Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
923 views
in Technique[技术] by (71.8m points)

postgresql - In Redshift/Postgres, how to count rows that meet a condition?

I'm trying to write a query that count only the rows that meet a condition.

For example, in MySQL I would write it like this:

SELECT
    COUNT(IF(grade < 70), 1, NULL)
FROM
    grades
ORDER BY
    id DESC;

However, when I attempt to do that on Redshift, it returns the following error:

ERROR: function if(boolean, integer, "unknown") does not exist

Hint: No function matches the given name and argument types. You may need to add explicit type casts.

I checked the documentation for conditional statements, and I found

NULLIF(value1, value2)

but it only compares value1 and value2 and if such values are equal, it returns null.

I couldn't find a simple IF statement, and at first glance I couldn't find a way to do what I want to do.

I tried to use the CASE expression, but I'm not getting the results I want:

SELECT 
    CASE
        WHEN grade < 70 THEN COUNT(rank)
        ELSE COUNT(rank)
    END
FROM
   grades

This is the way I want to count things:

  • failed (grade < 70)

  • average (70 <= grade < 80)

  • good (80 <= grade < 90)

  • excellent (90 <= grade <= 100)

and this is how I expect to see the results:

+========+=========+======+===========+
| failed | average | good | excellent |
+========+=========+======+===========+
|   4    |    2    |  1   |     4     |
+========+=========+======+===========+

but I'm getting this:

+========+=========+======+===========+
| failed | average | good | excellent |
+========+=========+======+===========+
|  11    |   11    |  11  |    11     |
+========+=========+======+===========+

I hope someone could point me to the right direction!

If this helps here's some sample info

CREATE TABLE grades(
  grade integer DEFAULT 0,
);

INSERT INTO grades(grade) VALUES(69, 50, 55, 60, 75, 70, 87, 100, 100, 98, 94);
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

First, the issue you're having here is that what you're saying is "If the grade is less than 70, the value of this case expression is count(rank). Otherwise, the value of this expression is count(rank)." So, in either case, you're always getting the same value.

SELECT 
    CASE
        WHEN grade < 70 THEN COUNT(rank)
        ELSE COUNT(rank)
    END
FROM
   grades

count() only counts non-null values, so typically the pattern you'll see to accomplish what you're trying is this:

SELECT 
    count(CASE WHEN grade < 70 THEN 1 END) as grade_less_than_70,
    count(CASE WHEN grade >= 70 and grade < 80 THEN 1 END) as grade_between_70_and_80
FROM
   grades

That way the case expression will only evaluate to 1 when the test expression is true and will be null otherwise. Then the count() will only count the non-null instances, i.e. when the test expression is true, which should give you what you need.

Edit: As a side note, notice that this is exactly the same as how you had originally written this using count(if(test, true-value, false-value)), only re-written as count(case when test then true-value end) (and null is the stand in false-value since an else wasn't supplied to the case).

Edit: postgres 9.4 was released a few months after this original exchange. That version introduced aggregate filters, which can make scenarios like this look a little nicer and clearer. This answer still gets some occasional upvotes, so if you've stumbled upon here and are using a newer postgres (i.e. 9.4+) you might want to consider this equivalent version:

SELECT
    count(*) filter (where grade < 70) as grade_less_than_70,
    count(*) filter (where grade >= 70 and grade < 80) as grade_between_70_and_80
FROM
   grades

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...