在php中通过ajax post发送文件和表单数据

<form id="myForm" enctype="multipart/form-data">
  <input type="file" name="file">
  <input type="text" name="name">
  <input type="submit" value="Submit">
</form>

var form = document.getElementById("myForm");
var formData = new FormData(form);

var xhr = new XMLHttpRequest();
xhr.open("POST", "upload.php", true);
xhr.onreadystatechange = function() {
  if (xhr.readyState === 4 && xhr.status === 200) {
    // 请求成功后的处理逻辑
    console.log(xhr.responseText);
  }
};
xhr.send(formData);

<?php
$file = $_FILES['file'];
$name = $_POST['name'];

// 处理文件上传逻辑
if ($file['error'] === UPLOAD_ERR_OK) {
  $tempName = $file['tmp_name'];
  $destination = 'uploads/' . $file['name'];
  move_uploaded_file($tempName, $destination);
}

// 处理表单数据逻辑
// ...

// 返回响应
echo "File and form data received successfully!";
?>