是否可以使用Selenium WebDriver进行截图?
(注:不含硒遥控器)
是否可以使用Selenium WebDriver进行截图?
(注:不含硒遥控器)
当前回答
Python
您可以使用Python web驱动程序从windows中捕获图像。使用下面的代码哪个页面需要捕获屏幕截图。
driver.save_screenshot('c:\foldername\filename.extension(png, jpeg)')
其他回答
你可以使用Webdriverclass对象创建一个webdriver支持的selenium对象,然后你可以截屏。
Ruby
time = Time.now.strftime('%a_%e_%Y_%l_%m_%p_%M_%S')
file_path = File.expand_path(File.dirname(__FILE__) + 'screens_shot')+'/'+time +'.png'
#driver.save_screenshot(file_path)
page.driver.browser.save_screenshot file_path
你可以在浏览器中截取网页可见部分的截图:
首先导入:
import java.io.File;
import com.google.common.io.Files;
Then
File src=((TakesScreenshot)driver).getScreenShotAs(OutputType.FILE);
Files.copy(src,new File("new path/pic.jpeg"));
另外,在Selenium4之后,你还可以截取webelement的截图:
WebElement element=driver.findElement(By.xpath("xpath
here"));
File src=element.getScreenShotAs(OutputType.FILE);
File.copy(src,new File("new path/pic.jpeg"));
Java
对于较新版本的Java, FileUtils函数不起作用。在这里,下面的功能可以完美地复制屏幕截图。
import java.io.File;
import org.openqa.selenium.OutputType;
import org.openqa.selenium.TakesScreenshot;
import org.openqa.selenium.WebDriver;
import org.openqa.selenium.io.FileHandler;
File scrFile = ((TakesScreenshot)driver).getScreenshotAs(OutputType.FILE);
File destfile = new File(destination folder /filename.extension);
FileHandler.copy(scrfile, destfile);
/**
* Take a screenshot and move to the given folder location.
*
* @param driver
* @param folderLocation
* @return screenShotFilePath
*/
public static String captureScreenshot(WebDriver driver, String folderLocation) {
// Variable to store screenshot's file path.
String screenShotFilePath = null;
// Generate unique id for screen shot name.
String uniqueId = UUID.randomUUID().toString().substring(31);
if (driver != null) {
// Generate screenshot as a file
File scrFile = ((TakesScreenshot) driver).getScreenshotAs(OutputType.FILE);
// New screenshot file path with having file name
screenShotFilePath = folderLocation + File.separator + uniqueId + ".png";
// Move file to the destination location.
FileUtils.moveFile(scrFile, new File(screenShotFilePath));
}
return screenShotFilePath;
}