FLUTTER - Checking if a string contains another one-CodePudding

I am working on an English vocabulary learning app. Some of the exercises given to the users are written quizzes. They have to translate French words into English words and vice versa.

To make the checking a little more sophisticated than just "1" or "0" (TypedWord == expectedWord), I have been working with similarities between strings and that worked well (for spelling mistakes for example).

I add also used the contains function, so that for example, if the user adds an article in front of the expected word, it doesn't consider it wrong. (Ex : Ecole (School is expected), but user writes "A school").

So I was checking with lines such as "if (typedWord.contains(word)==true) then...". It works fine for the article problem. But it prompts another issue :

Ex : A bough --> the expected French word is "branche". If user types "une branche", it considers it correct, which is great. But if user types "débrancher" (to unplug), it considers it correct as well as the word "branche" is a part of "débrancher"...

How could I keep this from happening ? Any idea of other ways to go about it ?

I read the three proposed answers which are really interesting. The thing is that some of the words are compound.... "Ex : kitchen appliance, garden tool" etc... so then I think the "space" functions might be problematic...

CodePudding user response：

In this case, separate the whole answer with the "space", then compare it with the correct word.

For an example:
User's answer: That is my school
Separate it with space, so that you will find an array of words:
that, is, my, school.

Then compare each word with your word. It will give you the correct answer.

The flutter code will be like below:

  usersAnswer?.split(" ").forEach((word){
    if(word == correctAnswer) 
      print("this is a correct answer");
  });

CodePudding user response：

You could check for whitespaces before and after the correct word: something like if (typedWord.contains(' ' word ' ')==true) then..., so that "débrancher" gets marked as wrong. This is kind of strict, though: if the sentence must be completed with some punctuation, it would be rejected by this check. You'll probably want some RegExp that allows punctuation but not whitespaces.

CodePudding user response：

You can split the string by space and check if the resulting array has the word you're looking for.

typedWord.split(' ').contains('debranche');

So if typedWord is 'une branchethesplit(' ') will turn it into this array: ['une', 'branche']. Now when you check if this array contains('branche') it will check if the exact string branche exists which in this case it does and returns true.

However if it's 'une debranche' the resulting array would be: ['une', 'debranche'] and because this array has no value equal to 'branche' it will return false. Remember that when you use split it turns the string into an array and by using contains on an array it checks whether or not an item of exactly the value you provide contains exists or not, whereas in a string it checks if part of that string matches the given value or not.