Question

我利用这一职能:

function end_incident() {
    var dataString =  name=Daniel&phone=01234123456 ;
    $.ajax({
        type: "POST",
        url: "http://www.example.co.uk/erc/end_incident.php",
        data: dataString,
        success: function(msg){ 
            alert( Success! +dataString);
        }
    });
};

将信息发送到<代码>end_incident.php,但Im无法查阅$_POST的变量。我试图这样做:

$name = $_POST[ name ];
$phone = $_POST[ phone ];

我做了一些错误?

Thanks for any help

Answer 1

Try sending the data as an object:

function end_incident() {
    $.ajax({
       type: "POST",
       url: "http://www.example.co.uk/erc/end_incident.php",
       data: { name: "Daniel", phone: "01234123456" },
       success: function(msg){ 
            alert( Success! );
       }
    });
};

Answer 2

如果是,请你申请的是,与你网站的同一来源是,如果是n t,你就会发现一个交叉点描述问题。仅靠这样做:

Getting "higher" access/priveledges within the browser, i.e. create an add-on/extension, or use Greasemonkey

Use a proxy through your own site to get the request for the file:

var getURL = "http://www.example.co.uk/erc/end_incident.php";
$.ajax({
   type: "POST",
   url: "/get_url.php?url=" + encodeURIComponent(getURL),
   data: { name: "Daniel", phone: "01234123456" },
   success: function(msg){ 
       alert( Success! );
   }
});

我建议将<代码>error功能添加到你的jax。它概括了多少人只关注success,从来没有处理过错误!

error: function()
{
   console.log(arguments);
}

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