<a class="LinkDetail" href="/settings/carsettings?xyz=L_11:1:*:2&carid=199&carnumber=4294967295" target="_top" tabindex="23"/>
In the above link, I need to locate element using /settings/carsettings and carid=199
Using a CSS locator. Can anyone let me know the syntax for the same? Also share the syntax for XPath too.
Show us what you tried please, so we can anylyse what you failed to achieve. If the following CSS Selector/XPath don't work, post your stacktrace and more HTML code to find the best locators.
CSS Selector
a[href*='settings/carsettings'][href*='carid=199']
XPath
.//a[contains(@href, 'settings/carsettings') and contains(@href, 'carid=199')]
What you require can be achieved using the following code :-
//get all <a> tags in the webpage to a list
List<WebElements> aTags = driver.findElements(By.tagName("a"));
int index = 0;
//iterate through list of <a> tags
for (WebElement aTag: aTags) {
//get the href attribute of each <a> tag
String href = aTag.getAttribute("href");
//see if the href contains /settings/carsettings and carid=199
if (href.contains("/settings/carsettings")&&href.contains("carid=199")) {
//if it contains break out of for loop. This esssentially gives the index
break;
}
index++;
}
//get the required <a> tag using the index
WebElement required = aTags.get(index);
Let me know if this helps you.
