php - 如何从 Swift 通过 PHP 将 UIImage 上传到 MySQL？

Question

我有一个应用程序，它通过我编写的一些 PHP 网络服务将一些数据上传到 MySQL 数据库。我已经成功上传了文本数据，但我正在努力尝试上传图片。基本上，用户使用他们的 iPhone 相机拍照，我将其转换为 UIImage，现在想上传它。

我查看了以下资源以寻求帮助，但它们似乎对我不起作用: How to Upload text, pdf, doc, zip file via PHP file upload to MySQL
https://www.youtube.com/watch?v=HqxeyS961Uk (这里的代码不会为我编译，我一个字母一个字母地复制它)
How to post a picture from iPhone app (in Swift) to MySQL database?
Saving image on db mysql through swift
Swift - uploading base64 encoded image to php and displaying the image
Getting a UIImage from MySQL using PHP and jSON

如果可能的话，我想在同一个查询中同时上传图片和文本，因为它们相互关联，而且我的 MySQL 表会为新行分配一个自动编号。

编辑感谢下面的答案。我有点不确定如何实现你们建议的更改。我目前上传文本数据的 Swift 代码是:

let url:NSURL = NSURL(string: "http://www.website.com.au/savenewexpense.php")!
        let request:NSMutableURLRequest = NSMutableURLRequest(URL: url)
        request.HTTPMethod = "POST"
        var dataDictionary:[String:AnyObject] = ["Date" : strDate, "Staff" : strStaff, "Amount" : dblAmount, "Category" : strCategory]
        var data:NSData = NSJSONSerialization.dataWithJSONObject(dataDictionary, options: nil, error: nil)!
        request.HTTPBody = data

        request.addValue("application/json", forHTTPHeaderField: "Content-Type")

        let task:NSURLSessionDataTask = NSURLSession.sharedSession().dataTaskWithRequest(request, completionHandler: { (data:NSData!, response:NSURLResponse!, error:NSError!) -> Void in

            let response:[String:String] = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers, error: nil) as! [String:String]
            if let result = response["result"] {
                if result == "success" {
                    NSLog("Text Data saved successfully")
                    if self.SendImageToServer() == true {
                        NSLog("Image Data saved successfully")
                        self.SaveSuccessful()
                    } else {
                        NSLog("Image Data failed to save")
                        self.DataFail()
                    }
                } else {
                    NSLog("Error saving data")
                    self.DataFail()
                }
            } else {
                NSLog("No response")
                self.NoResponse()
            }

        })

        task.resume()
    }

我的 PHP 代码是:

    // Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$json = file_get_contents('php://input');
$data = json_decode($json, true);

$stmt = $conn->prepare("INSERT INTO tblExpenses (Date, Staff, Amount, Category) VALUES (?, ?, ?, ?)");
$txtDate = $data['Date'];
$txtStaff = $data['Staff'];
$txtAmount = $data['Amount'];
$txtCategory = $data['Category'];
$stmt->bind_param("ssds", $txtDate, $txtStaff, $txtAmount, $txtCategory);
$stmt->execute();
echo '{"result" : "success"}';


$stmt->close();
$conn->close();

我应该如何在其中实现图像？

谢谢

Answer 1

你所要做的是以设置图像的边界为例，使用代码将图像的 NSDATA 转换为边界值

if let dta = data as? NSData {
                bodyData.appendData(String(format:"\r\n--\(boundary)\r\n").dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)!);
                bodyData.appendData(String(format:"Content-Disposition: form-data; name=\"photo\(i)\"; filename=\"image.png\"\r\n").dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)!);
                bodyData.appendData(String(format:"Content-Type: image/png\r\n\r\n").dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)!);

                bodyData.appendData(dta);
            }

这里你要做的是先设置图像名称，然后设置图像数据，首先使用 allowLossyConversion 转换数据，然后将 POST 数据与其他数据一起发送给服务

request.HTTPBody = bodyData

就用这个来发送图片