Python if 语句与selenium：有什么办法可以缩短吗？

Question

我正在制作一个硒脚本，用于搜索任何可见元素，打印它们是否显示，然后将它们附加到列表（ButtonDisplayed 或 ButtonNotDisplayed）。然而，这是非常重复的，因为我必须为我检查的每个元素创建一个 if else 语句。有没有办法让我缩短这段代码？

    if login.is_displayed():
        logInButtonDisplayed = "Login is displayed"
        print(logInButtonDisplayed)
        buttonDisplayed.append(logInButtonDisplayed)
    else:
        logInButtonNotDisplayed = "Login is not displayed"
        print(logInButtonNotDisplayed)
        buttonNotDisplayed.append(logInButtonNotDisplayed)

    if login.is_enabled():
        logInButtonEnabled = "Log In button is enabled"
        print(logInButtonEnabled)
        buttonEnabled.append(logInButtonEnabled)
    else:
        logInButtonDisabled = "Log In button is disabled"
        print(logInButtonDisabled)
        buttonDisabled.append(logInButtonDisabled)

Answer 1

您可以尝试使用 f 字符串和条件表达式。

def check_element_status(element, displayed_list, not_displayed_list, enabled_list, disabled_list):
    display_status = "displayed" if element.is_displayed() else "not displayed"
    enabled_status = "enabled" if element.is_enabled() else "disabled"

    display_message = f"{element.get_attribute('id')} is {'' if element.is_displayed() else 'not '}displayed"
    enabled_message = f"{element.get_attribute('id')} is {'' if element.is_enabled() else 'not '}enabled"

    print(display_message)
    print(enabled_message)

    displayed_list.append(display_message)
    not_displayed_list.append(display_message) if not element.is_displayed() else None
    enabled_list.append(enabled_message)
    disabled_list.append(enabled_message) if not element.is_enabled() else None

# Assuming you have login element
check_element_status(login, buttonDisplayed, buttonNotDisplayed, buttonEnabled, buttonDisabled)