Selenium：遍历元素列表

Question

我正在使用XPath/CSS和Selenium来定位网站上的元素。我想创建一个方法，在这个方法中，我遍历一个定位器列表（XPath/CSS），然后程序选择哪个有效。换句话说，它从定位器1开始-如果定位器存在，它将返回true并存在循环。否则，它将移动到列表中的下一个定位器。一旦用完所有CSS定位器，它就会转到XPath等等。

目前，我正在考虑如下实施：

public boolean iterate(WebDriver driver, By selectorType, String[] locator)
    {

        driver.get("URL");
        for(int selectorListCounter = 0; selectorListCounter < locator.length; selectorListCounter++) {

            try 
            {

                driver.findElement(By.(selectorType)).sendText();
                System.out.println("CSS Selector: " + CSS + " found");
                return true;
            } catch (Exception e)

            {
                System.out.println(CSS + " CSS Selector Not Present");
                return false;
            }


        }

然后，我计划为每种定位器类型调用此方法（一次用于XPath，一次用于CSS等）

这是最好的方式吗？

Answer 1

我认为下面的解决方案会对你有效，我不认为有必要为每种类型的定位器循环一次findElement（）不关心使用什么机制定位元素，只要它是有效的机制

 //Define as many locators as you like
 private static final By cssLaunchIcon() {return By.css("div.myDemoItemActions i.launchIcon");}
 private static final By cssLaunchIcon1() {return By.css("div.myDemoItemActions i.launchIcon.a");}
 private static final By cssLaunchIcon2() {return By.css("div.myDemoItemActions i.launchIcon.li");}
 private static final By xpathLauncIcon() {return By.xpath("//ul/li/a[text()='launch']");}
 private static final By idLaunchIcon() {return By.id("launchIcon");}

 //Initialise the non empty list of locators
 List<By> locators= Arrays.asList(cssLaunchIcon(), xpathLauncIcon(), idLaunchIcon(), cssLaunchIcon1(), cssLaunchIcon2());

 public booolean findElement(List<By> locators) {
    Iterator<By> locs = locators().iterator();
    boolean found = false;
    //while iterator has another locator && no locator found yet
    while(locs.hasNext && !found) {
    WebElement el = locs.next();
    try{
      if(driver.findElement(el).isDisplayed) {
        found = true
      }
    } catch(NoSuchElementException e) {
        System.out.println("Could not find " + el)
   }
   return found;
 }

Answer 2

事实上，我昨天这样做只是为了加快我的结果处理过程。而不是成功和失败，你有选择1和选择2。

此程序中每个元素的隐式等待时间为1秒。因此，while循环总共持续8秒（因为它每次生成两个数组）。这将通过将元素放入数组来检查元素是否存在，然后检查数组的大小，如果元素不存在，则数组将为空。但是，当其中一个条件失败时，意味着页面上存在一个元素。

然后在它的下面，我们设置布尔值来找出页面上存在的这些元素中的哪一个，并捕获不存在的元素的错误。

driver.manage().timeouts().implicitlyWait(1, TimeUnit.SECONDS);
int checkForElement=1;
success = false; failure = false;
        while (driver.findElements(By.className("successMessage")).size()==0 && driver.findElements(By.className("errormessage")).size()==0 && checkForElement<=4 )
        {
            checkForElement+=1;
        }
        checkForElement=1;//reset variable
        try
        {
            @SuppressWarnings("unused")//suppresses the warning so that the class is clean
            WebElement successful = driver.findElement(By.className("successMessage"));//not used practically, but logically used to look for the presence of an element without waiting the normal implicit wait time I would have at 6 seconds
            success = true;
        }
        catch(Exception e)
        {
            success = false;
        }
        try
        {
            @SuppressWarnings("unused")//suppresses the warning so that the class is clean
            WebElement failing = driver.findElement(By.className("errormessage"));//not used practically, but logically used to look for the presence of an element without waiting the normal implicit wait time I would have at 6 seconds
            failure = true;
        }
        catch(Exception e)
        {
            failure = false;
        }

        if(success)
        {
            //run success code
        }
        else if(failure)
        {
            //run failure code
        }