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Identify the error in this Selenium Java code snippet for switching to a new window:

medium📝 Debug Q14 of 15
Selenium Java - Handling Windows, Frames, and Alerts
Identify the error in this Selenium Java code snippet for switching to a new window:
Set windows = driver.getWindowHandles();
for (String win : windows) {
    if (win != driver.getWindowHandle()) {
        driver.switchTo().window(win);
        break;
    }
}
AUsing '!=' to compare strings instead of .equals()
BMissing driver.quit() after switching windows
CNot closing the previous window before switching
DUsing getWindowHandles() instead of getWindowHandle()
Step-by-Step Solution
Solution:
  1. Step 1: Check string comparison method

    In Java, strings must be compared with .equals(), not '!=' which compares references.
  2. Step 2: Identify impact of wrong comparison

    Using '!=' may cause the condition to fail unexpectedly, so the switch may not happen correctly.
  3. Final Answer:

    Using '!=' to compare strings instead of .equals() -> Option A
  4. Quick Check:

    Use .equals() for string comparison [OK]
Quick Trick: Always use .equals() to compare strings in Java [OK]
Common Mistakes:
  • Using '==' or '!=' for string comparison
  • Forgetting to break after switching
  • Confusing getWindowHandle() and getWindowHandles()

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