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You can keep existing approach. SelenideElement extends WebElement so that all existing conditional stuff will be working. Selenide does not have own cross-element conditions. Only the ones which are applied to a single element like .is(Condition.blah()). There is also SelenideWait waiter but it actually does nothing but runs FluentWait with some presets.


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The problem with using Xpath is understandable, but it's not easy to help without the HTML code. I'd recommend using another locator strategy. Is it only the row you want to interact with the one with the value "test"? If yes, in that case, if you're working in a table, you could search for all the rows elements and then select the one with that ...


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for some reason it does not recognize my saved user name and password so every time I open up a LinkedIn session By default, Selenium opens a "guest" window when running. code to recognize that element and click "Sign Up" You can use XPath //a[text()='Sign in'] to select the <a href="#">Sign in</a> link and then ...


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Chrome is currently on v91 and it needs a driver to match. Install Chrome in its default location and make sure it is updated. Then get the latest Chromedriver. There's a NuGet package for chromedriver if you prefer to manage it that way.


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You can use "Select" class for Selenium webdriver. WebElement select = driver.findElement(By.id("#enter your id")); List<WebElement> options = select.findElements(By.classname("x-menu-list-item")); for (WebElement option : options) { if("Lookup Contact".equals(option.getText().trim())) option.click(); } ...


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You can change the script, putting code in a loop: def submit_form(): driver.find_element_by_class_name("submit").click() time.sleep(2.5) driver.find_element_by_xpath("//input[@value='Continue >>']").click() time.sleep(2.5) driver.find_element_by_class_name("submit").click() time.sleep(2.5) driver....


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