$.ajax({
type: "POST",
url: "contacts.php",
data: dataString,
cache: false,
success: function(data, status, settings)
{
alert(The request URL and DATA);
}
,
error: function(ajaxrequest, ajaxOptions, thrownError)
{
}
});
如何提醒 Success 函数中的请求 URL 和 DATA 参数?
谢谢
最佳答案
你可以简单地;
success: function(data, textStatus, jqXHR)
{
alert(this.data + "," + this.url);
}
关于javascript - jQuery Ajax 成功函数参数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8971381/