使用Ajax的XMLHttpRequest向服务器发送POST请求，将数据传递给PHP脚本，然后旋转图像。

代码示例：

HTML：

旋转

JS：

const fileInput = document.getElementById("fileInput");
const rotateBtn = document.getElementById("rotateBtn");
const image = document.getElementById("image");
let rotationAngle = 0;

fileInput.addEventListener("change", () => {
  const file = fileInput.files[0];
  const reader = new FileReader();
  reader.readAsDataURL(file);
  reader.onload = () => {
    image.src = reader.result;
  };
});

rotateBtn.addEventListener("click", () => {
  rotationAngle += 90;
  rotateImage(rotationAngle);
  const data = new FormData();
  data.append("imageData", image.src);
  data.append("angle", rotationAngle);
  const xhr = new XMLHttpRequest();
  xhr.open("POST", "rotate-image.php");
  xhr.onload = () => {
    image.src = xhr.responseText;
  };
  xhr.send(data);
});

function rotateImage(angle) {
  image.style.transform = `rotate(${angle}deg)`;
}

PHP：