Question

我有一个图像选择器，可以从内置摄像头中检索图像：

- (void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info {

    [info objectForKey:UIImagePickerControllerEditedImage];
    UIImage *chosenImage = info[UIImagePickerControllerOriginalImage];
    self.imageView.image = chosenImage;
    NSData *image = UIImagePNGRepresentation(chosenImage);
    [self setImageDataToSend:image];
    [picker dismissViewControllerAnimated:YES completion:NULL];
}

然后我想用http

上传照片像这样：

 NSData *imageData = UIImagePNGRepresentation([UIImage imageNamed:@"image.png"]);  

    NSMutableURLRequest *request = [[NSMutableURLRequest alloc]
                                    initWithURL:[NSURL
                                                 URLWithString:@"http://******.co.uk/***/imageupload.php"]];

    [request setHTTPMethod:@"POST"];
    [request setValue:@"image/png"
   forHTTPHeaderField:@"Content-type"];
    [request setValue:[NSString stringWithFormat:@"%lu",
                       (unsigned long)[imageData length]]
   forHTTPHeaderField:@"Content-length"];
    [request setHTTPBody:imageData];

    [[NSURLConnection alloc] initWithRequest:request delegate:self];

这会选择一个名为＆＃34; image.png＆＃34;从应用程序。我希望它从图像选择器中获取NSData *图像

这里也是imageupload.php：

<?php
$handle = fopen("image.png", "wb"); // write binary

fwrite($handle, $HTTP_RAW_POST_DATA);

fclose($handle);

print "Received image file.";
?>

你知道有更好的方法吗？

Answer 1

您使用要发送的数据调用方法setImageDataToSend:，然后在此行中忽略它：

 NSData *imageData = UIImagePNGRepresentation([UIImage imageNamed:@"image.png"]);

您可能想要省略该行并进行更改：

 [request setHTTPBody:imageData];

为：

 [request setHTTPBody:[self imageDataToSend]];

上传图片选择器照片ios

1 个答案: