The method element.isDisplayed() is indicating a false value even though the element is visibly present on the screen

element.isDisplayed() is returning false even though the element is visible on the screen, causing issues with clicking on it. I attempted to use the code below, but without success.

 Actions cursor = new Actions(driver);
                    cursor.moveToElement(element);
                    cursor.click(element).build().perform();

I also tried the following code:

JavascriptExecutor executor = (JavascriptExecutor) driver;
                    executor.executeScript("arguments[0].click;", element);

However, this approach was also unsuccessful.

HTML 

<TD id=PanelTable class=menulevel1norm onmouseover="this.className='menulevel1hl';" onmouseout="this.className='menulevel1norm'" onclick="JAVASCRIPT CODE">&nbsp;Order&nbsp;</TD>
<DIV style="BORDER-BOTTOM: medium none; POSITION: absolute; BORDER-LEFT: medium none; VISIBILITY: hidden; BORDER-TOP: medium none; BORDER-RIGHT: medium none" name="actiondiv">
    <DIV myonclick="SOME JAVASCRIPT CODE HERE">Order Console</DIV>
    <DIV myonclick="SOME JAVASCRIPT CODE HERE">Order Release</DIV>
</DIV>
javascriptseleniumselenium-webdriver

Answer №1

One thing that you're doing incorrectly is running this JavaScript:

arguments[0].click;

An error won't be triggered because the .click field does exist (as long as what you provide to .executeScript is a WebElement instance.) However, .click is actually a method and needs to be executed. You should use:

arguments[0].click();

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