Find the average of two combined columns in sql

旧街凉风 提交于 2019-12-18 05:56:17

问题


I want to find the avg of the total of two columns. I want to count the total of col1 and the total of col2 then find the average(how many different rows they are in).

I have managed to come up with a solution in the this sqlfiddle (also see below) is this the best way? I initially thought I would need to use the avg function but couldn't work it out using this.

    CREATE TABLE test (
        id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
        uid INT,
        col1 INT,
        col2 INT
    ) DEFAULT CHARACTER SET utf8 ENGINE=InnoDB;

    INSERT INTO test (id, uid, col1, col2) VALUES
    (1,5,8,12),
    (2,1,2,3),
    (3,1,2,33),
    (4,5,25,50),
    (5,5,22,3);

    (
    SELECT ((sum(col1) + sum(col2))/count(*))
    FROM test
      WHERE uid=5
    )

回答1:


By definition, AVG(col1) = SUM(col1)/COUNT(*) and AVG(col2) = SUM(col2)/COUNT(*), therefore (SUM(col1)+SUM(col2))/COUNT(*) = AVG(col1) + AVG(col2).

Also, the commutativity of addition gives us (SUM(col1)+SUM(col2))/COUNT(*) = SUM(col1+col2)/COUNT(*) and hence AVG(col1+col2).




回答2:


To use the avg function,

SELECT avg(col1 + col2)
FROM test
WHERE uid=5;

SQLFIDDLE DEMO




回答3:


Is this what you are looking for?

SELECT avg(col1 + col2)
FROM test
where uid = 5
group by uid



回答4:


SELECT avg(col1 + col2) as avgtotal

FROM test WHERE uid=5




回答5:


i got my answer here , so i will add this note which may help others:

1.avg(col1+col2) as avg_col1_plus_col2,
2.avg(col1) + avg(col2) as avg_col1_plus_avg_col2,
3.avg(col1+col2)/2 as avgTotal1, 
4.avg(col1)/2+avg(col1)/2 as avgTotal2

sentence 1 is equal to sentence 2 as eggyal explained,grammar is ok but logically its not the result that we want, so we need to divide the average by columns numbers as in sentence 3 and 4.




回答6:


SELECT ((SUM(col1) + SUM(col2)) / (COUNT(col1) + COUNT(col2))) as a
FROM test
WHERE uid=5;


来源:https://stackoverflow.com/questions/14787561/find-the-average-of-two-combined-columns-in-sql

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