java 如何验证硒中某个字段的粗体外观
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How to verify bold appearance of a certain field in selenium
提问by Virendra Joshi
I am trying to automate a manual script (using selenium in java) to check the bold appearance of a certain field(label:which stands for mandatory field) on a web page . what can be the possible selenium java functions to verify the bold appearance of certain element(In class there is no information about the appearance)
我正在尝试自动化手动脚本（在 Java 中使用 selenium）来检查网页上某个字段（标签：代表必填字段）的粗体外观。什么是可能的 selenium java 函数来验证某些元素的粗体外观（在课堂上没有关于外观的信息）
采纳答案by Justin Ko
You can check the font-weightusing the style()method (assuming you are actually using Selenium-Webdriver).
您可以使用该方法检查字体粗细style()（假设您实际上使用的是 Selenium-Webdriver）。
So say you have HTML like:
所以说你有这样的 HTML：
<body>
<div id='1' style='font-weight:normal'>
<div id='2' style='font-weight:bold'>Field Label</div>
<div id='3'>Field</div>
</div>
</body>
You can do the following to check the font-weight of the field label div (the following is in Ruby, though similar should be possible in the other languages).
您可以执行以下操作来检查字段标签 div 的字体粗细（以下是在 Ruby 中，但在其他语言中应该类似）。
el = driver.find_element(:id, "2")
if el.style('font-weight') >= 700
puts 'text is bold'
else
puts 'text is not bold'
end
回答by Petr Jane?ek
With WebDriver (in Java), you can use getCssValue().
使用 WebDriver（在 Java 中），您可以使用getCssValue()。
import static org.junit.Assert.assertTrue;
(...)
// assuming elem is a healthy WebElement instance, your found element
String fontWeight = elem.getCssValue("font-weight");
assertTrue(fontWeight.equals("bold") || fontWeight.equals("700"));
(since 700is the same as bold)
With Selenium RC, see this technique, just use font-weight(or fontWeightdepending on the usage).
使用 Selenium RC，请参阅此技术，只需使用font-weight（或fontWeight取决于使用情况）。
回答by Rob Dennis
I really do like Justin Ko's suggestion to use the style("font-weight")method, however in the python bindings, the equivalent appears to be value_of_css_property("font-weight")
我真的很喜欢 Justin Ko 的使用该style("font-weight")方法的建议，但是在 python 绑定中，等效的似乎是value_of_css_property("font-weight")
>>> element = self.wd.find_element_by_id("some-id")
>>> element.value_of_css_property('font-weight')
u'700'
>>> element.style('font-weight')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'WebElement' object has no attribute 'style'
http://code.google.com/p/selenium/source/browse/py/selenium/webdriver/remote/webelement.py#132
http://code.google.com/p/selenium/source/browse/py/selenium/webdriver/remote/webelement.py#132
apologies that this is a separate answer versus a comment for that answer, but I apparently have too low of a karma threshold to comment there
很抱歉，这是一个单独的答案而不是对该答案的评论，但我显然没有在那里发表评论的业力阈值太低
回答by Bob Evans
I see that the element you mention has a compound class, why not try to locate all the elements with the label_required class, python example
我看到你提到的元素有一个复合类，为什么不尝试使用 label_required 类来定位所有元素，python 示例
els = driver.find_elements_by_css_selector(div[class*=label_required])
self.assertTrue(len(els) == [known value of required fields on page])
java(Note: I don't write java code, so syntax may be wrong):
java（注意：我不会写java代码，所以语法可能有误）：
import static org.junit.Assert.assertTrue;
WebDriver driver = new FirefoxDriver();
WebElement els=driver.findElements(By.cssSelector("div[class*=label_required]"));
assertTrue(els.length == [known value of required fields on page]);
