$('#saveNewData').click(function () {
            //保存数据的按钮被点击的时候，获得当前数据
            var type = $('select[name="type"] option:selected').val();
            var title = $('input[name="title"]').val();
            var imgSrc = $('input[name="imgSrc"]').val();
            var author = $('input[name="author"]').val();
            var createdAt = $('input[name="createdAt"]').val();
            var content = $('textarea[name="content"]').val();
          
            //封装数据
            var data = {
                type:type,
                title:title,
                imgSrc:imgSrc,
                author:author,
                createdAt:createdAt,
                content:content
            };
            //ajax提交数据
            $.ajax({
                type: "POST",
                url:'insert.php',
                data:data,
                datatype:'json',
                error: function(request) {
                    alert("保存失败");
                },
                success: function(msg) {
                    alert("保存成功");
                    alert(data);
                }
            });
        })

确定能够获得到表单元素的数据，html的地址栏提交的时候能显示所有提交数据
在insert.php中

$type =  $_POST['type'];
$title = $_POST['title'];
$imgSrc = $_POST['imgSrc'];
$author = $_POST['author'];
$createdAt = $_POST['createdAt'];
$content = $_POST['content'];

无法获得传过来的数据，提示
Notice: Undefined index: type in D:xampphtdocs8-1baiduNewsinsert.php on line 3

Notice: Undefined index: title in D:xampphtdocs8-1baiduNewsinsert.php on line 4

Notice: Undefined index: imgSrc in D:xampphtdocs8-1baiduNewsinsert.php on line 5

Notice: Undefined index: author in D:xampphtdocs8-1baiduNewsinsert.php on line 6

Notice: Undefined index: createdAt in D:xampphtdocs8-1baiduNewsinsert.php on line 7

Notice: Undefined index: content in D:xampphtdocs8-1baiduNewsinsert.php on line 8

第一次用php，以前写js和node数据交互的时候用的那样的数据传递形式，但是php不能获得，哪位大神给我看看代码，万分感激