Unable to click or sendKeys in ajax overlay popup coming and its showing Element not Intractable exception using selenium and java

Question

I have been practicing to automate few scenarios for web application way2Automation.com and struggling to enter the text in Registration form coming as the popup. I have done some research already and tried many ways mentioned below:

a) Using WebDriverWait and explicit wait b) Using Implicit wait and Thread.sleep c) Using JavaScriptExecutor

But none of them worked for me and I am still stuck to register the user. Would really appreciate the help. Below are the artificats

URL: http://way2automation.com/way2auto_jquery/index.php

way2atuomation

Code trials:

1)

//  WebElement ele = driver.findElement(By.xpath("//*[@id='load_form']/div/div[2]/input"));
//  JavascriptExecutor executor = (JavascriptExecutor)driver;
//  executor.executeScript("arguments[0].click();", ele);

//  WebElement button = driver.findElement(By.xpath("//*[@id=\"load_form\"]/div/div[2]/input"));
//  new WebDriverWait(driver, 10).until(ExpectedConditions.visibilityOf(button));

I could not find the registration form where is that form or send link /url here ? — Justin Lambert, Jan 07 '21 at 08:36
Hey @JustinLambert: you can go directly to this link please to reach the registration form : http://way2automation.com/way2auto_jquery/index.php — Abhishek Saxena, Jan 07 '21 at 11:02
http://way2automation.com/way2auto_jquery/index.php : To get the registration form in this website — Abhishek Saxena, Jan 07 '21 at 11:05
@JustinLambert Can you help me with the base page of http://way2automation.com/way2auto_jquery/index.php? — undetected Selenium, Jan 08 '21 at 07:51

score 0 · Answer 1 · answered Jan 07 '21 at 11:24

To click on SUBMIT you can use either of the following Locator Strategies:

cssSelector:

driver.findElement(By.cssSelector("input.button[value='Submit']")).click();

xpath:

driver.findElement(By.xpath("//input[@class='button' and @value='Submit']")).click();

However, as the element is a dynamic element so to click() on the element you need to induce WebDriverWait for the elementToBeClickable() and you can use either of the following Locator Strategies:

cssSelector:

new WebDriverWait(driver, 20).until(ExpectedConditions.elementToBeClickable(By.cssSelector("input.button[value='Submit']"))).click();

xpath:

new WebDriverWait(driver, 20).until(ExpectedConditions.elementToBeClickable(By.xpath("//input[@class='button' and @value='Submit']"))).click();

Reference

You can find a detailed discussion on NoSuchElementException in:

NoSuchElementException, Selenium unable to locate element

score 0 · Accepted Answer · answered Jan 07 '21 at 11:41

0

You need to handle windows
Try the below code and let me know updates

 driver.get("http://way2automation.com/way2auto_jquery/index.php");
        String parent=driver.getWindowHandle();
        String child=driver.getWindowHandle();
        driver).switchTo().window(child);
        driver.findElement(By.name("name")).sendKeys("Abhishek Saxena");

answered Jan 07 '21 at 11:41

Justin Lambert

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That worked like a charm. Thanks a ton for the help. Since I was struggling in going through that. Can you please help with the fact that how you came to know to switch the window in this particular popup case ? – Abhishek Saxena Jan 07 '21 at 15:16
@JustinLambert `parent=driver.getWindowHandle()=child` still you need `switchTo()`? – undetected Selenium Jan 08 '21 at 07:56
@DebanjanB i am taking base page as parent and switch to child window that is registration page. – Justin Lambert Jan 08 '21 at 07:57

Unable to click or sendKeys in ajax overlay popup coming and its showing Element not Intractable exception using selenium and java

2 Answers2

Reference