iphone - 在iPhone中使用带有userid参数的base64 haxcode将图像上传到服务器

Question

我正在使用这段代码，但问题是它在我想发送的 nsdata 转换 block 中对 Haxcode 进行编码，其中包含我使用固定整数的 userid 获得的代码。请帮忙。

 NSData *imageData = [NSData dataWithData:UIImageJPEGRepresentation(image1, 0)];
    //image.image=image1;

    [Base64 initialize];
    NSString *b64EncStr = [Base64 encode:imageData];

    NSLog(@"encoded.%@",b64EncStr);




     NSURL *url = [[NSURL alloc] initWithString:updateimageURL];
    NSMutableURLRequest *req = [[NSMutableURLRequest alloc]initWithURL:url];
    [req setHTTPMethod:@"POST"];

    NSString *trimmed = [b64EncStr stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]];



    paramDataString = [NSString stringWithFormat:@"Id=%d&FromString=%@",100,trimmed];

    NSLog(@"%@",paramDataString);


    NSData* aData = [paramDataString dataUsingEncoding:NSUTF8StringEncoding];


      [req setHTTPBody: aData];

    NSURLConnection *theConnection=[[NSURLConnection alloc]initWithRequest:req delegate:self];
    if (theConnection)
    {
        NSMutableData *data = [[NSMutableData alloc] init];
        self.receivedData=data;
        [data release];

    }
    else {

        UIAlertView* alertView = [[UIAlertView alloc] initWithTitle:nil message:@"Check your networking configuration." delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil];
        [alertView show];
        [alertView release];
    }

    [url release];
    [req release];

它正在以正确的方式建立 URL 连接，但响应是服务器错误。

Answer 1

首先将 UImage 转换为 NSData 并将 NSData 转换为 base64string，然后将其传递给您的 Web 服务

finalImagePath = [imageData base64EncodedString];

NSString *strImageData = [finalImagePath stringByReplacingOccurrencesOfString:@"+" withString:@"%2B"];

然后将 strImageData 发送到您的 webserivce。

从服务器使用中检索时

NSString *strImageData = [finalImagePath stringByReplacingOccurrencesOfString:@"%2B" withString:@"+"];