CODEWITHSUNDEEP
javaselenium-webdriverappium
sandy
sandy

Reputation: 117

Element not found with webdriver

I am automating a android app with Appium using Java. My scenario is, I need to click either button 1 or button 2, whichever is present

Appium error log:[debug] [AndroidBootstrap] [BOOTSTRAP LOG] [debug] Failed to locate element. Clearing Accessibility cache and retrying. [debug] [AndroidBootstrap] [BOOTSTRAP LOG] [debug] Finding '//android.widget.ImageButton[@resource-id='net.ilius.android.meetic:id/profileMailPremiumButton']' using 'XPATH' with the contextId: '' multiple: false

if (driver.findElementByXPath("//android.widget.ImageButton[@resource-id='net.ilius.android.meetic:id/profileMailPremiumButton']")
        .isDisplayed()) {
     driver.findElementByXPath("//android.widget.ImageButton[@resource-id='net.ilius.android.meetic:id/profileMailPremiumButton']")
        .click();
} else {
    driver.findElementById("net.ilius.android.meetic:id/profileMailButton").click();
}

Upvotes: 1

Views: 442

Answers (1)

Pritam Maske
Pritam Maske

Reputation: 2770

If you use the isDisplayed() and element is not present on the UI it will throw the exception - element not found.

So instead of that first check whether that element is exist or not by using the findElements : driver.findElements(selector).isEmpty()

if it is empty it means element is not available now you can go to the else block

Use this piece of code : `

if (!driver.findElements(By.xPath("//android.widget.ImageButton[@resource-id='button1']")).isEmpty()) {
     driver.findElementByXPath("//android.widget.ImageButton[@resource-id='button1']").click();
} else {
    driver.findElementById("button2").click();
}`

Upvotes: 1

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