Java Selenium:遍历元素列表

Java Selenium:遍历元素列表,java,css,selenium,xpath,Java,Css,Selenium,Xpath,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[] locato

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;
            }


        }
public boolean iterate(WebDriver驱动程序,按selectorType,字符串[]定位器)
{
获取驱动程序(“URL”);
对于(int-selectorListCounter=0;selectorListCounter
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)


这是最好的方法吗?

我认为下面的解决方案适合您,我认为没有必要为每种类型的定位器循环一次。
findElement()
不关心您使用什么机制来定位元素,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;
 }
//定义任意多个定位器
私有静态final By cssLaunchIcon(){return By.css(“div.myDemoItemActions i.launchIcon”);}
私有静态final By cssLaunchIcon1(){return By.css(“div.myDemoItemActions i.launchIcon.a”);}
私有静态final By cssLaunchIcon2(){return By.css(“div.myDemoItemActions i.launchIcon.li”);}
xpathLauncIcon()的私有静态final({return By.xpath(“//ul/li/a[text()='launch']”;}
idLaunchIcon()的私有静态final{return By.id(“launchIcon”);}
//初始化定位器的非空列表
列表定位器=Arrays.asList(cssLaunchIcon(),xpathLauncIcon(),idLaunchIcon(),cssLaunchIcon1(),cssLaunchIcon2());
public booolean findElement(列表定位器){
迭代器locs=定位器()。迭代器();
布尔值=false;
//而迭代器有另一个定位器&&尚未找到定位器
while(locs.hasNext&&!找到){
WebElement el=locs.next();
试一试{
if(显示驱动程序findElement(el){
找到=真
}
}捕获(无接触元素例外e){
System.out.println(“找不到”+el)
}
发现退货;
}

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

隐式等待时间是该程序中每个元素检查1秒。因此while循环总共持续8秒(因为它每个元素生成两个数组)。此操作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,如果其中一个条件失败,则表示页面上存在一个元素

然后在下面设置布尔值,找出页面上出现了哪些元素,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
        }
driver.manage().timeouts().implicitlyWait(1,TimeUnit.SECONDS);
int checkForElement=1;
成功=错误;失败=错误;

while(driver.findElements(By.className(“successMessage”)).size()=0&&driver.findElements(By.className(“errormessage”)).size()==0&&CHECKFORELEMENT您能告诉我,为什么需要这个吗?其中一个答案有用吗?或者您决定使用另一个解决方案吗?我发现此代码有两个问题..1.一旦找不到元素,代码就会崩溃(NoSuchElementException)。我正在使用try/catch来解决这个问题。2.传递多个CSS如何(或多个XPath)选择器?代码仅适用于1 CSS/1 XPath..1.我忽略了。2.您可以添加任意多个定位器,只要它们具有不同的名称。我将进行快速编辑。进一步编辑定位器以消除任何歧义。