点击样式=“显示:块;";在Selenium WebDriver中 [英] clicking on an style="display: block;" in selenium webdriver
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问题描述
我试图单击样式为 style ="display:block;" 的元素,但无法这样做.
i am trying to click on an element for which style is style="display: block;" but not able to to do so.
下面是html
代码
<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
代码:
driver.get("idfystaging.com/users/sign_in");
driver.findElement(By.name("user[login]")).sendKeys("rustam1@yopmail.com");
driver.findElement(By.name("user[password]")).sendKeys("Password123");
driver.findElement(By.name("commit")).click();
driver.switchTo().frame("upload_iframe_1");
driver.findElement(By.cssSelector("body")).sendKeys(Keys.ESCAPE);
JavascriptExecutor executor= (JavascriptExecutor)driver;
executor.executeScript("document.getElementById('qEgULQ').style.display='block';");click();
错误日志:
Cannot read property 'style' of null.
推荐答案
您正在尝试将Javascript
应用于具有id="qEgULq"
的元素,而需要具有class="qEgULq"
的元素:
You're trying to apply Javascript
to element with id="qEgULq"
while you need element with class="qEgULq"
:
只需替换
executor.executeScript("document.getElementById('qEgULQ').style.display='block';");
使用
executor.executeScript("document.querySelector('div.clear.new_Passport_form.qEgULQ.some-margin.togglable').style.display='block';");
还请注意,您提供的代码中的click();
并未绑定到元素
Also note that your click();
from provided piece of code doesn't bound to element
更新
如果某些类名是动态生成的或仅在某些事件上分配的,则可以避免将类名用作定位符:
If some of class names are generated dynamically or assigned just on some events, you can avoid using class names as locator:
WebElement targetDiv = driver.findElement(By.xpath("//div[text()='+ Add Passport']/following-sibling::div"));
executor.executeScript("arguments[0].style.display='block';", targetDiv);
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