Question

我试图在Alamofire库的帮助下将一组少量图像（图像阵列）和参数上传到服务器。问题是我想上传具有相同ID但名称不同的图像，我与此完全相反。我还提到了一些在堆栈溢出时回答的问题，但这些问题仅适用于单个图像上传。请帮朋友。提前谢谢。

上传功能

let parameters = [
        "defect_id" : "10",
        "file_name" : self.dateMashup(1)
    ]
    Alamofire.upload(
        multipartFormData: { multipartFormData in
            var count = 1
            for img in self.photoArray{
                let imgdata = UIImagePNGRepresentation(img)
                // the name should be as array other wise want work
                let randsome = self.dateMashup(count)
                multipartFormData.append(imgdata!,withName: "image[]", fileName: "image\(randsome).png", mimeType: "image/png")
                count += 1
            }

            for (key, value) in parameters {
                multipartFormData.append(value.data(using: .utf8)!, withName: key)
            }
    },
        to: "http://localhost/Source2/ImageDir/uploadImages.php",

        encodingCompletion: { encodingResult in
            switch encodingResult {
            case .success(let upload, _, _):
                upload.responseString { response in
                    debugPrint(response)
                }
            case .failure(let encodingError):
                print(encodingError)
            }
    })

PHP文件

<?php 

$server = "localhost";
$dbuser = "root";
$dbpassword = "";
$dbname = "lmg_new";

$con = mysqli_connect($server, $dbuser, $dbpassword, $dbname);

$defect_id=$_POST["defect_id"];
$file_name = $_POST["file_name"];
$created = date('Y-m-d H:i:s');

// $defect_id = 1;
// $file_name = "sadas.png";
// $created = date('Y-m-d H:i:s');

$sql = "INSERT INTO `files`(`defect_id`, `file_name`, `created`) VALUES (defect_id, file_name, created)";

$result= mysqli_query($con, $sql) or die("query fail:". mysqli_error($con));

// $response = array();

$response["success"]= true;
if($result){
    while($row = mysqli_fetch_assoc($result))
    {
        $flag[]=$row;   
    }
    $response['data'] = $flag;
}

if (empty($_FILES["image"])){
    $response['File1'] = "NOFILE";
}else{
    foreach ($_FILES["image"]["tmp_name"] as $index => $tmp_name) {
        $filePath = "images/" . basename($_FILES["image"]["name"][$index]);
        if (move_uploaded_file($tmp_name, $filePath)) {
            $response['filePath'][$index] = $filePath;
        }
        $response['status'] = "Success";
    }
}
echo json_encode($response);
?>

使用alamofire将图像数组和参数上传到服务器

0 个答案: