解决 ajax传值到PHP后台，如何获取值

前端代码

<script type="text/javascript">

var filename=$("input[name=filename]").val();

var crowd_file =$('#crowd_file')[0].files[0];

  var formData =new FormData();

  formData.append("crowd_file",crowd_file);

  formData.append("filename", filename);

$.ajax({

    url:'http://****/test/index.php?type=bb',

    dataType:'json',

    type:'POST',

    async:false,

    data: formData,

    processData:false,  // jQuery不要去处理发送的数据

    contentType:false,  // jQuery不要去设置Content-Type请求头

    success:function(data){

        console.log(data);

        // if (data.status == 'ok') {

            //    alert('上传成功！');

        // }

        },

        error:function(response){

                console.log(response);

            }

        });

</script>

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formData 传值后台，接收方法。

<?php

header('Access-Control-Allow-Origin:*');

header('Access-Control-Allow-Methods:OPTIONS, GET, POST'); // 允许option，get，post请求

header('Access-Control-Allow-Headers:x-requested-with'); // 允许x-requested-with请求头

header('Content-type: application/json');

header("Content-type:text/html;charset=utf-8");

$data = $_POST;

$data['files'] = $_FILES;

echo $data;

?>