ajax php如何实现文字和文件的同时上传？

Question

新手，我想通过ajax php实现文字和文件的同时上传。问下该如何实现？我尝试的方法是获取input的file然后放在data里ajax传出去但是失败了。html:<form action="" method="post"        enctype="multipart/form-data">            <h4>曲名</h4>            <input type="text" name="name" id="uploadname" placeholder="请输入曲名">            <h4>歌手</h4>            <input type="text" name="singer" id="uploadsinger" placeholder="请输入歌手">            <h4>请选择.gp3/.gp4/.gp5/.gpx文件：</h4>            <input type="file" name="file" class="uploadbutton1" id="uploadfile">            <input type="submit" name="submit" value="Submit" id="upload-newtab-button"/>        </form>js部分：$("#upload-newtab-button").click(function(){            $.ajax(              {                  url:"upload.php",                  type: "POST",                  processData:false,                  contentType:false,                  data:{file:$("#uploadfile").files[0],name:$("#uploadname").val(),singer:$("#uploadsinger").val()}                success:function(data){                                            alert(data);                  }              }              );           });php部分：<?php$upload_path = $_SERVER['DOCUMENT_ROOT']."./uploads/";$dest_file = $upload_path.basename($_FILES['myfile']['name']);if(move_uploaded_file($_FILES['file']['tmp_name'],$dest_file)){    echo "文件已上传至服务器根目录的upload目录下";}else{    echo "上传错误".$_FILES['myfile']['error'];}$servername = "localhost";$username = "root";$password = "root";$dbname = "guitartabs";$name=$_POST["name"];$singer=$_POST["singer"];         // 创建连接$conn = new mysqli($servername, $username, $password, $dbname);// Check connectionif ($conn->connect_error) {    die("连接失败: " . $conn->connect_error);   } $sql_insert = "insert into tabs(name,singer) values ('$name','$singer')";$res_insert = $conn->query($sql_insert);if($res_insert) { echo "注册成功！"; } else { echo "系统繁忙，请稍候"; }                  $conn->close();?>

Answer 1

使用FormData

var formdata = new FormData();

formdata.append("name",name);

formdata.append("file" , $("#fileupload")[0].files[0]);

$.ajax({

    url: "/upload/file",

    type: 'post',

    data:formdata,

    processData : false, //不处理发送的数据，因为data值是Formdata对象，不需要对数据做处理

    contentType : false,

    success: function(ret) {

        // code

    }

})

Answer 2

$sql_insert = "insert into tabs(name,singer) values ('$name','$singer')";去掉單引號