How to get the first result of rows with duplicate dates?

How to get the first result of rows with duplicate dates?

Hi,
I have a table that contains the following fields:
  1. An ID
  2. An event
  3. A date-time for each event
It looks like this:
ID
Event
Event Date
1122334
Event 1
2016-10-01 10:00:00
1122334
Event 2
2016-10-01 10:00:00
1122335
Event 1
2018-06-23 11:43:22
1122336
Event 2
2019-02-02 08:04:13
1122336
Event 1
2019-02-03 06:07:17
1122336
Event 2
2019-02-06 13:37:44


My goal is to find the first event for a given ID, based on the date so that I have a list of unique IDs and their corresponding events and dates like so:
ID
Event
Event Date
1122334
Event 1
2016-10-01 10:00:00
1122335
Event 2
2018-06-23 11:43:22
1122336
Event 2
2019-02-02 08:04:13

The following query gets me close, but due to the fact that there are duplicate values in the date-time column (to the second), I still get duplicate entries:

  2. SELECT
  3. "Table_1"."Event"
  4. FROM (SELECT
  5. "Table_1"."ID" as "ID",
  6. MIN("Table_1"."Event Date") AS "Event Date"
  7. FROM  "Table_1" 
  8. GROUP BY  "ID" 
  9. ) AS  "First Action"
  10. LEFT JOIN "Table_1" ON "Table_1"."ID"  = "First Action"."ID"
  11. AND "Table_1"."Event Date"  = "First Action"."Event Date" 
This results in:

ID
Event
Event Date
1122334
Event 1
2016-10-01 10:00:00
1122334
Event 2
2016-10-01 10:00:00
1122335
Event 1
2018-06-23 11:43:22
1122336
Event 2
2019-02-02 08:04:13

For record 1122334 there are still two entries because the date is duplicated.  My logical solution would be to select only the first row, but I am not sure how to do this.

I can't change the data in the underlying table, and I tried various implementations of DISTINCT and COUNT(*) that didn't yield good results.

I would appreciate any assistance in getting the correct SQL query to get the results.