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Скачать или смотреть How to Remove Elements from an Array of Objects Based on Another Array in JavaScript

  • vlogize
  • 2025-09-20
  • 0
How to Remove Elements from an Array of Objects Based on Another Array in JavaScript
JS remove elements from array of objects if element value is in another arrayjavascriptarrays
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Описание к видео How to Remove Elements from an Array of Objects Based on Another Array in JavaScript

Learn how to effortlessly exclude specific elements from an array of objects in JavaScript based on values found in another array.
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This video is based on the question https://stackoverflow.com/q/62657719/ asked by the user 'awariat' ( https://stackoverflow.com/u/4086818/ ) and on the answer https://stackoverflow.com/a/62657848/ provided by the user 'Sudhir Ojha' ( https://stackoverflow.com/u/6682274/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: JS remove elements from array of objects if element value is in another array

Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l...
The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license.

If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com.
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How to Remove Elements from an Array of Objects Based on Another Array in JavaScript

In the world of coding, managing and manipulating arrays is a common task that developers face on a regular basis—especially in JavaScript. A frequent scenario is when you want to filter out certain elements from an array of arrays (or objects) if the first value is found in another array. In this post, we will tackle the problem of excluding specific items from an array based on values from a second array.

Understanding the Problem

Imagine you have the following two arrays:

Main array (oldArray):

[[See Video to Reveal this Text or Code Snippet]]

Exclusion array (toExcludeArray):

[[See Video to Reveal this Text or Code Snippet]]

Your goal is to create a new array that excludes all elements from the oldArray whose first value is listed in the toExcludeArray. The desired result would look like this:

[[See Video to Reveal this Text or Code Snippet]]

The Solution

To achieve this, JavaScript's filter() method comes in handy. This method creates a new array with all elements that pass the test implemented by the provided function. Additionally, includes() can be used to check if an element exists in the second array. Below, we'll break down the solution step by step.

Step-by-Step Guide

Initialize the Arrays:
Start by defining your original array and the array with elements you want to exclude.

[[See Video to Reveal this Text or Code Snippet]]

Filter the Main Array:
Use the filter() method to iterate over each element in oldArray. Inside the filter function:

Use the includes() method on toExcludeArray.

Access the first element of each sub-array using el[0] and check if it’s not in toExcludeArray.

[[See Video to Reveal this Text or Code Snippet]]

Output the Result:
Finally, log or return the filtered array to see the result.

[[See Video to Reveal this Text or Code Snippet]]

Full Code Example

Here is the complete code that incorporates all the steps above:

[[See Video to Reveal this Text or Code Snippet]]

Conclusion

By using the filter() method in combination with includes(), you can efficiently exclude specific elements from an array based on another array's contents. This pattern is extremely useful in many data manipulation tasks in JavaScript, so understanding how to implement it will serve you well in your coding journey.

Now you can confidently filter your arrays and manage your data as needed!

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