What is the best way to verify and handle an undefined array in order to display an error message?

Currently working on a ternary statement to display the certification of a movie (e.g. PG, R...).

If the array length is zero or undefined, I'm aiming to output an error message stating "No Certification Available For This Movie". While I have successfully managed to display the error message when the array is empty, I'm facing a challenge in handling undefined arrays. I constantly receive a console error reading

TypeError: Cannot read properties of undefined (reading 'release_dates')
.

Below is my attempt so far:

 const movieRating = rating.results;

 //Fetching US ratings
 const findUSRating = movieRating.find((item) => item.iso_3166_1 === 'US');

//Applying filter to remove null certifications from US ratings.
 const array = findUSRating.release_dates.filter((item) => item.certification);

//Ternary statement & my current hurdle.
console.log(array.length > 0 || array[index] !== undefined ? array[0].certification : `No Certification Available For This Movie`);

Answer №1

Issue: Cannot access properties of undefined (reading 'release_dates')

It seems that the variable findUSRating is not defined. This error occurs because the object within movieRating does not have the property with iso_3166_1 set to US.

For instance, consider the following:

const movieRating = [{
    iso_3166_1: 'Canada'
}, {
    iso_3166_1: 'Australia'
}];

 //Finding US ratings
 const findUSRating = movieRating.find((item) => item.iso_3166_1 === 'US');

//Filtering US ratings to remove null certifications. 
 const array = findUSRating.release_dates.filter((item) => item.certification);
 
 console.log(array);

To resolve this issue, you can utilize the Optional chaining (?.) operator.

Check out the Live Demo below:

const movieRating = [{
    iso_3166_1: 'US',
  release_dates: [{
    certification: 'A'
  }]
}, {
    iso_3166_1: 'Australia',
  release_dates: [{
    certification: 'B'
  }]
}];

 //Finding US ratings
 const findUSRating = movieRating?.find((item) => item.iso_3166_1 === 'US');

//Filtering US ratings to remove null certifications. 
 const array = findUSRating?.release_dates.filter((item) => item.certification);
 
console.log(array?.length > 0 || array !== undefined ? array[0].certification : `No certification Available For This Movie`);

Answer №2

The problem lies within this specific line of code

 const array = findUSRating.release_dates.filter((item) => item.certification);

Attempting to access findUSRating.release_dates when findUSRating is undefined.

A potential solution is to add a conditional check before accessing the property

const array = (findUSRating && findUSRating.release_dates) ? findUSRating.release_dates.filter((item) => item.certification) : [];

You may encounter a similar issue if movieRating is undefined or not an array. It's advisable to implement a similar approach to avoid errors

 const findUSRating = (movieRating && movieRating.length) ? movieRating.find((item) => item.iso_3166_1 === 'US') : [];

Answer №3

By utilizing try catch, you can effectively handle any runtime errors that may occur.

try {
  const movieRating = rating.results;

   //Locating US ratings
   const findUSRating = movieRating.find((item) => item.iso_3166_1 === 'US');

  //Filtering US ratings to eliminate null certifications.
   const array = findUSRating.release_dates.filter((item) => item.certification);

  //Ternary statement & where I am struggling.
  console.log(array.length > 0 || array[index] !== undefined ?    array[0].certification : `No certification Available For This Movie`);
} catch(err) 
{ 
  console.log("Error") 
}

Answer №4

@Quinn. Feel free to modify the code snippet below.

//Removing null certifications from US ratings. 
const array = findUSRating.release_dates.filter((item) => item.certification);

const array = findUSRating ? findUSRating.release_dates.filter((item) => item.certification) : [];

     or

const array = findUSRating?.release_dates?.filter((item) => item.certification);

Hope this adjustment is useful.

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