CODEWITHSUNDEEP
javastringannotationscharacterwhitespace
Pratik Jaiswal
Pratik Jaiswal

Reputation: 299

Java trim character and whitespaces

Reading the annotation on top of Java TestNG test and I have annotation as: 

@TestInfo(id={ " C26603", " C10047" })

where TestInfo is just the interface that has id() as String array:

public String[] id() default {};

and C26603 and C10047 are just test ids that I assign.

Here is how test structure looks like (for example):

CASE 1:

@TestInfo(id={ " C26603", " C10047" })
public void testDoSomething() {
     Assert.assertTrue(false);
}

Similarly more cleaner case would be:

CASE 2:

@TestInfo(id={ "C26603", "C10047" })

As you can see this case 2 is more clear than the case 1. This case 2 does not have white spaces in test ids.

How do I fetch these ids and make sure that they don't have that C character in beginning and just a pure number? For example, I just want 26603 for my first id and 10047 for 2nd one. There are some spaces in the id array (inside quotes). I want to trim everything of that (like white spaces) and just get the id. I am currently applying for loop to process each id and once I get the pure number, I want to make a 3rd party API call (API expects pure number as input and so removal of C as initial character and other white spaces is important).

Here is what I have tried:

TestInfo annotation = method.getAnnotation(TestInfo.class);
if(annotation!=null) {
        for(String test_id: annotation.id()) {
            //check if id is null or empty
            if (test_id !=null && !test_id.isEmpty()) {
         //remove white spaces and check if id = "C1234" or id = "1234"
                    if(Character.isLetter(test_id.trim().charAt(0))) {
                                test_id = test_id.substring(1);
                    }
                    System.out.println(test_id);
                    System.out.println(test_id.trim());
            }
      }
}

Above code gives me C26603 and not 26603 for the case 1. It works for the case 2.

CASE 3:

@TestInfo(id={ " 26603", " 10047" })

For this case, there is no C as beginning character of test id, so the function should be smart enough to just trim white spaces and go ahead.

Upvotes: 2

Views: 1347

Answers (2)

jmcg
jmcg

Reputation: 1567

I highly encourage you to debug your method. You will learn a lot.

If you take a look at your if statement here:

if(Character.isLetter(test_id.trim().charAt(0))) {
    test_id = test_id.substring(1);
}

When your test_id = " C1234", your condition is true. However, your problem becomes the substring.

ANSWER: trim it!

test_id = test_id.trim().substring(1);

Upvotes: 2

Robby Cornelissen
Robby Cornelissen

Reputation: 97247

The simplest approach would be to just remove everything that is not a digit, using the regular expression non-digit character class (\D):

test_id = test_id.replaceAll("\\D", "");

Upvotes: 5

Related Questions