SELECT DISTINCT field1, field2, field3, ...... FROM table
I am trying to accomplish the following sql statement but I want it to return all columns is this possible? Something like:
SELECT DISTINCT field1, * from table
You're looking for a group by:
select *
from table
group by field1
Which can occasionally be written with a distinct on statement:
select distinct on field1 *
from table
On most platforms, however, neither of the above will work because the behavior on the other columns is unspecified. (The first works in MySQL, if that's what you're using.)
You could fetch the distinct fields and stick to picking a single arbitrary row each time.
On some platforms (e.g. PostgreSQL, Oracle, T-SQL) this can be done directly using window functions:
select *
from (
select *,
row_number() over (partition by field1 order by field2) as row_number
from table
) as rows
where row_number = 1
On others (MySQL, SQLite), you'll need to write subqueries that will make you join the entire table with itself (example), so not recommended.
From the phrasing of your question, I understand that you want to select the distinct values for a given field and for each such value to have all the other column values in the same row listed. Most DBMSs will not allow this with neither DISTINCT
nor GROUP BY
, because the result is not determined.
Think of it like this: if your field1
occurs more than once, what value of field2
will be listed (given that you have the same value for field1
in two rows but two distinct values of field2
in those two rows).
You can however use aggregate functions (explicitely for every field that you want to be shown) and using a GROUP BY
instead of DISTINCT
:
SELECT field1, MAX(field2), COUNT(field3), SUM(field4), .... FROM table GROUP BY field1
If I understood your problem correctly, it's similar to one I just had. You want to be able limit the usability of DISTINCT to a specified field, rather than applying it to all the data.
If you use GROUP BY without an aggregate function, which ever field you GROUP BY will be your DISTINCT filed.
If you make your query:
SELECT * from table GROUP BY field1;
It will show all your results based on a single instance of field1.
For example, if you have a table with name, address and city. A single person has multiple addresses recorded, but you just want a single address for the person, you can query as follows:
SELECT * FROM persons GROUP BY name;
The result will be that only one instance of that name will appear with its address, and the other one will be omitted from the resulting table. Caution: if your fileds have atomic values such as firstName, lastName you want to group by both.
SELECT * FROM persons GROUP BY lastName, firstName;
because if two people have the same last name and you only group by lastName, one of those persons will be omitted from the results. You need to keep those things into consideration. Hope this helps.
That's a really good question. I have read some useful answers here already, but probably I can add a more precise explanation.
Reducing the number of query results with a GROUP BY statement is easy as long as you don't query additional information. Let's assume you got the following table 'locations'.
--country-- --city--
France Lyon
Poland Krakow
France Paris
France Marseille
Italy Milano
Now the query
SELECT country FROM locations
GROUP BY country
will result in:
--country--
France
Poland
Italy
However, the following query
SELECT country, city FROM locations
GROUP BY country
...throws an error in MS SQL, because how could your computer know which of the three French cities "Lyon", "Paris" or "Marseille" you want to read in the field to the right of "France"?
In order to correct the second query, you must add this information. One way to do this is to use the functions MAX() or MIN(), selecting the biggest or smallest value among all candidates. MAX() and MIN() are not only applicable to numeric values, but also compare the alphabetical order of string values.
SELECT country, MAX(city) FROM locations
GROUP BY country
will result in:
--country-- --city--
France Paris
Poland Krakow
Italy Milano
or:
SELECT country, MIN(city) FROM locations
GROUP BY country
will result in:
--country-- --city--
France Lyon
Poland Krakow
Italy Milano
These functions are a good solution as long as you are fine with selecting your value from the either ends of the alphabetical (or numeric) order. But what if this is not the case? Let us assume that you need a value with a certain characteristic, e.g. starting with the letter 'M'. Now things get complicated.
The only solution I could find so far is to put your whole query into a subquery, and to construct the additional column outside of it by hands:
SELECT
countrylist.*,
(SELECT TOP 1 city
FROM locations
WHERE
country = countrylist.country
AND city like 'M%'
)
FROM
(SELECT country FROM locations
GROUP BY country) countrylist
will result in:
--country-- --city--
France Marseille
Poland NULL
Italy Milano
Great question @aryaxt -- you can tell it was a great question because you asked it 5 years ago and I stumbled upon it today trying to find the answer!
I just tried to edit the accepted answer to include this, but in case my edit does not make it in:
If your table was not that large, and assuming your primary key was an auto-incrementing integer you could do something like this:
SELECT
table.*
FROM table
--be able to take out dupes later
LEFT JOIN (
SELECT field, MAX(id) as id
FROM table
GROUP BY field
) as noDupes on noDupes.id = table.id
WHERE
//this will result in only the last instance being seen
noDupes.id is not NULL
For SQL Server you can use the dense_rank and additional windowing functions to get all rows AND columns with duplicated values on specified columns. Here is an example...
with t as (
select col1 = 'a', col2 = 'b', col3 = 'c', other = 'r1' union all
select col1 = 'c', col2 = 'b', col3 = 'a', other = 'r2' union all
select col1 = 'a', col2 = 'b', col3 = 'c', other = 'r3' union all
select col1 = 'a', col2 = 'b', col3 = 'c', other = 'r4' union all
select col1 = 'c', col2 = 'b', col3 = 'a', other = 'r5' union all
select col1 = 'a', col2 = 'a', col3 = 'a', other = 'r6'
), tdr as (
select
*,
total_dr_rows = count(*) over(partition by dr)
from (
select
*,
dr = dense_rank() over(order by col1, col2, col3),
dr_rn = row_number() over(partition by col1, col2, col3 order by other)
from
t
) x
)
select * from tdr where total_dr_rows > 1
This is taking a row count for each distinct combination of col1, col2, and col3.
Found this elsewhere here but this is a simple solution that works:
WITH cte AS /* Declaring a new table named 'cte' to be a clone of your table */
(SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY val1 DESC) AS rn
FROM MyTable /* Selecting only unique values based on the "id" field */
)
SELECT * /* Here you can specify several columns to retrieve */
FROM cte
WHERE rn = 1
Add GROUP BY to field you want to check for duplicates your query may look like
SELECT field1, field2, field3, ...... FROM table GROUP BY field1
field1 will be checked to exclude duplicate records
or you may query like
SELECT * FROM table GROUP BY field1
duplicate records of field1 are excluded from SELECT
SELECT DISTINCT FIELD1, FIELD2, FIELD3 FROM TABLE1 works if the values of all three columns are unique in the table.
If, for example, you have multiple identical values for first name, but the last name and other information in the selected columns is different, the record will be included in the result set.
SELECT DISTINCT * FROM table
does not work for you? – ypercubeᵀᴹdistinct
by definition. If you are trying to just selectDISTINCT field1
but somehow return all other columns what should happen for those columns that have more than one value for a particularfield1
value? You would need to useGROUP BY
and some sort of aggregation on the other columns for example. – Martin Smith