Week 3 (Beginning Jan 24th) Learning Week Cycle 3 - emcsquared2/2022-Learning GitHub Wiki

Week 3 (Beginning Jan 24th) Learning Week Cycle 4

GCSE Computer Science

The Use of SQL to search for Data

Takeaway learning

SQL is used to create, delete, modify and manipulate records in a database The basic commands include:

• SELECT which fields to be returned. * Can be used for all fields.
• FROM which table
• WHERE records meet a condition

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MIMO mobile app

#####Methods in Classes

Takeaway learning

Adding a method in a class is like creating a regular function without the word "function"

class VirtualPet{
 constructor(name) {
   this.name = name
 }

 eat() {
 console.log("nom nom");
 }
}

//create a new object
var pet = new VirtualPet("Tom");
pet.eat();
//OUTPUT "nom nom"

#####Instances

Takeaway learning

Every time we create an object from a class we are creating an "instance" of that class.

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Professor Steve

Playing with Objects

Takeaway learning

Objects are lists like arrays except instead of being numbered they have labels. Revision of object properties

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Codewars Exercise (Level 8)

In this simple assignment you are given a number and have to make it negative. But maybe the number is already negative?

My Solution

const makeNegative = (num) => num>0 ? -num: num;

I did see another solution but I'm not sure about it because it returns a negative 0 value. Still it is a good learning point around the Absolute value Math function.

function makeNegative(num) {
    return -Math.abs(num);
  }

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JS Practice

Given a string of digits, output all the contiguous substrings of length n in that string in the order that they appear.

My solution

The solution required a particular structure to this answer in the form of a Class. The Class creates an object with each slice of numbers returned as numbers within an array. To start with I just worked out how to simply do this by returning a string in a simple function...

str = "123456"
n = 3;

for (let i=0; i<=str.length-n; i++){
 return str.substring(i,n+i) 
};

That worked but wasn't what would pass the tests which also required some constraints on the inputs. I had to go to the community to get a solution to help. Here is my solution in the end:

class Series {
  constructor(series) {
    this.series = series;
  }

  slices(sliceLength) {
    if (!this.series) throw new Error('series cannot be empty');
    if (sliceLength > this.series.length) throw new Error ('slice length cannot be greater than series length');
    if (sliceLength === 0) throw new Error ('slice length cannot be zero')
    if (sliceLength < 0) throw new Error ('slice length cannot be negative')

    const slices = []
      for (let i=0; i<=this.series.length-sliceLength; i++){
        slices.push((this.series.substring(i,sliceLength+i)).split('').map(Number))
      };
    return slices;
  };
};

const series1 = new Series('12345')
console.log(series1)
console.log(series1.slices(3))

//OUTPUT [ [ 1, 2, 3 ], [ 2, 3, 4 ], [ 3, 4, 5 ] ]

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Udacity (Object Orientated JavaScript - Free Course)

Object.keys() and Object.values() Methods

Takeaway learning

So to start with I wanted to see if I could freestyle using a for loop and get the keys and values:

const mealPlanner = {
breakfast: "cereal",
lunch: "sandwich",
dinner:"fish and chips"
};

const keyArray = [];

for (key in mealPlanner){
  keyArray.push(key);
};

console.log(keyArray);
//OUPUT [ 'breakfast', 'lunch', 'dinner' ]

const valueArray = [];

for (key in mealPlanner){
  valueArray.push(mealPlanner[key])
};

console.log(valueArray);
//OUTPUT [ 'cereal', 'sandwich', 'fish and chips' ]

I'm really pleased that I can write these bits of code now with no trouble at all and correct myself as I go. The equivalent using the Object Method is...

console.log(Object.keys(mealPlanner));
console.log(Object.values(mealPlanner));

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Code Signal

Given a divisor and a bound, find the largest integer N such that: N is divisible by divisor. N is less than or equal to bound. N is greater than 0.

My Solutions

function solution(divisor, bound) {
  let int = 0;

  for (let i=bound; i>1; i--){
   if (i%divisor===0){
       int = i;
       break;
   };   
  };
  return int;
};

I've realised I have done a very convoluted way of solving this!! A much simpler version...

function solution(divisor, bound) {
return bound - (bound % divisor);
}