How to mass unblock your Twitter block list

I had several thousand (?) accounts blocked because of twitter advertising.

Thanks to this I could no longer tell whether or not twitter “bugs” were blocking people without my consent. So I decided to unblock most of the accounts (most of them companies). The problem though, is this was taking forever to do manually. So I came up with a way to do it automatically (for the most part).

Step 1 — Grab the block list

Go here:

Then right-click and inspect an element in the block list.

In Brave’s HTML inspector, right-click on the very first element just above the list of <div></div> elements, the one that represents the <div> containing the list:

Select "Store as global variable" from the menu.

This should bring up the console and store the list in a temp1 variable.

Step 2 — Create the unblock function

Paste the following into the console to create unblockFn and press enter:

var unblockFn = function () {
  Array.from(temp1.children).forEach((e) => {
    let button = e.firstElementChild.firstElementChild.firstElementChild.children[1].firstElementChild.children[1].firstElementChild;
    if (button.classList.contains('r-jwli3a')) {

This code will need slight adjustments if twitter updates this page, but the idea will remain the same.

Step 3 — Run the unblock function while scrolling the page

Type unblockFn() and hit Enter.

Everything you see should be unblocked.

If you have a huge block list, you will need to scroll a bit and then run the function again. Scroll some more, run it again. Repeat until all accounts are unblocked.

Then you can re-block whatever accounts you want.

2 thoughts on “How to mass unblock your Twitter block list”

  1. Is it possible to only mass unblock people who follow a certain account? I’m on stan twitter, and I do chain blocks, but I want to unblock only the people who follow a certain celebrity. Is that possible?

    • I don’t know what stan twitter is, and yes it’s probably possible but you’d have to figure out the code for it (which would probably be considerably more complex).

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