Selenium Webdriver, how to validate a string in a url

Question

Any help to validate a parameter in the url using selenium web driver + java

getCurrentUrl() = https://www.loca.com/usa/test/#/xxx

My url string is as above, and based on the url I need to enter a value on the loaded page, like lets say - I have zipcode field to enter zipcode value.

If url string contains 'usa' Enter zipcode as 12345 if url string contains canada Enter zipcode as Ab1233

Answer 1

If you are using Java, you can do something like:

String zipcode = "";
String currentUrl = "https://www.loca.com/usa/test/#/xxx"
//This will check to make sure USA is a part of the URL, and doesn't exist as another part of the URL.
if(currentUrl.contains("/usa/")){
   zipcode = "12345";
//Same here with Canada
} else if(currentUrl.contains("/canada/"){
   zipcode = "ab1233"; 
//You can continue pattern for any other countries
} else if(currentUrl.contains(...){
...

Just put the URL into a string, and then use the String method contains to check and see if USA in the string. If it is, use zip code, otherwise, do something else.

Answer 2

I think an if then else is going to get pretty messy if you have more than just a few of them. I prefer to use a switch . I think it makes the code easier to read but maybe that's a personal preference.

String currentUrl = "https://www.loca.com/usa/test/#/xxx";
String patternString1 = ".com\\/(\\w+)\\/";
Matcher matcher = Pattern.compile(patternString1).matcher(currentUrl);
if (matcher.find())
{
    String zipcode = "";
    switch (matcher.group(1))
    {
        case "usa":
            zipcode = "12345";
            break;
        case "canada":
            zipcode = "Ab1233";
            break;
        default:
            // handle an unexpected country
            break;
    }
}
else
{
    // handle the case of a bad URL
}

// do something with zipcode
System.out.println(zipcode);