When you are loading an imagefrom a Stream, You have to keep the stream open for the lifetime of the image, see this MSDN Image.FromStream. I think the exception is caused because the memory stream gets closed even before the image gets disposed. You can change your code like this: byte[] bitmapData = new byte[imageText.Length]; … Read more
You can use css to solve the problem. div { position: relative; display: inline; } .imtip { position: absolute; bottom: 0; right: 0; } <div> <img src=”https://i.stack.imgur.com/Blaly.png” /> <img src=”http://www.w3schools.com/favicon.ico” class=”imtip” /> </div> Basically, I’ve done more or less what ZDYN said, just that you need to include a display: inline in the container otherwise … Read more
You shouldn’t make anything synchronous (not even AJAX) calls but instead simply put your code in the appropriate callback: function loadSprite(src, callback) { var sprite = new Image(); sprite.onload = callback; sprite.src = src; } Then use it like this: loadSprite(“https://stackoverflow.com/questions/8645143/sprites/sheet1.png”, function() { // code to be executed later }); If you want to pass … Read more
The problem is that you are returning the toString() of the call to Base64.encodeBase64(bytes) which returns a byte array. So what you get in the end is the default string representation of a byte array, which corresponds to the output you get. Instead, you should do: encodedfile = new String(Base64.encodeBase64(bytes), “UTF-8”);
There’s a method on every control called DrawToBitmap. You don’t need to p/invoke to do this. Control c = new TextBox(); System.Drawing.Bitmap bmp = new System.Drawing.Bitmap(c.Width, c.Height); c.DrawToBitmap(bmp, c.ClientRectangle);
Let’s see in steps what should happen: JSP is basically a view technology which is supposed to generate HTML output. To display an image in HTML, you need the HTML <img> element. To let it locate an image, you need to specify its src attribute. The src attribute needs to point to a valid http:// … Read more
Use Pillow, it’s a fork of PIL that is still in active development, and supports python3. Here I use a dict generator to map the exif data to a dict from PIL import Image, ExifTags img = Image.open(“/path/to/file.jpg”) exif = { ExifTags.TAGS[k]: v for k, v in img._getexif().items() if k in ExifTags.TAGS }
You can always use System.Resources.ResourceManager which returns the cached ResourceManager used by this class. Since chan1 and chan2 represent two different images, you may use System.Resources.ResourceManager.GetObject(string name) which returns an object matching your input with the project resources Example object O = Resources.ResourceManager.GetObject(“chan1”); //Return an object from the image chan1.png in the project channelPic.Image = … Read more
Use img.shape It provides you the shape of img in all directions. ie number of rows, number of columns for a 2D array (grayscale image). For 3D array, it gives you number of channels also. So if len(img.shape) gives you two, it has a single channel. If len(img.shape) gives you three, third element gives you … Read more
If I’m guessing right, what you get back is an unsigned integer of the form 0xAARRGGBB, so int b = (argb)&0xFF; int g = (argb>>8)&0xFF; int r = (argb>>16)&0xFF; int a = (argb>>24)&0xFF; would extract the color components. However, a quick look at the docs says that you can just do Color c = new … Read more