What is the proper way to search for a specific string within a JavaScript array during an iteration?

Question

What is the proper way to search for a specific string within a JavaScript array during an iteration?

I am working with an array that is continuously updated with new string elements every 2 seconds. The code snippet below showcases how the updates are processed:

//tick world
setInterval(function(){
  doTradeUpdate();
},5000);

function doTradeUpdate(){
  var randyManu = Math.floor(Math.random() * 10);
  switch(randyManu){
  case 0:
  //new duro mine
  countPush(manufacture,"duro-mine");
  break;

  case 1:
  //new e-plant
    countPush(manufacture,"e-plant");
    break;
//etc
}

function countPush(arrayz,addable){
  console.log(arrayz);
  console.log("Attempting to add: " + addable);
  if(arrayz.length == 0){
    arrayz.push(addable);
  }
  else{
    if (arrayz.indexOf(addable)  > 0){
       console.log("FOUND");
    }
    else{
        console.log("NOT FOUND");
        arrayz.push(addable);
    }
  }
}

However, an issue arises where sometimes the code outputs "FOUND" and other times "NOT FOUND" for the same array element, like "e-plant". Consequently, multiple duplicate entries can exist within the array. What could be causing this inconsistency in matching elements?

This script is solely responsible for handling the array updates.

Thank you!

G

javascript arrays

Answer 1

Answer №1

implement

if (arrayList.indexOf(newValue) !== -1)
{
  console.log("VALUE FOUND");
}
else
{
  console.log("VALUE NOT FOUND");
  arrayList.push(newValue);
}

For a more optimized approach, consider using (!== -1) instead of (> 0).
It's worth noting that if the element is in position 0 it will return "VALUE NOT FOUND". You can enhance the code further as shown below

function updateArray(arrayList, newValue)
{
console.log(arrayList);
console.log("Attempting to add: " + newValue);
if(Array.isArray(arrayList))
{
  if(arrayList.indexOf(newValue) !== -1) {
    console.log('Element exists in the list');
  }
  else {
    arrayList.push(newValue);
  }
}    
}

There are more efficient ways to check if a variable is an array, this is just a simple example!
Also, consider storing the array in a common outer namespace instead of passing it every time if it remains the same in both function calls.

Answer 2

implement

if (arrayList.indexOf(newValue) !== -1)
{
  console.log("VALUE FOUND");
}
else
{
  console.log("VALUE NOT FOUND");
  arrayList.push(newValue);
}

For a more optimized approach, consider using (!== -1) instead of (> 0).
It's worth noting that if the element is in position 0 it will return "VALUE NOT FOUND". You can enhance the code further as shown below

function updateArray(arrayList, newValue)
{
console.log(arrayList);
console.log("Attempting to add: " + newValue);
if(Array.isArray(arrayList))
{
  if(arrayList.indexOf(newValue) !== -1) {
    console.log('Element exists in the list');
  }
  else {
    arrayList.push(newValue);
  }
}    
}

There are more efficient ways to check if a variable is an array, this is just a simple example!
Also, consider storing the array in a common outer namespace instead of passing it every time if it remains the same in both function calls.

Answer 3

Answer №2

Give this a shot

 let targetIndex = arrayz.indexOf(addable);
if (targetIndex > -1) 
{
    console.log("Item is FOUND");
} 
else 
{
    console.log("Item is NOT FOUND");
    arrayz.push(addable);
}

Answer 4

Give this a shot

 let targetIndex = arrayz.indexOf(addable);
if (targetIndex > -1) 
{
    console.log("Item is FOUND");
} 
else 
{
    console.log("Item is NOT FOUND");
    arrayz.push(addable);
}

Answer 5

Answer №3

In the event that the element cannot be located, the function will return -1, causing your validation to fail as it mistakenly checks if the index is greater than 0.

To rectify this, ensure that you check:

(arrayz.indexOf(addable) !== -1)

Answer 6

In the event that the element cannot be located, the function will return -1, causing your validation to fail as it mistakenly checks if the index is greater than 0.

To rectify this, ensure that you check:

(arrayz.indexOf(addable) !== -1)

Answer 7

Answer №4

indexOf() will return -1 if the element is not found. If it returns 0, it means the element is found in the first position of the array, which may not be the desired outcome.

It is recommended to modify the code from

if (arrayz.indexOf(addable) > 0)

to

if (arrayz.indexOf(addable) > -1)

Alternatively, you can combine your conditions to simplify the logic:

function countPush(arrayz, addable) {
    console.log(arrayz);
       console.log("Attempting to add: " + addable);

       // Same as nested if statements, due to short-circuiting
       if (arrayz.length === 0 || arrayz.indexOf(addable) === -1) {
           console.log("NOT FOUND");
           arrayz.push(addable);
       } else {
           console.log("FOUND");
       }
}

Answer 8

indexOf() will return -1 if the element is not found. If it returns 0, it means the element is found in the first position of the array, which may not be the desired outcome.

It is recommended to modify the code from

if (arrayz.indexOf(addable) > 0)

to

if (arrayz.indexOf(addable) > -1)

Alternatively, you can combine your conditions to simplify the logic:

function countPush(arrayz, addable) {
    console.log(arrayz);
       console.log("Attempting to add: " + addable);

       // Same as nested if statements, due to short-circuiting
       if (arrayz.length === 0 || arrayz.indexOf(addable) === -1) {
           console.log("NOT FOUND");
           arrayz.push(addable);
       } else {
           console.log("FOUND");
       }
}

Answer 9

Answer №5

give this a shot

(arrayz.indexOf(addable) !== -1)?(console.log("ITEM FOUND")):((console.log("NOT FOUND")) || (arrayz.push(addable))); console.log(arrayz);

Answer 10

give this a shot

(arrayz.indexOf(addable) !== -1)?(console.log("ITEM FOUND")):((console.log("NOT FOUND")) || (arrayz.push(addable))); console.log(arrayz);

What is the proper way to search for a specific string within a JavaScript array during an iteration?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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