Search for a vacant string array

Question

Search for a vacant string array

Trying to determine if a string array is empty, I attempted the following:

var array = ['',''];
//Other code
if (array == ['',''])
    //Do stuff

To my surprise, I discovered that

['',''] == ['','']

returns false. Can someone explain why this happens? What is the best approach to check for an empty string array without having to compare each item individually?

javascript arrays

Answer 1

Answer №1

In addition to considering the use of Array#toString, I recommend utilizing Array#join with an empty string as a separator before testing the outcome. This method is advantageous because it can handle any number of elements within the array.

var check = array.join('') ? 'not all empty strings' : 'all empty strings';

Answer 2

In addition to considering the use of Array#toString, I recommend utilizing Array#join with an empty string as a separator before testing the outcome. This method is advantageous because it can handle any number of elements within the array.

var check = array.join('') ? 'not all empty strings' : 'all empty strings';

Answer 3

Answer №2

['', ''] == ['', ''] yields false because arrays in JavaScript are treated as objects with reference semantics. When comparing two arrays, the comparison is actually between their unique reference IDs, which will be distinct for each array. Therefore, even if the arrays appear identical, they are different references.

To verify that an array only contains empty strings, you can utilize Array.prototype.every like this:

myArray = ['']
console.log(myArray.every(el => el === '')) // true
myArray = []
console.log(myArray.every(el => el === '')) // true
myArray = ['test']
console.log(myArray.every(el => el === '')) // false

If you are working in an environment without ES6 support, you can replace el => el === '' with

function(el) { return el === '' }

.

Answer 4

['', ''] == ['', ''] yields false because arrays in JavaScript are treated as objects with reference semantics. When comparing two arrays, the comparison is actually between their unique reference IDs, which will be distinct for each array. Therefore, even if the arrays appear identical, they are different references.

To verify that an array only contains empty strings, you can utilize Array.prototype.every like this:

myArray = ['']
console.log(myArray.every(el => el === '')) // true
myArray = []
console.log(myArray.every(el => el === '')) // true
myArray = ['test']
console.log(myArray.every(el => el === '')) // false

If you are working in an environment without ES6 support, you can replace el => el === '' with

function(el) { return el === '' }

.

Answer 5

Answer №3

Here is a solution for you:

const newArray = ["item1", "item2", "item3"].reduce(function (previousValue, currentValue) {
  return previousValue + currentValue;
});

if(newArray === "") {
console.log('The array is empty');
}

Answer 6

Here is a solution for you:

const newArray = ["item1", "item2", "item3"].reduce(function (previousValue, currentValue) {
  return previousValue + currentValue;
});

if(newArray === "") {
console.log('The array is empty');
}

Answer 7

Answer №4

Another example that should work:

let array = ["hello"];
console.log(String.valueOf(array[0]) === String.valueOf('hello'));

Answer 8

Another example that should work:

let array = ["hello"];
console.log(String.valueOf(array[0]) === String.valueOf('hello'));

Answer 9

Answer №5

Following Nina's response, this approach should also yield the desired outcome!

 let lst1 = ['']
 const flag1 = !!lst1.join(''); // false
 console.log(flag1)
            
 let lst2 = ['Text']
 const flag2 = !!lst2.join(''); // true
 console.log(flag2)

Answer 10

Following Nina's response, this approach should also yield the desired outcome!

 let lst1 = ['']
 const flag1 = !!lst1.join(''); // false
 console.log(flag1)
            
 let lst2 = ['Text']
 const flag2 = !!lst2.join(''); // true
 console.log(flag2)

Search for a vacant string array

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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