What is the process for selecting a cssSelector() in Chrome Browser using Chrome Developer Tools?
Could you provide an example for the given code snippet below?
WebElement searchBox = driver.findElement(By.cssSelector("selector"));
searchBox.click();
How do you determine the appropriate value for the SELECTOR field in Chrome Developer Tools?-F12 in Chrome. Please illustrate with an example.
Appreciate your assistance
If you are working with Selenium as indicated by your tags, it seems like you are attempting to utilize
driver.findElements(By.className(".."));
or a similar method, it's important to have a good understanding of CSS Selectors.
For instance, if we examine the website logo, we might find:
<div id="hlogo">
<a href="/">
Website Name
</a>
</div>
In this scenario, we would use:
driver.findElements(By.id("hlogo"));
Alternatively, we can utilize a cssSelector like this:
driver.findElements(By.cssSelector("#hlogo"));
From what I gather from your inquiry, this is what you require:
$$("your_selector")
Here are a couple of examples:
$$("#id_of_element")
$$("[type="text"]")
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