Extra java input validation for strings

Question

I want to make this so that short inputs can still be detected, such as "Londo" and "Lon", but want to keep it small and use it without basically copying and pasting the code, any tips? thank you.

if (Menu.answer1.equals("London"))
        {
         if (location.equals("London")) {
             System.out.print(location + " ");
             System.out.print(date + " ");
             System.out.print(degrees + "C ");
             System.out.print(wind + "MPH ");
             System.out.print(winddirection + " ");
             System.out.print(weather + " ");
             System.out.println("");
         }

shoud Lonely be treated as London?

fantaghirocco
– fantaghirocco

2019-03-05 11:52:45 +00:00
Commented Mar 5, 2019 at 11:52 — fantaghirocco
– fantaghirocco, Commented Mar 5, 2019 at 11:52

Nicholas K · Accepted Answer · 2019-03-05 11:50:28Z

3

You can use startsWith()

String city = "London";
if (city.startsWith("Lon")) {
    // do something
}

answered Mar 5, 2019 at 11:50

Nicholas K

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1 Comment

user10762593 Over a year ago

To state the obvious: on its own, (1) that allows the user to type Lonxxxxxx and get London data, and (2) if you ever need to add a town named "Lonsomethingelse", you will need to adjust your code.

Ruslan · Accepted Answer · 2019-03-05 11:54:29Z

1

Also if you need to check some substring, you can use contains method:

Menu.answer1 = "London";
Menu.answer1.contains("ondo"); // true

edited Mar 5, 2019 at 11:54

answered Mar 5, 2019 at 11:52

Ruslan

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1 Comment

Vishwa Ratna Over a year ago

agree, nice concept though, i like it.

Luis Iñesta · Accepted Answer · 2019-03-05 11:54:27Z

0

If you want to check against a fixed set of alternatives, you may use a list of valid inputs using contains:

List<String> londonNames = Arrays.asList("London", "Londo", "Lon");
if (londonNames.contains(Menu.answer1)) {
 ...
}

answered Mar 5, 2019 at 11:54

Luis Iñesta

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Comments

Darshan Mehta · Accepted Answer · 2019-03-05 11:58:09Z

0

You can use (case-insensitive) regex to do the same, e.g.:

(?)Lon[a-z]{0,3} where

(?) = case insensitivity
Lon = Initial 3 characters
[a-z]{0,3} = any number of alphabets between 0 and 3

Here's an example:

String regex = "(?)Lon[a-z]{0,3}";
System.out.println("London".matches(regex));
System.out.println("Lond".matches(regex));
System.out.println("Lon".matches(regex));

answered Mar 5, 2019 at 11:58

Darshan Mehta

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Comments

user10762593 · Accepted Answer · 2019-03-05 12:41:05Z

If the underlying problem is that the user can enter one of several names, and you want to allow abbreviations, then a fairly standard approach is to have a table of acceptable names.

Given the user input, loop through the table testing "does the table entry start with the string typed by the user?" (like one of the previous answers here). If yes, then you have a potential match.

Keep looking. If you get a second match then the user input was ambiguous and should be rejected.

As a bonus, you can collect all names that match, and then use them in an error message. ("Pick one of London, Lonfoo, Lonbar").

This approach has the advantage (compared to a long chain of if-then-else logic) of not requiring you to write more code when all you want to do is have more data.

It automatically allows the shortest unique abbreviation, and will adjust when a once-unique abbreviation is no longer unique because of newly-added names.

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