如何让Selenium 2.0等待页面加载?


当前回答

隐式和显式的等待更好。

但是如果你在Java中处理一个异常,那么你可以使用这个来等待页面重新加载:

Thead.sleep(1000);

其他回答

如果有人使用硒化物:

public static final Long SHORT_WAIT = 5000L; // 5 seconds
$("some_css_selector").waitUntil(Condition.appear, SHORT_WAIT);

更多条件可以在这里找到: http://selenide.org/javadoc/3.0/com/codeborne/selenide/Condition.html

最简单的方法就是等待一些元素出现在加载页面上。

如果你想在页面加载后点击一些按钮,你可以使用等待,然后点击:

await().until().at.most(20, TimeUnit.Seconds).some_element.isDisplayed(); // or another condition
getDriver().find(some_element).click;

如果你想等待一个特定的元素加载,你可以在RenderedWebElement上使用isdisplay()方法:

// Sleep until the div we want is visible or 5 seconds is over
long end = System.currentTimeMillis() + 5000;
while (System.currentTimeMillis() < end) {
    // Browsers which render content (such as Firefox and IE) return "RenderedWebElements"
    RenderedWebElement resultsDiv = (RenderedWebElement) driver.findElement(By.className("gac_m"));

    // If results have been returned, the results are displayed in a drop down.
    if (resultsDiv.isDisplayed()) {
      break;
    }
}

(例子来自《5分钟入门指南》)

driver.asserts().assertElementFound("Page was not loaded",
By.xpath("//div[@id='actionsContainer']"),Constants.LOOKUP_TIMEOUT);

The best way to wait for page loads when using the Java bindings for WebDriver is to use the Page Object design pattern with PageFactory. This allows you to utilize the AjaxElementLocatorFactory which to put it simply acts as a global wait for all of your elements. It has limitations on elements such as drop-boxes or complex javascript transitions but it will drastically reduce the amount of code needed and speed up test times. A good example can be found in this blogpost. Basic understanding of Core Java is assumed.

http://startingwithseleniumwebdriver.blogspot.ro/2015/02/wait-in-page-factory.html