How to click on sing up button using java in selenium i am able to open page but not able to click

+1 vote
package TestNG;

import org.openqa.selenium.By;

import org.openqa.selenium.WebDriver;

import org.openqa.selenium.WebElement;

import org.openqa.selenium.chrome.ChromeDriver;

import org.testng.annotations.AfterMethod;

import org.testng.annotations.BeforeMethod;

import org.testng.annotations.Test;

public class Basic {

WebDriver driver;

@BeforeMethod

public void setup() {

System.setProperty("webdriver.chrome.driver","D:\\selenium\\chromedriver.exe");

driver = new ChromeDriver();

driver.manage().window().maximize();

driver.get("https://www.instagram.com/");

}

@AfterMethod

public void after() {

driver.quit();

}

@Test

public void  demo() throws InterruptedException {

Thread.sleep(6000);

WebElement clickOnsingButton=driver.findElement(By.xpath("//body/div[@id='react-root']/section/main/article/div/div/p/a[1]"));

clickOnsingButton.click();

Thread.sleep(6000);

}

}
Sep 17, 2020 in Selenium by anonymous
• 410 points
1,165 views

2 answers to this question.

0 votes
Best answer

For a software testing services company while automating the application we avoid the usage of the Xpaths. We give preferences to the Classname>=Id>=Linktext>=Css and then last preference is given to the Xpath. Please try the code below.

public void clickCancelBtn() {    
webElement = driver.findElement(By.cssselector("button.btn.btn-outline-primary.bold"));
webElement.click(); 
}

Also, Form seems to be written using jquery. We have to look into the other solution if it not working at your end. Please do share your code snippet or error that it throws if it is not working at your end.

answered Sep 17, 2020 by Vishal

selected Sep 17, 2020 by Jordan
Thank you it work for me
Glad to help!
0 votes

If you know for sure that the element is present, you could try this to simulate the click - if.Click() isn't working

driver.findElement(By.name("submit")).sendKeys(Keys.RETURN);

or

driver.findElement(By.name("submit")).sendKeys(Keys.ENTER);
answered Sep 17, 2020 by Jean

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