如何从下拉列表中选择第一个元素.回报率/水豚/硒

Question

所以这段代码工作得很好,直到我最近更新了我的selenium webdriver:

When /^I search for (.*)$/ do |term|
    term = " " if term == "blank"
    step "I fill in search with #{term}"
    within(".navbar-search") do
        page.find(:css, "li:first").click
    end
end

我更新了,现在我收到以下错误:

 An invalid or illegal string was specified (Selenium::WebDriver::Error::UnknownError)
  ./features/step_definitions/search_steps.rb:5:in `block (2 levels) in <top (required)>'
  ./features/step_definitions/search_steps.rb:4:in `/^I search for (.*)$/'
  features/search_friend.feature:13:in `When I search for <term>'

这是黄瓜功能:

@javascript
Scenario Outline: The search bar
    Given I login
    And I have a contact named ABC
    And I have a contact named DEF
    And I have a contact named GHI
    When I search for <term>
    Then I should see the message <message>

    Examples:
    | term | message |
    | ... some examples ... | |

Answer 1

我猜你不仅更新了Webdriver而且更新了Capybara.

Capybara 2.1现在使用驱动程序的CSS选择器实现.对于Selenium Webdriver,它意味着使用浏览器的实现.:first伪选择器不是标准的,浏览器不支持,因此您的CSS选择器无效.

之前它有用,因为Capybara使用Nokogiri将CSS选择器转换为XPath.Nokogiri支持:first伪选择器.

因此,您应该将无效的CSS选择器更改为有效的类似:

• page.first(:css, 'li').click
• page.find(:css, 'li', match: :first).click(之前的变体与此变体之间的区别在于此变体等待元素出现在页面上,但第一个不等待.请参阅Capybara README的此部分以获取更多信息match)
• page.find(:css, "li:first-child").click
• page.find('li:first-child').click