Gmail Time Ago Feature

This would be a great feature for me too, but AFAIK it doesn't exist.

My solution to cope with such a problem is to use the service called Boomerang. What I do is that I empty my inbox as much as possible multiple times a day (at least once). For any email that would take me longer than 5 minutes to write a response to, I delay it: I ask for it to be "boomeranged" when I'll have time to deal with it (perhaps tomorrow morning very early, so that I handle it first thing in the morning, as it appears at the top of my inbox).

Often enough, I find myself delaying emails that I know don't need a quick answer (for e.g. someone invites you to a party in 2 months...), so I used to procrastinate and end up not answering. But if I boomerang it right when I'll have to reply (say a week before the party), I force myself to answer the email at the best time for me and the recipient.

Note: I'm not affiliated with Boomerang, just a happy user.

It is possible to specify time range with accuracy up to a second, because the search operators after:, before:, newer:, older: accept Unix timestamps. Using a tool such as Epoch Converter, you can find out that

• 2014-10-04 at 15:00 (using GMT for example) has timestamp 1412434800
• Add two hours (7200 seconds), or use the same tool again: 1412442000

Searching for after:1412434800 before:1412442000 will return the messages within this 2-hour range.

Since converting to Unix time looks like a chore, I made a bookmarklet that makes this substitution in place.

For example, after typing

after:2015/07/26 10:00 before:2015/07/27 11:30

in the search box (which would not work as is), invoke the bookmarklet and the string will be replaced by

after:1437919200 before:1438011000

(The bookmarklet interprets the given timestamps in your local time.)

The page I linked has both the source and a draggable link with the bookmarklet. For completeness, the source code is also posted below.

inp = document.querySelectorAll('input');
for (i in inp) {
  if (/(after|before|newer|older):/.test(inp[i].value)) {
    str = inp[i].value;
    times = str.match(/\d[\d\/: ]*\d/g);
    for (j in times) {
      if (Date.parse(times[j])) {
        str = str.replace(times[j], Date.parse(times[j])/1000);
      }
    }
    inp[i].value = str;
  } 
}