Google Chrome 扩展程序：如何获取打开标签的所有网址？

Question

I'm using chome.tabs.query to get all detail from all tabs are opening in Chrome.

chrome.tabs.query({ currentWindow: true }, function (tabs) {
  tabs.forEach(function (tab) {
    console.log(tab);
  });
});

However, in the results I received, I could not find the URL of the tab page. The information is the same as

{
  active: false
  audible: false
  autoDiscardable: true
  discarded: false
  groupId: -1
  height: 748
  highlighted: false
  id: 134
  incognito: false
  index: 1
  mutedInfo: {muted: false}
  pinned: false
  selected: false
  status: "complete"
  width: 1440
  windowId: 4
}

Is there a way that allows me to get all the information of the open tabs according to its URL?

My Chrome version: 89.04

Answer 1

Have you add "tabs" permission in manifest.json?

From https://developer.chrome.com/docs/extensions/reference/tabs/#type-Tab

URL property is only present if the extension's manifest includes the "tabs" permission and may be an empty string if the tab has not yet committed.

You can use several methods of tabs API without "tabs" permission but if you want to retreive this kind of information from a query you have to include tabs permission in manifest.