ruby+watir随机而不重复获取Menu菜单的元素

测试用例是类似上面的Menu菜单，共9个

先看看元素定义（yaml）：

#频道切换-美食
channel_0_link: div(:class,'navMenuBg').li(:id,'num_2').link(:href,'http://beijing.xxxx.com/xxxshi')
channel_0_link_on: div(:class,'navMenuBg').li(:id,'num_2').span(:class,'curCorner')
#频道切换-娱乐
channel_1_link: div(:class,'navMenuBg').li(:id,'num_4').link(:href,'http://beijing.xxxx.com/xxxxian')
channel_1_link_on: div(:class,'navMenuBg').li(:id,'num_4').span(:class,'curCorner')
#频道切换-生活服务
channel_2_link: div(:class,'navMenuBg').li(:id,'num_5').link(:href,'http://beijing.xxxx.com/xxxxxhuo')
channel_2_link_on: div(:class,'navMenuBg').li(:id,'num_5').span(:class,'curCorner')
#频道切换-商品div(:class,'navMenuBg').li(:id,'num_6').
channel_3_link: link(:index,21)
channel_3_link_on: span(:class,'curCorner')
#频道切换-酒店div(:class,'navMenuBg').li(:id,'num_7').
channel_4_link: link(:index,22)
channel_4_link_on: span(:class,'curCorner')
#频道切换-旅游div(:class,'navMenuBg').li(:id,'num_8').
channel_5_link: link(:index,23)
channel_5_link_on: span(:class,'curCorner')
#频道切换-抽奖
channel_6_link: div(:class,'navMenuBg').li(:id,'num_9').link(:href,'http://www.xxxx.com/xxxxjiang')
channel_6_link_on: div(:class,'navMenuBg').li(:id,'num_9').span(:class,'curCorner')
#频道切换-促销
channel_7_link: div(:class,'navMenuBg').li(:id,'num_10').link(:href,'http://www.xxxx.com/xxxxxiao')
channel_7_link_on: div(:class,'navMenuBg').li(:id,'num_10').span(:class,'curCorner')
#频道切换-往期团购div(:class,'navMenuBg').li(:id,'num_12').
channel_8_link: link(:index,26)
channel_8_link_on: span(:class,'curCorner')

测试用例：使用循环，随机获取9个Menu菜单，每个都必须点击到，并验证

  def channel
    @b.goto URL
    channel = 0
    while channel <= 8
        times = rand(9).to_s
        AutoTest("channel_#{times}_link").click
        sleep 1
        assert_true(AutoTest("channel_#{times}_link_on").exists?)
        channel += 1
    end
  end

脚本中循环9次，每次都取一个随机值，随机数rand()是从0开始，所以我在元素定义时从0开始对这9个Menu菜单的元素进行编码，如：channel_0_link。

但是这里有个问题rand()函数中取的有重复值，即有些Menu菜单被点击2次或者多次，这就与我们的要求相驳。我几乎找遍了API，没有找到按顺序或者随机而不重复的方法。下一步我决定使用另外一种随机的方法来解决，其实随机播放分为两种，random和shuffle

  def channel_food
    @b.goto URL
    linkid=[0,1,2,3,4,5,6,7,8]
    linkid.shuffle.each{
    |i|
    times = i 
    AutoTest("channel_#{times}_link").click
    sleep 1
    assert_true(AutoTest("channel_#{times}_link_on").exists?)
    }
  end

以上的代码，可以实现随机获取9个Menu菜单，每个都必须点击到，并验证的需求。each的方法是从数组中获取数据；shuffle的方法是对获取的值进行重新排列，在洗牌程序中也是使用这种方法来做的（不会产生重复）。

    a=[1,2,3,4,5,6,7,8,9]
    a.shuffle.each{
    |i|
    b = i
    puts b
    }

大家可以去试试

参考：http://blog.sina.com.cn/s/blog_6a55d9950100v4xu.html