带有WebElements的java Selenium Webdriver列表

Question

您好，我必须从页面中保存相同的<a>，并已获得为class=my_img我将此元素保存在列表中，在我尝试进入列表的第一个元素后，返回后获得第二个元素，但selenium给我此错误

Exception in thread "main" org.openqa.selenium.StaleElementReferenceException: Element not found in the cache - perhaps the page has changed since it was looked up

这是我的密码

    List <WebElement> Element = drivers.findElements(By.cssSelector(".my_img"));
    System.out.println("Megethos"+Element.size());
    System.out.println("Pame stous epomenous \n");
    for (i = 1; i < Element.size(); i++) {
        drivers.manage().timeouts().implicitlyWait(35, TimeUnit.SECONDS);
        System.out.println(i+" "+Element.size());
        System.out.println(i+" "+Element.get(i));
        action.click(Element.get(i)).perform();
        Thread.sleep(2000);
        System.out.println("go back");
        drivers.navigate().back();
        Thread.sleep(6000);
        drivers.navigate().refresh();
        Thread.sleep(6000);

    }

Answer 1

如果主要目的是单击链接并返回上一页，则最好获取页面中所有“a”元素的“href”属性，并导航到每个元素。当您导航回原始DOM更改时，您所遵循的方式将始终导致StaleElementReferenceExeception

下面是我建议的方法：

List<WebElement> linkElements = driver.findElements(By.xpath("//a[@class='my_img']"));
System.out.println("The number of links under URL is: "+linkElements.size());

//Getting all the 'href' attributes from the 'a' tag and putting into the String array linkhrefs
String[] linkhrefs = new String[linkElements.size()];
int j = 0;
for (WebElement e : linkElements) {
    linkhrefs[j] = e.getAttribute("href");
    j++;
}

// test each link
int k=0;
for (String t : linkhrefs) {
    try{
        if (t != null && !t.isEmpty()) {
            System.out.println("Navigating to link number "+(++k)+": '"+t+"'");
            driver.navigate().to(t);
            String title;
            title = driver.getTitle();
            System.out.println("title is: "+title);

            //Some known errors, if and when, found in the navigated to page.
            if((title.contains("You are not authorized to view this page"))||(title.contains("Page not found"))
                            ||(title.contains("503 Service Unavailable"))
                            ||(title.contains("Problem loading page")))
            {
            System.err.println(t + " the link is not working because title is: "+title);
            } else {
                System.out.println("\"" + t + "\"" + " is working.");
            }
        }else{
            System.err.println("Link's href is null.");
        }
    }catch(Throwable e){

        System.err.println("Error came while navigating to link: "+t+". Error message: "+e.getMessage());
    }

    System.out.println("++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++");
}

Answer 2

你的行动。click（）和/或navigate（）调用将导致页面重新加载，从而导致列表中的WebElement不再有效。将findElements（）调用放入循环中：

List <WebElement> Element = drivers.findElements(By.cssSelector(".my_img"));
for (i = 1; i < Element.size(); i++) {
    Element = drivers.findElements(By.cssSelector(".my_img"));
    drivers.manage().timeouts().implicitlyWait(35, TimeUnit.SECONDS);
    System.out.println(i+" "+Element.size());
    System.out.println(i+" "+Element.get(i));
    action.click(Element.get(i)).perform();
    Thread.sleep(2000);
    System.out.println("go back");
    drivers.navigate().back();
    Thread.sleep(6000);
    drivers.navigate().refresh();
    Thread.sleep(6000);

}