将responseObject base64编码的字符串转换为UIImage

Question

我想用PHP（在服务器端）向我的iOS应用程序发送图像，以便我可以在UIImageView中显示它。

我的服务器端代码：

<?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; 

?> 

我收到的图像与此代码在我的iOS应用：

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); 
    }]; 

当我运行的代码 - 它击中了不良区有错误读作我发送的base64编码“字符串”（图片）：

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.} 

我在做什么inco rrectly？

Answer 1

即使它不是JSON，也已指定AFJSONResponseSerializer。当然，你已经重写了acceptableContentTypes，但是这并不能阻止它在响应中试图解析JSON。我想用AFHTTPResponseSerializer，然后丢失acceptableContentTypes。

顺便说一句，我不会使用image/jpeg作为base64编码的响应，因为它是文本而不是jpeg。如果你要返回一个原始的base64字符串，你可以使用application/text或类似的东西。

或者，更好的办法是改变你的PHP实际返回JSON（因为这样可以更容易解析响应）并保持AFJSONResponseSerializer（但是一旦你修复了头文件就会丢失acceptableContentTypes），然后你可以抓住base64字符串从response[@"image"]。

<?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; 

?>