php - 将 responseObject base64 编码字符串转换为 UIImage

Question

我正在尝试使用 PHP(在服务器端)将图像发送到我的 iOS 应用程序，以便我可以在 UIImageView 中显示它。

我的服务器端代码是:

<?php

header("Content-Type: image/jpeg"); //if your data is format jpeg

$username = $_POST['username'];
$count = $_POST['count'];
$base64string =     base64_encode(file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo $base64string;

?>

我在我的 iOS 应用程序中收到带有此代码的图像:

NSString * uploadURL = @"http://192.168.1.4/getimage.php";

NSLog(@"uploadImageURL: %@", uploadURL);

NSString *queryStringss = [NSString stringWithFormat:@"%@", uploadURL];
                queryStringss = [queryStringss stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
manager.responseSerializer=[AFJSONResponseSerializer serializerWithReadingOptions:NSJSONReadingAllowFragments];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"text/html"];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"text/plain"];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"image/jpeg"];

NSUserDefaults *userdefaults = [NSUserDefaults standardUserDefaults];

NSString *usernameEncoded = [marker.title stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

NSDictionary *params = @{@"username": usernameEncoded, @"count": [object valueForKey:@"count"]};

[manager POST:queryStringss parameters:params success:^(AFHTTPRequestOperation * _Nonnull operation, id  _Nonnull responseObject) {
       NSLog(@"Success: %@ ***** %@", operation.responseString, responseObject);
       NSData *decodedData = [[NSData alloc] initWithBase64EncodedString:responseObject options:0];
       image.image = [UIImage imageWithData:decodedData scale:300/2448];
           [self.view addSubview:image];
       }
   failure:^(AFHTTPRequestOperation *operation, NSError *error) {
       NSLog(@"Error: %@ ***** %@", operation.responseString, error);
   }];

当我运行代码时 - 它遇到了一个错误 block ，错误显示为我发送的 base64 编码“字符串”(图像):

2015-12-03 01:19:15.655 sneek[6261:1952572] Error: /9j/4AAQSkZJRgABAQAASABIAAD/4QBYRXhpZgAATU0AKgAAAAgAAgESAAMAAAABAAYAAIdpAAQAAAABAAAAJgAAAAAAA6ABAAMAAAABAAEAAKACAAQAAAABAAAMwKADAAQAAAABAAAJkAAAAAD/7QA4UGhvdG9zaG9wIDMuMAA4QklNBAQAAAAAAAA4QklNBCUAAAAAABDUHYzZjwCyBOmACZjs+EJ+/8AAEQgJkAzAAwEiAAIRAQMRAf/EAB8AAAEFAQEBAQEBAAAAAAAAAAABAgMEBQYHCAkKC//EALUQAAIBAwMCBAMFBQQEAAABfQECAwAEEQUSITFBBhNRYQcicRQygZGhCCNCscEVUtHwJDNicoIJChYXGBkaJSYnKCkqNDU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6g4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tr

... (very long) ...

Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.}

我做错了什么？

Answer 1

您已指定 AFJSONResponseSerializer，即使它不是 JSON。当然，您已经覆盖了 acceptableContentTypes，但这并不能阻止它尝试解析响应中的 JSON。

我会使用 AFHTTPResponseSerializer，然后丢失 acceptableContentTypes。

顺便说一句，我不会将 image/jpeg 用于 base64 编码的响应，因为它是文本，而不是 jpeg。如果您要返回原始 base64 字符串，您可以使用 application/text 或类似的东西。

或者，更好的是，将您的 PHP 更改为实际返回 JSON(因为这样可以更轻松地解析响应)并继续使用 AFJSONResponseSerializer(但丢失一次 acceptableContentTypes您修复 header )，然后您可以从 response[@"image"] 中获取 base64 字符串。

<?php

header("Content-Type: application/json");

$username = $_POST['username'];
$count = $_POST['count'];
$base64string =     base64_encode(file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo json_encode(array("image" => $base64string));

?>

或者，使用 AFImageResponseSerializer 并更改 PHP 以返回图像:

<?php

header("Content-Type: image/jpeg"); //if your data is format jpeg

$username = $_POST['username'];
$count = $_POST['count'];
$contents = file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo $contents;

?>