點擊style =「display：block;」在硒webdriver

Question

我想點擊一個元素的風格是style =「display：block;」但無法這樣做。下面

是html代碼

<div class="fl f18 dtoggler pointer underline some-padding new_data_entry" data-div-id="eBWJcg" data-div-ref="new_Passport_form.eBWJcg">+ Add Passport</div> 
<div class="clear new_Passport_form qEgULQ some-margin togglable" style="display: none;"> 

Selenium代碼：

driver.get("idfystaging.com/users/sign_in"); 
driver.findElement(By.name("user[login]")).sendKeys("rustam1‌​@yopmail.com"); 
driver.findElement(By.name("user[password]")).sendKeys("Pass‌​word123"); 
driver.findElement(By.name("commit")).click();  
driver.switchTo().frame("upload_iframe_1"); 
driver.findElement(By.cssSelector("body")).sendKeys(Keys.ESC‌​APE); 
JavascriptExecutor executor= (JavascriptExecutor)driver; 
executor.executeScript("document.getElementById('qEgULQ').st‌​yle.display='block';‌​");click(); 

錯誤日誌：

Cannot read property 'style' of null. 

Answer 1

qEgULQ是類，而不是ID。嘗試

executor.executeScript("document.getElementsByClassName('qEgULQ').st‌​yle.display='block';‌​"); 

Answer 2

你想申請Javascript到元素與id="qEgULq"，而你需要用元素class="qEgULq"：

只需更換

executor.executeScript("document.getElementById('qEgULQ').st‌​yle.display='block';‌​"); 

與

executor.executeScript("document.querySelector('div.clear.new_Passport_form.qEgULQ.some-margin.togglable').st‌​yle.display='block';‌​"); 

還要注意，你的click();從如果某些類名是動態生成或分配的只是一些事件的代碼提供一塊不綁定元素

更新

，你能避免使用類名作爲定位：

WebElement targetDiv = driver.findElement(By.xpath("//div[text()='+ Add Passport']/following-sibling::div")); 
executor.executeScript("arguments[0].st‌​yle.display='block';‌​", targetDiv);