How to select unique records by SQL

There are 4 methods you can use:

1. DISTINCT
2. GROUP BY
3. Subquery
4. Common Table Expression (CTE) with ROW_NUMBER()

Consider the following sample TABLE with test data:

/** Create test table */
CREATE TEMPORARY TABLE dupes(word text, num int, id int);

/** Add test data with duplicates */
INSERT INTO dupes(word, num, id)
VALUES ('aaa', 100, 1)
      ,('bbb', 200, 2)
      ,('ccc', 300, 3)
      ,('bbb', 400, 4)
      ,('bbb', 200, 5)     -- duplicate
      ,('ccc', 300, 6)     -- duplicate
      ,('ddd', 400, 7)
      ,('bbb', 400, 8)     -- duplicate
      ,('aaa', 100, 9)     -- duplicate
      ,('ccc', 300, 10);   -- duplicate

Option 1: SELECT DISTINCT

This is the most simple and straight forward, but also the most limited way:

SELECT DISTINCT word, num 
FROM    dupes
ORDER BY word, num;

/*
word|num|
----|---|
aaa |100|
bbb |200|
bbb |400|
ccc |300|
ddd |400|
*/

Option 2: GROUP BY

Grouping allows you to add aggregated data, like the min(id), max(id), count(*), etc:

SELECT  word, num, min(id), max(id), count(*)
FROM    dupes
GROUP BY word, num
ORDER BY word, num;

/*
word|num|min|max|count|
----|---|---|---|-----|
aaa |100|  1|  9|    2|
bbb |200|  2|  5|    2|
bbb |400|  4|  8|    2|
ccc |300|  3| 10|    3|
ddd |400|  7|  7|    1|
*/

Option 3: Subquery

Using a subquery, you can first identify the duplicate rows to ignore, and then filter them out in the outer query with the WHERE NOT IN (subquery) construct:

/** Find the higher id values of duplicates, distinct only added for clarity */
    SELECT  distinct d2.id
    FROM    dupes d1
        INNER JOIN dupes d2 ON d2.word=d1.word AND d2.num=d1.num
    WHERE d2.id > d1.id

/*
id|
--|
 5|
 6|
 8|
 9|
10|
*/

/** Use the previous query in a subquery to exclude the dupliates with higher id values */
SELECT  *
FROM    dupes
WHERE   id NOT IN (
    SELECT  d2.id
    FROM    dupes d1
        INNER JOIN dupes d2 ON d2.word=d1.word AND d2.num=d1.num
    WHERE d2.id > d1.id
)
ORDER BY word, num;

/*
word|num|id|
----|---|--|
aaa |100| 1|
bbb |200| 2|
bbb |400| 4|
ccc |300| 3|
ddd |400| 7|
*/

Option 4: Common Table Expression with ROW_NUMBER()

In the Common Table Expression (CTE), select the ROW_NUMBER(), partitioned by the group column and ordered in the desired order. Then SELECT only the records that have ROW_NUMBER() = 1:

WITH CTE AS (
    SELECT  *
           ,row_number() OVER(PARTITION BY word, num ORDER BY id) AS row_num
    FROM    dupes
)
SELECT  word, num, id 
FROM    cte
WHERE   row_num = 1
ORDER BY word, num;

/*
word|num|id|
----|---|--|
aaa |100| 1|
bbb |200| 2|
bbb |400| 4|
ccc |300| 3|
ddd |400| 7|
*/

I am not sure if the accepted answer works. It does not work on postgres 12 at least. DISTINCT keyword is supposed to be applied to all the columns in the select query and not just to the column next to which DISTINCT keyword is written. So, basically, it means that every row returned in the result will be unique in terms of the combination of the select query columns. In OP's question, the below two result rows are already distinct, as they have different values for column1 and column 3.

1 item1 data1
2 item1 data2

Now, to answer the question, atleast in postgres, there is a DISTINCT ON keyword. This will achieve what the OP requires.

select DISTINCT ON(column2) column1, column3 from Table1;