Currently I'm trying to capture a screenshot using the Selenium WebDriver. But I can only obtain the whole page screen shot. However, what I wanted is just to capture a part of the page or perhaps just on specific element based on ID or any specific element locator. (For example, I wish to capture the picture with image id = "Butterfly")
Is there any way to capture a screenshot by selected item or element?
We can get the element screenshot by cropping entire page screenshot as below:
driver.get("http://www.google.com");
WebElement ele = driver.findElement(By.id("hplogo"));
// Get entire page screenshot
File screenshot = ((TakesScreenshot)driver).getScreenshotAs(OutputType.FILE);
BufferedImage fullImg = ImageIO.read(screenshot);
// Get the location of element on the page
Point point = ele.getLocation();
// Get width and height of the element
int eleWidth = ele.getSize().getWidth();
int eleHeight = ele.getSize().getHeight();
// Crop the entire page screenshot to get only element screenshot
BufferedImage eleScreenshot= fullImg.getSubimage(point.getX(), point.getY(),
eleWidth, eleHeight);
ImageIO.write(eleScreenshot, "png", screenshot);
// Copy the element screenshot to disk
File screenshotLocation = new File("C:\\images\\GoogleLogo_screenshot.png");
FileUtils.copyFile(screenshot, screenshotLocation);
Here is a Python 3 version using Selenium webdriver and Pillow.
This program captures the screenshot of the whole page and crop the element based on its location. The element image will be available as image.png. Firefox supports saving element image directly using element.screenshot_as_png('image_name').
from selenium import webdriver
from PIL import Image
driver = webdriver.Chrome()
driver.get('https://www.google.co.in')
element = driver.find_element_by_id("lst-ib")
location = element.location
size = element.size
driver.save_screenshot("shot.png")
x = location['x']
y = location['y']
w = size['width']
h = size['height']
width = x + w
height = y + h
im = Image.open('shot.png')
im = im.crop((int(x), int(y), int(width), int(height)))
im.save('image.png')
Update
Now chrome also supports individual element screenshots. So you may directly capture the screenshot of the web element as given below.
from selenium import webdriver
driver = webdriver.Chrome()
driver.get('https://www.google.co.in')
image = driver.find_element_by_id("lst-ib").screenshot_as_png
# or
# element = driver.find_element_by_id("lst-ib")
# element.screenshot_as_png("image.png")
The AShot framework from Yandex can be used for taking screenshots in Selenium WebDriver scripts for
full web pages
web elements
This framework can be found on https://github.com/yandex-qatools/ashot.
The code for taking the screenshots is very straightforward:
ENTIRE PAGE
Screenshot screenshot = new AShot()
.shootingStrategy(new ViewportPastingStrategy(1000))
.takeScreenshot(driver);
ImageIO.write(screenshot.getImage(), "PNG", new File("c:\\temp\\results.png"));
SPECIFIC WEB ELEMENT
Screenshot screenshot = new AShot()
.takeScreenshot(driver, driver.findElement(By.xpath("(//div[#id='ct_search'])[1]")));
ImageIO.write(screenshot.getImage(), "PNG", new File("c:\\temp\\div_element.png"));
See more details and more code samples on this article.
In Node.js, I wrote the following code which works but it is not based on selenium's official WebDriverJS, but based on SauceLabs's WebDriver: WD.js and a very compact image library called EasyImage.
I just wanna emphasize that you cannot really take the screenshot of an element but what you should do is to first, take the screenshot of the whole page, then select the part of the page you like and crop that specific part:
browser.get(URL_TO_VISIT)
.waitForElementById(dependentElementId, webdriver.asserters.isDisplayed, 3000)
.elementById(elementID)
.getSize().then(function(size) {
browser.elementById(elementID)
.getLocation().then(function(location) {
browser.takeScreenshot().then(function(data) {
var base64Data = data.replace(/^data:image\/png;base64,/, "");
fs.writeFile(filePath, base64Data, 'base64', function(err) {
if (err) {
console.log(err);
}
else {
cropInFile(size, location, filePath);
}
doneCallback();
});
});
});
});
And the cropInFileFunction, goes like this:
var cropInFile = function(size, location, srcFile) {
easyimg.crop({
src: srcFile,
dst: srcFile,
cropwidth: size.width,
cropheight: size.height,
x: location.x,
y: location.y,
gravity: 'North-West'
},
function(err, stdout, stderr) {
if (err) throw err;
});
};
For everyone asking for code in C#, below is a simplified version of my implementation.
public static void TakeScreenshot(IWebDriver driver, IWebElement element)
{
try
{
string fileName = DateTime.Now.ToString("yyyy-MM-dd HH-mm-ss") + ".jpg";
Byte[] byteArray = ((ITakesScreenshot)driver).GetScreenshot().AsByteArray;
System.Drawing.Bitmap screenshot = new System.Drawing.Bitmap(new System.IO.MemoryStream(byteArray));
System.Drawing.Rectangle croppedImage = new System.Drawing.Rectangle(element.Location.X, element.Location.Y, element.Size.Width, element.Size.Height);
screenshot = screenshot.Clone(croppedImage, screenshot.PixelFormat);
screenshot.Save(String.Format(#"C:\SeleniumScreenshots\" + fileName, System.Drawing.Imaging.ImageFormat.Jpeg));
}
catch (Exception e)
{
logger.Error(e.StackTrace + ' ' + e.Message);
}
}
I wasted a lot of time on taking screenshot and I want to save yours.
I have used chrome + selenium + c# the result was totally horrible. Finally i wrote a function :
driver.Manage().Window.Maximize();
RemoteWebElement remElement = (RemoteWebElement)driver.FindElement(By.Id("submit-button"));
Point location = remElement.LocationOnScreenOnceScrolledIntoView;
int viewportWidth = Convert.ToInt32(((IJavaScriptExecutor)driver).ExecuteScript("return document.documentElement.clientWidth"));
int viewportHeight = Convert.ToInt32(((IJavaScriptExecutor)driver).ExecuteScript("return document.documentElement.clientHeight"));
driver.SwitchTo();
int elementLocation_X = location.X;
int elementLocation_Y = location.Y;
IWebElement img = driver.FindElement(By.Id("submit-button"));
int elementSize_Width = img.Size.Width;
int elementSize_Height = img.Size.Height;
Size s = new Size();
s.Width = driver.Manage().Window.Size.Width;
s.Height = driver.Manage().Window.Size.Height;
Bitmap bitmap = new Bitmap(s.Width, s.Height);
Graphics graphics = Graphics.FromImage(bitmap as Image);
graphics.CopyFromScreen(0, 0, 0, 0, s);
bitmap.Save(filePath, System.Drawing.Imaging.ImageFormat.Png);
RectangleF part = new RectangleF(elementLocation_X, elementLocation_Y + (s.Height - viewportHeight), elementSize_Width, elementSize_Height);
Bitmap bmpobj = (Bitmap)Image.FromFile(filePath);
Bitmap bn = bmpobj.Clone(part, bmpobj.PixelFormat);
bn.Save(finalPictureFilePath, System.Drawing.Imaging.ImageFormat.Png);
Surya's answer works great if you don't mind involving disk IO. If you'd rather not, then this method may be better for you
private Image getScreenshot(final WebDriver d, final WebElement e) throws IOException {
final BufferedImage img;
final Point topleft;
final Point bottomright;
final byte[] screengrab;
screengrab = ((TakesScreenshot) d).getScreenshotAs(OutputType.BYTES);
img = ImageIO.read(new ByteArrayInputStream(screengrab));
//crop the image to focus on e
//get dimensions (crop points)
topleft = e.getLocation();
bottomright = new Point(e.getSize().getWidth(),
e.getSize().getHeight());
return img.getSubimage(topleft.getX(),
topleft.getY(),
bottomright.getX(),
bottomright.getY());
}
If you prefer you can skip declaring screengrab and instead doing
img = ImageIO.read(
new ByteArrayInputStream(
((TakesScreenshot) d).getScreenshotAs(OutputType.BYTES)));
which is cleaner, but I left it in for clarity. You can then save it as a file or put it in a JPanel to your heart's content.
I think most of the answers here are over-engineered. The way i did it is through 2 helper methods, the first to wait for an element based on any selector; and the second to take a screenshot of it.
Note: We cast the WebElement to a TakesScreenshot instance, so we only capture that element in the image specifically. If you want the full page/window, you should cast driver instead.
Edit: I forgot to say that i'm using Java and Selenium v3 (but should be the same for v4)
WebDriver driver = new FirefoxDriver(); // define this somewhere (or chrome etc)
public <T> T screenshotOf(By by, long timeout, OutputType<T> type) {
return ((TakesScreenshot) waitForElement(by, timeout))
.getScreenshotAs(type);
}
public WebElement waitForElement(By by, long timeout) {
return new WebDriverWait(driver, timeout)
.until(driver -> driver.findElement(by));
}
And then just screenshot whatever u want like this :
long timeout = 5; // in seconds
/* Screenshot (to file) based on first occurence of tag */
File sc = screenshotOf(By.tagName("body"), timeout, OutputType.FILE);
/* Screenshot (in memory) based on CSS selector (e.g. first image in body
who's "src" attribute starts with "https") */
byte[] sc = screenshotOf(By.cssSelector("body > img[href^='https']"), timeout, OutputType.BYTES);
Python 3
Tried with Selenium 3.141.0 and chromedriver 73.0.3683.68, this works,
from selenium import webdriver
chromedriver = '/usr/local/bin/chromedriver'
chromeOptions = webdriver.ChromeOptions()
chromeOptions.add_argument('window-size=1366x768')
chromeOptions.add_argument('disable-extensions')
cdriver = webdriver.Chrome(options=chromeOptions, executable_path=chromedriver)
cdriver.get('url')
element = cdriver.find_element_by_css_selector('.some-css.selector')
element.screenshot_as_png('elemenent.png')
No need to get a full image and get a section of a fullscreen image.
This might not have been available when Rohit's answer was created.
public void GenerateSnapshot(string url, string selector, string filePath)
{
using (IWebDriver driver = new ChromeDriver())
{
driver.Navigate().GoToUrl(url);
var remElement = driver.FindElement(By.CssSelector(selector));
Point location = remElement.Location;
var screenshot = (driver as ChromeDriver).GetScreenshot();
using (MemoryStream stream = new MemoryStream(screenshot.AsByteArray))
{
using (Bitmap bitmap = new Bitmap(stream))
{
RectangleF part = new RectangleF(location.X, location.Y, remElement.Size.Width, remElement.Size.Height);
using (Bitmap bn = bitmap.Clone(part, bitmap.PixelFormat))
{
bn.Save(filePath, System.Drawing.Imaging.ImageFormat.Png);
}
}
}
driver.Close();
}
}
If you are looking for a JavaScript solution, here's my gist:
https://gist.github.com/sillicon/4abcd9079a7d29cbb53ebee547b55fba
The basic idea is the same, take the screen shot first, then crop it.
However, my solution will not require other libraries, just pure WebDriver API code. However, the side effect is that it may increase the load of your testing browser.
Here is an extension function for C#:
public static BitmapImage GetElementImage(this IWebDriver webDriver, By by)
{
var elements = webDriver.FindElements(by);
if (elements.Count == 0)
return null;
var element = elements[0];
var screenShot = (webDriver as ITakesScreenshot).GetScreenshot();
using (var ms = new MemoryStream(screenShot.AsByteArray))
{
Bitmap screenBitmap;
screenBitmap = new Bitmap(ms);
return screenBitmap.Clone(
new Rectangle(
element.Location.X,
element.Location.Y,
element.Size.Width,
element.Size.Height
),
screenBitmap.PixelFormat
).ToBitmapImage();
}
}
Now you can use it to take the image of any element like this:
var image = webDriver.GetElementImage(By.Id("someId"));
Consider using needle - tool for automated visual comparison
https://github.com/bfirsh/needle ,
which has built-in functionality that allows to take screenshots of specific elements (selected by CSS selector). The tool works on Selenium's WebDriver and it's written in Python.
Below the function for taking snapshot a specific element in Selenium. Here the driver is a type of WebDriver.
private static void getScreenshot(final WebElement e, String fileName) throws IOException {
final BufferedImage img;
final Point topleft;
final Point bottomright;
final byte[] screengrab;
screengrab = ((TakesScreenshot) driver).getScreenshotAs(OutputType.BYTES);
img = ImageIO.read(new ByteArrayInputStream(screengrab));
topleft = e.getLocation();
bottomright = new Point(e.getSize().getWidth(), e.getSize().getHeight());
BufferedImage imgScreenshot=
(BufferedImage)img.getSubimage(topleft.getX(), topleft.getY(), bottomright.getX(), bottomright.getY());
File screenshotLocation = new File("Images/"+fileName +".png");
ImageIO.write(imgScreenshot, "png", screenshotLocation);
}
c# code:
public Bitmap MakeElemScreenshot( IWebDriver driver, WebElement elem)
{
Screenshot myScreenShot = ((ITakesScreenshot)driver).GetScreenshot();
Bitmap screen = new Bitmap(new MemoryStream(myScreenShot.AsByteArray));
Bitmap elemScreenshot = screen.Clone(new Rectangle(elem.Location, elem.Size), screen.PixelFormat);
screen.Dispose();
return elemScreenshot;
}
using System.Drawing;
using System.Drawing.Imaging;
using OpenQA.Selenium;
using OpenQA.Selenium.Firefox;
public void ScreenshotByElement()
{
IWebDriver driver = new FirefoxDriver();
String baseURL = "www.google.com/"; //url link
String filePath = #"c:\\img1.png";
driver.Navigate().GoToUrl(baseURL);
var remElement = driver.FindElement(By.Id("Butterfly"));
Point location = remElement.Location;
var screenshot = (driver as FirefoxDriver).GetScreenshot();
using (MemoryStream stream = new MemoryStream(screenshot.AsByteArray))
{
using (Bitmap bitmap = new Bitmap(stream))
{
RectangleF part = new RectangleF(location.X, location.Y, remElement.Size.Width, remElement.Size.Height);
using (Bitmap bn = bitmap.Clone(part, bitmap.PixelFormat))
{
bn.Save(filePath, ImageFormat.Png);
}
}
}
}
If you get an exception java.awt.image.RasterFormatException in chrome, or you want to scroll a element into view then capture a screenshot.
Here is a solution from #Surya answer.
JavascriptExecutor jsExecutor = (JavascriptExecutor) driver;
Long offsetTop = (Long) jsExecutor.executeScript("window.scroll(0, document.querySelector(\""+cssSelector+"\").offsetTop - 0); return document.querySelector(\""+cssSelector+"\").getBoundingClientRect().top;");
WebElement ele = driver.findElement(By.cssSelector(cssSelector));
// Get entire page screenshot
File screenshot = ((TakesScreenshot)driver).getScreenshotAs(OutputType.FILE);
BufferedImage fullImg = ImageIO.read(screenshot);
// Get the location of element on the page
Point point = ele.getLocation();
// Get width and height of the element
int eleWidth = ele.getSize().getWidth();
int eleHeight = ele.getSize().getHeight();
// Crop the entire page screenshot to get only element screenshot
BufferedImage eleScreenshot= fullImg.getSubimage(point.getX(), Math.toIntExact(offsetTop),
eleWidth, eleHeight);
ImageIO.write(eleScreenshot, "png", screenshot);
// Copy the element screenshot to disk
File screenshotLocation = new File("c:\\temp\\div_element_1.png");
FileUtils.copyFile(screenshot, screenshotLocation);
This is my version, in C#, I was basically get most part from Brook's answer and modified it to fit my purpose
public static byte[] GetElementImage(this IWebElement element)
{
var screenShot = MobileDriver.Driver.GetScreenshot();
using (var stream = new MemoryStream(screenShot.AsByteArray))
{
var screenBitmap = new Bitmap(stream);
var elementBitmap = screenBitmap.Clone(
new Rectangle(
element.Location.X,
element.Location.Y,
element.Size.Width,
element.Size.Height
),
screenBitmap.PixelFormat
);
var converter = new ImageConverter();
return (byte[]) converter.ConvertTo(elementBitmap, typeof(byte[]));
}
}
To take a screenshot for a specific element you can now just use this:
public void takeCanvasScreenshot(WebElement element, String imageName) {
File screenshot = element.getScreenshotAs(OutputType.FILE);
try {
FileUtils.copyFile(screenshot, new File("src/main/resources/screenshots/" + imageName + ".png"));
} catch (IOException e) {
e.printStackTrace();
}
}
For C#, below code can work .
try
{
IWebElement transactions = driver.FindElement(By.XPath(".//*[#id='some element']"));
Screenshot screenshot = ((ITakesScreenshot)driver).GetScreenshot();
string title = "some title";
screenshot.SaveAsFile(title, ScreenshotImageFormat.Jpeg);
}
catch (Exception)
{
// handle if element not found
}
I'm using a modified version of #Brook's answer and is working fine even for elements that needs the page to be scrolled.
public void TakeScreenshot(string fileNameWithoutExtension, IWebElement element)
{
// Scroll to the element if necessary
var actions = new Actions(_driver);
actions.MoveToElement(element);
actions.Perform();
// Get the element position (scroll-aware)
var locationWhenScrolled = ((RemoteWebElement) element).LocationOnScreenOnceScrolledIntoView;
var fileName = fileNameWithoutExtension + ".png";
var byteArray = ((ITakesScreenshot) _driver).GetScreenshot().AsByteArray;
using (var screenshot = new System.Drawing.Bitmap(new System.IO.MemoryStream(byteArray)))
{
var location = locationWhenScrolled;
// Fix location if necessary to avoid OutOfMemory Exception
if (location.X + element.Size.Width > screenshot.Width)
{
location.X = screenshot.Width - element.Size.Width;
}
if (location.Y + element.Size.Height > screenshot.Height)
{
location.Y = screenshot.Height - element.Size.Height;
}
// Crop the screenshot
var croppedImage = new System.Drawing.Rectangle(location.X, location.Y, element.Size.Width, element.Size.Height);
using (var clone = screenshot.Clone(croppedImage, screenshot.PixelFormat))
{
clone.Save(fileName, ImageFormat.Png);
}
}
}
The two ifs were necessary (at least for the chrome driver) because the size of the crop exceeded in 1 pixel the screenshot size, when scrolling was needed.
I believe this isn't going to work for you as you use C# and my solution includes a Java library, however maybe others will find it helpful.
For capturing custom screenshots you can use the Shutterbug library. The specific call for this purpose would be:
Shutterbug.shootElement(driver, element).save();
I followed the sample code from #codeslord, but for some reason I had to access my screenshot data differently:
# Open the Firefox webdriver
driver = webdriver.Firefox()
# Find the element that you're interested in
imagepanel = driver.find_element_by_class_name("panel-height-helper")
# Access the data bytes for the web element
datatowrite = imagepanel.screenshot_as_png
# Write the byte data to a file
outfile = open("imagepanel.png", "wb")
outfile.write(datatowrite)
outfile.close()
(using Python 3.7, Selenium 3.141.0 and Mozilla Geckodriver 71.0.0.7222)
I have an application to capture video of the screen and save to a file. I give the user the ability to pick between 480, 720, and "Full Screen" video sizes. A 480 will record in a small box on the screen, 720 will record in a larger box, and of course, "Full Screen" will record in an even larger box. However, this full screen box is NOT the actual screen resolution. It is the app window size, which happens to be around 1700x800. The Video Tool works perfectly for the 480 and 720 options, and will also work if "Full Screen" is overwridden to be the entire screen of 1920x1080.
My question: Are only certain sizes allowed? Does it have to fit a certain aspect ratio, or be an "acceptable" resolution? My code, below, is modified from the xuggle CaptureScreenToFile.java file (the location of the problem is noted by comments):
public void run() {
try {
String parent = "Videos";
String outFile = parent + "example" + ".mp4";
file = new File(outFile);
// This is the robot for taking a snapshot of the screen. It's part of Java AWT
final Robot robot = new Robot();
final Rectangle customResolution = where; //defined resolution (custom record size - in this case, 1696x813)
final Toolkit toolkit = Toolkit.getDefaultToolkit();
final Rectangle fullResolution = new Rectangle(toolkit.getScreenSize()); //full resolution (1920x1080)
// First, let's make a IMediaWriter to write the file.
final IMediaWriter writer = ToolFactory.makeWriter(outFile);
writer.setForceInterleave(false);
// We tell it we're going to add one video stream, with id 0,
// at position 0, and that it will have a fixed frame rate of
// FRAME_RATE.
writer.addVideoStream(0, 0, FRAME_RATE, customResolution.width, customResolution.height); //if I use fullResolution, it works just fine - but captures more of the screen than I want.
// Now, we're going to loop
long startTime = System.nanoTime();
while (recording) {
// take the screen shot
BufferedImage screen = robot.createScreenCapture(fullResolution); //tried capturing using customResolution, but did not work. Instead, this captures full screen, then tries to trim it below (also does not work).
// convert to the right image type
BufferedImage bgrScreen = convertToType(screen, BufferedImage.TYPE_3BYTE_BGR); //Do I need to convert after trimming?
BufferedImage trimmedScreen = bgrScreen.getSubimage((int)customResolution.getX(), (int)customResolution.getY(), (int)customResolution.getWidth(), (int)customResolution.getHeight());
// encode the image
try{
//~~~~Problem is this line of code!~~~~ Error noted below.
writer.encodeVideo(0, trimmedScreen, System.nanoTime() - startTime, TimeUnit.NANOSECONDS); //tried using trimmedScreen and bgrScreen
} catch (Exception e) {
e.printStackTrace();
}
// sleep for framerate milliseconds
Thread.sleep((long) (1000 / FRAME_RATE.getDouble()));
}
// Finally we tell the writer to close and write the trailer if
// needed
writer.close();
} catch (Throwable e) {
System.err.println("an error occurred: " + e.getMessage());
}
}
public static BufferedImage convertToType(BufferedImage sourceImage, int targetType) {
BufferedImage image;
// if the source image is already the target type, return the source image
if (sourceImage.getType() == targetType)
image = sourceImage;
// otherwise create a new image of the target type and draw the new image
else {
image = new BufferedImage(sourceImage.getWidth(), sourceImage.getHeight(), targetType);
image.getGraphics().drawImage(sourceImage, 0, 0, null);
}
return image;
}
Error:
java.lang.RuntimeException: could not open stream com.xuggle.xuggler.IStream#2834912[index:0;id:0;streamcoder:com.xuggle.xuggler.IStreamCoder#2992432[codec=com.xuggle.xuggler.ICodec#2930320[type=CODEC_TYPE_VIDEO;id=CODEC_ID_H264;name=libx264;];time base=1/50;frame rate=0/0;pixel type=YUV420P;width=1696;height=813;];framerate:0/0;timebase:1/90000;direction:OUTBOUND;]: Operation not permitted
Note: The file is successfully created, but has size of zero, and cannot be opened by Windows Media Player, with the following error text:
Windows Media Player cannot play the file. The Player might not support the file type or might not support the codec that was used to compress the file.
Sorry for the wordy question. I'm interested in learning WHAT and WHY, not just a solution. So if anyone can explain why it isn't working, or point me towards material to help, I'd appreciate it. Thanks!
Try to have the dimension even numbers 1696x812