Home > Web Front-end > JS Tutorial > How Can I Accurately View the Current State of a JavaScript Object in the Browser Console?

How Can I Accurately View the Current State of a JavaScript Object in the Browser Console?

Mary-Kate Olsen
Release: 2024-12-07 19:21:13
Original
617 people have browsed it

How Can I Accurately View the Current State of a JavaScript Object in the Browser Console?

Getting the Current State of an Object with Console.log

In browsers like Safari, console.log presents the object's last state, not its state at the time of calling. This can be inconvenient when debugging or analyzing object changes.

Avoiding Object Cloning

Conventional methods involve cloning the object before using console.log to retrieve the state at a specific point. However, this approach is inefficient and can be avoided.

Introducing console.dir()

Instead of cloning, consider using console.dir(). Unlike console.log(), console.dir() displays a directory of the object's properties at the time of calling. This provides a precise representation of the object's current state.

Example:

const test = { a: true };
console.dir(test); // { a: true }
test.a = false;
console.dir(test); // { a: false }
Copy after login

JSON Parsing for Object Exploration

Another technique is to convert the object to JSON and then parse it back into an object. This creates a new object with the same properties as the original, allowing for exploration within the console.

console.log(JSON.parse(JSON.stringify(obj))); // Browsable object representation
Copy after login

By utilizing console.dir() or JSON parsing, developers can effectively obtain the current state of an object in the console without the need for object cloning or external tools.

The above is the detailed content of How Can I Accurately View the Current State of a JavaScript Object in the Browser Console?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template