上传图片到php服务器出错

Question

我在尝试将swift中的图像上载到PHP服务器时遇到问题。在php处理文件之前，一切看起来都很好。在那一刻，我得到了错误。

swft代码的相关部分是：

func myImageUploadRequest(image: UIImage, realfilename: String)
{

    let myUrl = NSURL(string: "http://www.ignistudios.com/boda/postImage.php");

    let request = NSMutableURLRequest(URL:myUrl!);
    request.HTTPMethod = "POST";

    let param = [
        "firstName"  : "username",
        "lastName"    : "lastname",
        "userId"    : "9"
    ]

    let boundary = generateBoundaryString()

    request.setValue("multipart/form-data; boundary=\(boundary)", forHTTPHeaderField: "Content-Type")


    let imageData = UIImageJPEGRepresentation(image, 1)
    print(imageData.debugDescription)

    if(imageData==nil)  { return; }

    request.HTTPBody = createBodyWithParameters(param, filePathKey: "file", realfilename: realfilename, imageDataKey: imageData!, boundary: boundary)

    let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
        data, response, error in

        if error != nil {
            print("error=\(error)")
            return
        }

        // You can print out response object
        print("******* response = \(response)")

        // Print out reponse body
        let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
        print("****** response data = \(responseString!)")



    }

    task.resume()

}


func createBodyWithParameters(parameters: [String: String]?, filePathKey: String?, realfilename: String?, imageDataKey: NSData, boundary: String) -> NSData {
    var body = NSMutableData();

    if parameters != nil {
        for (key, value) in parameters! {
            body.appendString("--\(boundary)\r\n")
            body.appendString("Content-Disposition: form-data; name=\"\(key)\"\r\n\r\n")
            body.appendString("\(value)\r\n")
        }
    }

    let filename = realfilename

    let mimetype = "image/jpg"

    body.appendString("--\(boundary)\r\n")
    body.appendString("Content-Disposition: form-data; name=\"\(filePathKey!)\"; filename=\"\(filename)\"\r\n")
    body.appendString("Content-Type: \(mimetype)\r\n\r\n")
    body.appendData(imageDataKey)
    body.appendString("\r\n")



    body.appendString("--\(boundary)--\r\n")

    return body
}




func generateBoundaryString() -> String {
    return "Boundary-\(NSUUID().UUIDString)"
}

php是

<?php
$uploaddir = '/fotos/';
$uploadfile = $uploaddir . basename($_FILES['file']['name']);
echo "<p>";
if (move_uploaded_file($_FILES['file']['tmp_name'], $uploadfile)) {
  echo "File is valid, and was successfully uploaded.\n";
} else {
  echo "Upload failed";
}
echo "</p>";
echo '<pre>';
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
?>

最后，我得到的错误是：

******响应数据=

上载失败

Here is some more debugging info:Array
(
    [file] => Array
        (
            [name] => Optional(\"boda20160428_135709.jpg\")
            [type] => 
            [tmp_name] => 
            [error] => 1
            [size] => 0
        )
)
任何提示都将不胜感激。
                

Answer 1

正确答案是图像太大，上传前需要压缩图像，使用这个

 UIImageJPEGRepresentation(image,0.2)

当图像足够小时，您将获得您的tmp名称

Answer 2

看起来Marc B评论了答案，这是一个PHP文件上传大小的问题。

然而，在这种情况下，值得分别测试链中的每个环节。例如，测试和审查createBodyAnd Parameters的输出。创建一个测试并在那里运行它，以检查标头是否正确形成。尤其要确保字符串编码正确。我不熟悉NSMutableData上的appendString方法。

应该可以更直接地将输出输入服务器，以消除其他潜在问题。