Home Carbon Ads JSON.parse() throws “unexpected token” error for valid JSON

JSON.parse() throws “unexpected token” error for valid JSON

Although your JSON string is valid JSON.parse is likely to throw error. This is because JSON.parse cannot parse some special characters that are \n, \t, \r and \f.

You need to escape these special characters for before passing JSON string to JSON.parse function.

Here is a function that takes a JSON string and escapes the special characters:

        function escapeSpecialChars(jsonString) {

            return jsonString.replace(/\n/g, "\\n")
                .replace(/\r/g, "\\r")
                .replace(/\t/g, "\\t")
                .replace(/\f/g, "\\f");

        }
Narayan Prusty
Best Photo Straightening Online ToolGenerate Unique Number in JavaScript
Comments: 3
  1. Tyler Collier
    4 years ago

    Don’t forget about \", as mentioned on this stackoverflow answer.

    ReplyCancel
  2. Gajapathi
    5 years ago

    Thanks for this lovely post

    ReplyCancel
  3. Gajapathi
    5 years ago

    Very Helpful, after wasting my 2 hrs of time, i found this, and ended my serach for JSON.parse error

    ReplyCancel

Leave a Reply

To create code blocks or other preformatted text, indent by four spaces:

    This will be displayed in a monospaced font. The first four
    spaces will be stripped off, but all other whitespace
    will be preserved.
    
    Markdown is turned off in code blocks:
     [This is not a link](http://example.com)

To create not a block, but an inline code span, use backticks:

Here is some inline `code`.

Narayan Prusty

I am a full-stack web developer. I specialize in Blockchain and JavaScript. This is my personal blog where I write about programming and technologies that I learn and feel interesting to share.

5 years ago 3 Comments Web Development
Share this
Related Articles