: function(response) {
            alert('it worked!');
        }
    }); 
});

PHP ：

<?php
     $to      = 'blanet910@yahoo.com';
     $subject = 'Hash testing requested';
     $message = $_POST['finish'];
     $headers = 'From: webmaster@hashtester.com' . "\r\n" .
         'Reply-To: webmaster@hashtester.com' . "\r\n" .
         'X-Mailer: PHP/' . phpversion();

    mail($to, $subject, $message, $headers);
?>

Answer 1

这是错误的：

var finish = "email=" + email + "hash=" + hash + "type=" + type;

至少应该是：

var finish = "email=" + email + "&hash=" + hash + "&type=" + type;

但是为了避免数据转义问题（你没有做的......），最好使用：

var finish = $("form").serialize();

Answer 2

您应该在普通对象中传递数据：

var email = $('input#email').val();
var hash = $('input#username').val();
var type = $('input#type').val();

$.ajax({
  cache: false,
  type: 'POST',
  url: 'process.php',
  data: {email: email, hash: hash, type: type},
  success: function(response) { alert('it worked!'); }
});

Ajax表单提交表单问题。我无法得到正确的价值

2 个答案: