How to select unique records by SQL

When I perform SELECT * FROM table I got results like below:

1 item1 data1
2 item1 data2
3 item2 data3
4 item3 data4

A开发者_C百科s you can see, there are dup records from column2 (item1 are dupped). So how could I just get result like this:

1 item1 data1
2 item2 data3
3 item3 data4

Only one record are returned from the duplicate, along with the rest of the unique records.

With the distinct keyword with single and multiple column names, you get distinct records:

SELECT DISTINCT column 1, column 2, ...
FROM table_name;

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|
*/

If you only need to remove duplicates then use DISTINCT. GROUP BY should be used to apply aggregate operators to each group

GROUP BY v DISTINCT

It depends on which rown you want to return for each unique item. Your data seems to indicate the minimum data value so in this instance for SQL Server.

SELECT item, min(data)
FROM  table
GROUP BY item

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;

just use inner join because group by won't work with multiple columns saying not contained in either an aggregate function.

SELECT a.*
FROM yourtable a
INNER JOIN 
  (SELECT yourcolumn,
    MIN(id) as id
  FROM yourtable 
  GROUP BY yourcolumn
) AS b
  ON a.yourcolumn= b.yourcolumn
  AND a.id = b.id;

To get all the columns in your result you need to place something as:

SELECT distinct a, Table.* FROM Table

it will place a as the first column and the rest will be ALL of the columns in the same order as your definition. This is, column a will be repeated.

I find that if I can't use DISTINCT for any reason, then GROUP BY will work.

I think person, who want to operate query, which want to write query like asterix(*) instead of column(s). if so than this question helps to him answer.

how to select rows based on distinct values of A COLUMN only

SELECT 
  Eff_st 
FROM 
  (
    SELECT 
      EFF_ST, 
      ROW_NUMBER() over(PARTITION BY eff_st) XYZ - 
    FROM 
      ABC.CODE_DIM
  ) 
WHERE 
  XYZ = 1 
ORDER BY 
  EFF_STFETCH FIRST 5 row only