我想用Python创建一个伪验证API，用于处理图像（工人的ID）。它应该基于图像中的某些模式（如圆形、方形等，类似于Snapchat著名的系统）。如何在用户通过普通的HTML文件输入表单发送图像时运行Python脚本？我是说用户将照片发送到服务器，服务器运行Python脚本并给出结果？提前感谢。
我在网上查找这个问题时，没有找到任何解决方案。似乎这是一个非常昂贵的操作，可能不是最优的，但尽管如此，我还是想这样做。
回答：
你可以使用PHP和Python的组合来创建这个解决方案，有几种库可以操作图像以达到预期的结果，在员工ID验证的情况下，Python脚本应该能够识别它所看到的内容，假设公司里有一个同事和一个摄像设备或类似设备拍摄了ID的照片：
客户端应用程序需要不断读取摄像头或图片，以下是摄像头和本地保存图片的代码：
Python条码读取器：（涉及Numpy是个好主意，可以增强cv2条码识别）
##import cv2##img = cv2.imread('/home/jbsidis/Pictures/sofiaID.png')####barcode_detector = cv2.barcode_BarcodeDetector()##### 'retval' is boolean mentioning whether barcode has been detected or not##retval, points, _ = barcode_detector.detectAndDecode(img)##### copy of original image##img2 = img.copy()##### proceed further only if at least one barcode is detected:##if retval:## points = points.astype(np.int)## for i, point in enumerate(points):## img2 = cv2.drawContours(img2,[point],0,(0, 255, 0),2)######print(retval,points,barcode_detector.detectAndDecode(img))import sysimport cv2import timecamera = cv2.VideoCapture(0)if (camera.isOpened() == False): print("Can not open camera #0.") sys.exit(0)print("Camera ready")doAgain = Truewhile doAgain: ret, image = camera.read() if ret: qrCodeDetector = cv2.QRCodeDetector() text, points, _ = qrCodeDetector.detectAndDecode(image) if points is not None: print(text) cv2.imwrite("./result.jpg",image) else: print("QR code not detected") cv2.imshow("Image", image) key = cv2.waitKey(1) & 0xFF if key == 27: cv2.destroyAllWindows() doAgain = Falsecamera.release()服务器端应用程序（PHP）：
<?php//server side for the python process$target_dir = "uploads/"; //add a directory where to save the file in the server//this receives the data from the HTML formif ($_SERVER["REQUEST_METHOD"] == "POST") { $file = $_FILES["picture"]; if ($file["type"] != "image/jpeg" && $file["type"] != "image/png" && $file["type"] != "image/gif") { echo "This file is not allowed! IMages only please"; exit(); } if ($file["size"] > 1000000) { echo "Error: File size is too large."; exit(); } // here we are guessing that you have storage in your server or php server $filename = basename($file["name"]); $target_path = $target_dir . $filename; move_uploaded_file($file["tmp_name"], $target_path); echo "THanks for uploading the image, Barcode is being recognized!";} else { echo "No file was uploaded. Try again please";}if (move_uploaded_file($file["tmp_name"], $target_path)) { // this will run the script the python script we created, you have to modify to if this will read the png or the image file or a camera, in this case it must be the camera $python_script = "python /securepathtothescriptorsomeonecouldfindit/to/your/python/script.py " . $target_path; $output = shell_exec($python_script); echo "Barcode generated successfully!";} else { echo "There was an error processing your image, please contact support";}?>客户端（访问您网站或Web应用程序的用户）：
<!DOCTYPE html><html><head> <title>User will upload Picture</title> <style> .card {width: 900px;background-color: #f9f9f9;border: 1px solid #ddd;border-radius: 10px;box-shadow: 0 0 10px rgba(0, 0, 0, 0.1);padding: 20px;margin: 40px auto;} </style></head><body> <div class="card"> <h2>Upload Picture</h2> <form action="action.php" method="post" enctype="multipart/form-data"> <input type="file" name="picture" accept="image/*"> <button type="submit">Upload</button> </form> <div id="image-preview"></div> </div> <script> const fileInput = document.querySelector('input[type="file"]'); fileInput.addEventListener('change', (event) => { const file = event.target.files[0]; const reader = new FileReader(); reader.onload = (event) => { const imageDataUrl = event.target.result; document.getElementById('image-preview').innerHTML = `<img src="${imageDataUrl}" alt="Uploaded Image width=200 height=550">`; }; reader.readAsDataURL(file); }); </script> </div></body></html>结果：
要测试这个解决方案，您可以安装XAMPP或类似的工具，在本地运行PHP服务器并离线测试，或者您可以使用支持PHP的公共托管进行测试。
