问题
String strPrimaryNav = "MEN";
String strSecondaryNav = "Shoes";
String strTertiaryNav = "Golf";
driver.findElement(By.linkText(strPrimaryNav)).click();
WebElement weSecNav = driver.findElement(By.className("secondaryButton").linkText(strSecondaryNav));
Mouse mouse = ((HasInputDevices) driver).getMouse();
mouse.mouseDown(((Locatable)weSecNav).getCoordinates());
//just trying with for loop as the tertiary popup disappears quickly and hence getting ElementNotVisible Exception
for (int i = 0 ; i< 2; i++){
mouse.mouseDown(((Locatable)weSecNav).getCoordinates());
//mouse.mouseMove(((Locatable)weSecNav).getCoordinates(), 0, 0 );
//WebElement weTerNav = driver.findElement(By.className("tertiaryButton").linkText(strTertiaryNav));
WebElement weTerNav = driver.findElement(By.linkText(strTertiaryNav));
boolean isSecDisplayed = ((RenderedWebElement)weTerNav).isDisplayed();
System.out.println("isDisplayed: " + isSecDisplayed);
System.out.println(" " + ((RenderedWebElement)weSecNav).getAttribute("href"));
System.out.println(" " + ((RenderedWebElement)weSecNav).getValueOfCssProperty("action"));
weTerNav.click();
}
I was trying the below code using selenium 2 but, the tertiary popup not stays long to click it and hence getting ElementNotVisible exception at Tertiary click.
回答1:
You can at least check that the element is visible before you send your click:
Selenium s = new WebDriverBackedSelenium( driver, URL );
s.waitForCondition( "!( document.getElementById('.....') === null )", "20000" );
s.click( .... );
This avoids the exception. I'm not sure there is a way to make the popup stay any longer than it should.
来源:https://stackoverflow.com/questions/5981656/how-to-handle-mouseover-in-selenium-2-api