Question

我使用Selenium WebDriver我需要对第一个帐户项执行和操作（单击），执行某些操作，然后返回并在下一个项目上重复该过程。它最多可能包含90个帐户项目，因此我试图避免对所有90个帐户进行硬编码。任何想法如何绕过帐户列表？

＆＃13;

<ul class="account-list">
  <li class="account-item">one item</li>
  <li class="account-item">two items</li>
  <li class="account-item">three items</li>
</ul>

＆＃13;

Answer 1

你不能迭代Selenium中的元素列表，因为当DOM更改时，它们的对象引用将变为无效（StaleElement异常），但是只要列表的大小不是这样，你就可以通过索引号来完成它改变

elements = @driver.find_elements(css: ".account-item>a")    
elements.size.times do |i|
  @driver.find_element(css: ".account-list > li:nth-child(#{i})>a").click
  # Do something
  @driver.navigate.back
end

Answer 2

你可以通过import java.util.Scanner; public class Fibonacci { public static void main(String args[]) { Scanner scan = new Scanner(System.in); System.out.println("This is a Fibonacci sequence generator"); System.out.println("Choose what you would like to do"); System.out.println("1. Find the nth Fibonacci number"); System.out.println("2. Find the smallest Fibonacci number that exceeds user given value"); System.out.println("3. Find the two Fibonacci numbers whose ratio is close enough to the golden number"); System.out.print("Enter your choice: "); int choice = scan.nextInt(); int xPre = 0; int xCurr = 1; int xNew = 0; switch (choice) { case 1: System.out.print("Enter the target index to generate (>1): "); int index = scan.nextInt(); for (int i = 2; i <= index; i++) { xNew = xPre + xCurr; xPre = xCurr; xCurr = xNew; } System.out.println("The " + index + "th number Fibonacci number is " + xNew); break; case 2: System.out.print("Enter the target value (>1): "); int value = scan.nextInt(); } } }获取所有元素，然后按照以下方式迭代元素来实现：

css

如何使用Ruby中的Selenium获取<ul>中的<li>元素列表？

2 个答案: