Array problems with localStorage

Question

Array problems with localStorage

Navigation

#1 by (4 votes)
#2 by (0 votes)

2

var x2=x1+c;
var x2i=x2.toFixed(2);
var y2=y1+ci;
var y2i=y2.toFixed(2);//52.23
var arrayX2=new Array();
arrayX2.push(y2i);
var arrayX2=new Array();
arrayX2.push(y2i);

By entering the element the array alert gives me 1 (CORRECT)

alert(arrayX2.length);

localStorage.setItem(x2i,arrayX2);

Now comes the problem, putting the array in the variable of the localStorage then I look at the size of the array of the localStorage variable and it gives me 5, and it's like I put the character number to character as if it were '5' , then '2' then '.' then a '2' and then a '3' (52.23)

alert(localStorage.getItem(x2i).length));

I've tried to do this:

localStorage.setItem(x2i,JSON.stringify(arrayX2));

But now the lenght gives me 9;

I do not know how to put the array correctly in the localStorage variable.

If someone can tell me how to solve this, I would appreciate it. Greetings.

javascript

asked by Andrés 26.09.2018 в 11:33

source

2 answers

R: change the colors of a DCA Reading of strings in C

score 4 · Answer 1

The localStorage does not store objects. If you put:

var miarray = [1,2,3];
localStorage.setItem('primer_array', miarray);

Implicitly happens

localStorage.setItem('primer_array', miarray.toString());

That stores the string 1,2,3

For the same reason, it is correct to insert it using JSON.stringify as samples in your second attempt:

localStorage.setItem(x2i,JSON.stringify(arrayX2));

So to get it back "rehydrated", you have to retrieve it as

var array_guardado = JSON.parse(localStorage.getItem(x21));

On the other hand, in your logic you say that the element you want to store is defined in the snippet

var x2=x1+c;
var x2i=x2.toFixed(2);
var y2=y1+ci;
var y2i=y2.toFixed(2);//52.23
var arrayX2=new Array(); // arrayX2 = []
arrayX2.push(y2i); // arrayX2 = ["53.23"]
var arrayX2=new Array(); // arrayX2 = []
arrayX2.push(y2i); // arrayX2 = ["53.23"]

What are the last two lines for?

Why do you redefine arrayX2 as an empty array? Redefining an already declared variable is an antipattern. Think:

var x = 'hola';
console.log(x);
// imprime 'hola';
var x;
console.log(x);
// sigue imprimiento 'hola' cuando debiera imprimir 'undefined'

Will not you want the final array to contain x2i e y2i Finally, it is not recommended to use the Array constructor new Array() but the literal [] to declare a new array.

Think that

var miarray = [] when interpreted in the browser engine, call ARRAY_INIT .

Whereas var miarray = new Array() calls new (constructor) Array (identifier) and then goes through the chain of prototypes going up the stack until reaching window.Array , and on that prototype calls the method apply .

This happens because on the way you could have added methods to the prototype Array (it's a bad idea, but it's done) or worse, having overwritten the constructor for example:

Array =  function() {
  this.nombre = 'array';
};

That would generate that

var miaarray = new Array();

Out an object with property nombre and without methods like push , shift , unshift , pop , etc. You would have destroyed the constructor. And yet, after doing that

var miarray = [];

It would remain intact and functioning.

score 0 · Answer 2

The problem I see in your code is in how you use the "localStorage", to tell you that the "localStorage" and "sessionStorage" can only save string, not another type of data, so you can not store an "array" directly, otherwise you have to transform it into a string and from here save it. The JSON.stringify () method is for JSON not for arrays.

I'll show you an example:

var array = [12.45, 55.55]

//array.toString() => "12.45,55.55"

localStorage.setItem('test',array.toString())

var value = localStorage.getItem('test')

var data = value.split(',')

//data => ["12.45","55.55"]

Remember to transform the Strings to Number, in this case Float.

I hope it helped you.

Greetings