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Captura HTML5 Cámara frontal

Estoy tratando de capturar una imagen de la cámara trasera, el código funciona como se esperaba, el teléfono se abre en la parte delantera o trasera, pero al usar la cámara trasera obtengo el valor $_FILES["fileToUpload"]["tmp_name"] vacío. al archivo php. Si uso la cámara frontal, se proporciona un valor de $_FILES["fileToUpload"]["tmp_name"] al archivo php como se esperaba.

 <!DOCTYPE html> <html> <body> <form action="upload.php" method="post" enctype="multipart/form-data"> Select image to upload: <input type="file" name="fileToUpload" id="fileToUpload" accept="image/*" capture="environment"> <input type="submit" value="Upload Image" name="submit"> </form> </body> </html>

 <?php $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); echo $target_file.'<br>'; $uploadOk = 1; $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION)); // Check if image file is a actual image or fake image if(isset($_POST["submit"])) { $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); echo 'check '.$check; if($check !== false) { echo "File is an image - " . $check["mime"] .$_FILES["fileToUpload"]["tmp_name"]; $uploadOk = 1; } else { echo "File is not an image."; $uploadOk = 0; } } // Check if $uploadOk is set to 0 by an error if ($uploadOk == 0) { echo "Sorry, your file was not uploaded."; // if everything is ok, try to upload file } else { if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) { echo "The file ". htmlspecialchars( basename( $_FILES["fileToUpload"]["name"])). " has been uploaded."; } else { echo "Sorry, there was an error uploading your file."; } } ?>

about 4 years ago · Juan Pablo Isaza
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