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Vistasforeach Statements with a function
ok I am trying to answer a question and learning to code. regarding an array and using foreach statements the function must remain as this "function animalNames" must stay somewhere but apparently, I am doing something wrong because I get it returned as undefined. even through it produces the correct array back could someone look at it and let me know what i have done wrong.
attached is a picture of the code and array and question that i answered. this is how i wrote my function.
const displayNames = [];
zooAnimals.forEach(function animalNames(element){
var display = "name: " + element.animal_name + ", " + "scientific: " + element.scientific_name
displayNames.push(display);
})
console.log(displayNames); again i get the correct array back and the data looks correct...but animalNames comes back as undefined...i cannot remove this portion i am to keep it there but i do not know what to do with it.
2 Respuestas
Responde la pregunta0
I believe you are following this challenge. In this case, make sure to read the instructions properly:
The zoos want to display both the scientific name and the animal name in front of the habitats. Use animalNames to populate and return the displayNames array with only the animal name and scientific name of each animal. displayNames will be an array of strings, and each string should follow this pattern: "name: {name}, scientific: {scientific name}"
So from the instructions you are expected to create a function animalNames() that creates an array named displayNames and returns this array from within your function animalNames(). The array displayNames contains strings of the sturcture name: animal_name, scientific: scientific_name.
// you could set zooAnimals as a required parameter in your function
// but as it is defined in the same file you can access it directly as well
function animalNames() {
const displayNames = [];
// you have to do the forEach on zooAnimals here
zooAnimals.forEach(animal => displayNames.push(`name: ${animal.animal_name}, scientific: ${animal.scientific_name}`))
// return an array
return displayNames;
}
The way you did it, the function animalsName() is an anonymous function that only exists within the zooAnimals.forEach() call. Therefore, it is undefined.
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try this, it defines a function animalNames as separate function
const displayNames = [];
const zooAnimals = [
{animal_name:"Jackel",scientific_name:"Canine"}
]
function animalNames({animal_name, scientific_name}){
return `name: ${animal_name}, scientific: ${scientific_name}`
}
zooAnimals.forEach((element)=>{
displayNames.push(animalNames(element));
})
console.log(displayNames);