我有两个具有以下模式的表。
CREATE TABLE Product (
id INT AUTO_INCREMENT,
name VARCHAR(64) UNIQUE,
unit_price VARCHAR(6),
PRIMARY KEY (id)
);
CREATE TABLE Sale (
id INT AUTO_INCREMENT,
product_id INT, created_at DATETIME,
units INT, PRIMARY KEY (id),
FOREIGN KEY (product_id) REFERENCES Product(id)
);
INSERT INTO Product(name, unit_price)
VALUES
('Methadone', 30),
('Codeine', 20),
('Morphine', 40),
('Fentanyl', 10);
INSERT INTO Sale(product_id, created_at, units)
VALUES
(1, '2004-01-06 08:27:25', 8),
(4, '2004-03-05 05:21:15', 24),
(2, '2005-01-06 08:26:55', 3),
(4, '2005-03-31 01:55:25', 6),
(1, '2005-11-09 11:33:20', 2),
(4, '2006-02-15 10:46:30', 4),
(3, '2006-04-14 07:40:50', 1),
(4, '2008-01-06 08:27:25', 5),
(4, '2008-01-10 21:31:45', 3),
(1, '2008-01-28 16:07:30', 9),
(3, '2008-03-05 05:21:00', 4),
(2, '2008-03-08 18:25:50', 5),
(4, '2008-05-02 02:15:20', 1),
(4, '2008-05-05 15:19:40', 3),
(4, '2008-06-28 23:08:55', 2),
(4, '2008-07-02 12:14:00', 2),
(3, '2008-08-29 09:07:50', 1),
(2, '2008-10-22 16:56:20', 5),
(3, '2008-10-26 06:01:25', 2),
(4, '2008-12-19 13:49:55', 4),
(2, '2008-12-23 02:55:30', 4),
(1, '2009-02-15 10:43:45', 4),
(2, '2009-04-14 07:37:50', 6),
(3, '2009-06-11 04:31:25', 3),
(4, '2009-08-08 01:25:30', 12),
(1, '2010-11-09 11:33:20', 8),
(1, '2011-01-14 10:22:35', 3),
(3, '2011-03-13 07:16:10', 3),
(4, '2011-05-10 04:10:15', 4),
(4, '2011-07-07 01:04:35', 4),
(4, '2011-09-02 21:58:10', 3),
(2, '2011-10-30 18:51:45', 3),
(1, '2011-12-27 15:45:20', 1),
(4, '2012-02-23 12:43:25', 2),
(4, '2012-04-21 09:37:30', 3),
(4, '2012-06-18 06:31:20', 1),
(2, '2012-08-15 03:25:40', 4),
(2, '2013-01-14 10:23:20', 4),
(4, '2013-02-01 05:01:35', 1),
(4, '2013-03-13 07:16:55', 1),
(4, '2013-03-31 01:55:10', 2),
(4, '2013-05-10 04:11:15', 2),
(3, '2013-05-27 22:48:45', 3),
(1, '2013-07-07 01:05:20', 3),
(2, '2013-07-24 19:42:20', 4),
(4, '2013-09-02 21:59:40', 4),
(3, '2013-09-20 16:36:10', 1),
(4, '2013-10-30 18:54:00', 2),
(4, '2013-11-17 13:29:45', 1),
(4, '2013-12-27 15:48:20', 3);
我想找到按年份降序排列的每年销量最高的产品，以及销量最高的产品的总销售额。
我写了以下查询。
SELECT EXTRACT(YEAR FROM S.created_at) as year,
GROUP_CONCAT(DISTINCT P1.name
ORDER BY P1.name
SEPARATOR ',')as top,
MAX(S.units*P1.unit_price) as sale
FROM Sale S
INNER JOIN Product P1 ON S.product_id=P1.id
GROUP BY year
ORDER BY year desc
预期输出：
2013 Codeine,Fentanyl,Morphine 160
2012 Codeine 80
2011 Methadone,Morphine 120
2010 Methadone 240
2009 Codeine,Fentanyl,Methadone,Morphine 120
2008 Codeine,Morphine 280
2006 Fentanyl,Morphine 40
2005 Codeine,Fentanyl,Methadone 60
2004 Fentanyl,Methadone 240
原始输出：
2013 Codeine,Fentanyl,Methadone,Morphine 120
2012 Codeine,Fentanyl 80
2011 Codeine,Fentanyl,Methadone,Morphine 120
2010 Methadone 240
2009 Codeine,Fentanyl,Methadone,Morphine 120
2008 Codeine,Fentanyl,Methadone,Morphine 270
2006 Fentanyl,Morphine 40
2005 Codeine,Fentanyl,Methadone 60
2004 Fentanyl,Methadone 240
该查询应该只返回最畅销的产品，但要返回所有产品。谁能提供帮助吗？
答案 0 :(得分：2)
这是MySQL <8.0的解决方案。首先，按年份和产品汇总销售额，并使用汇总的相关子查询对最畅销的产品进行过滤；然后，按年份汇总：
select
year_at,
group_concat(name order by name) products,
total_sales
from (
select
year(s.created_at) year_at,
p.name,
sum(s.units * p.unit_price) total_sales
from Product p
inner join Sale s on s.product_id = p.id
group by year_at, p.id, p.name
having total_sales = (
select sum(s1.units * p1.unit_price) total_sales1
from Product p1
inner join Sale s1 on s1.product_id = p1.id
where year(s1.created_at) = year_at
group by s1.product_id
order by total_sales1 desc
limit 1
)
) t
group by year_at, total_sales
order by year_at desc
year_at | products | total_sales ------: | :---------------------------------- | ----------: 2013 | Codeine,Fentanyl,Morphine | 160 2012 | Codeine | 80 2011 | Methadone,Morphine | 120 2010 | Methadone | 240 2009 | Codeine,Fentanyl,Methadone,Morphine | 120 2008 | Codeine,Morphine | 280 2006 | Fentanyl,Morphine | 40 2005 | Codeine,Fentanyl,Methadone | 60 2004 | Fentanyl,Methadone | 240
答案 1 :(得分：2)
您似乎是在追求...
SELECT GROUP_CONCAT(a.name) names,a.total, a.year
FROM
( SELECT p.name
, SUM(p.unit_price*s.units) total
, YEAR(created_at) year
FROM product p
JOIN sale s
ON s.product_id = p.id
GROUP
BY year
, name
) a
JOIN
( SELECT MAX(total) total
, year
FROM
( SELECT p.name
, SUM(p.unit_price*s.units) total
, YEAR(created_at) year
FROM product p
JOIN sale s
ON s.product_id = p.id
GROUP
BY year
, name
) n
GROUP
BY year) b
ON b.year = a.year AND b.total = a.total
GROUP BY a.total, a.year
ORDER BY year DESC;
+-------------------------------------+-------+------+
| names | total | year |
+-------------------------------------+-------+------+
| Fentanyl,Codeine,Morphine | 160 | 2013 |
| Codeine | 80 | 2012 |
| Morphine,Methadone | 120 | 2011 |
| Methadone | 240 | 2010 |
| Morphine,Methadone,Fentanyl,Codeine | 120 | 2009 |
| Morphine,Codeine | 280 | 2008 |
| Morphine,Fentanyl | 40 | 2006 |
| Codeine,Methadone,Fentanyl | 60 | 2005 |
| Methadone,Fentanyl | 240 | 2004 |
+-------------------------------------+-------+------+
答案 2 :(得分：1)
这是我的Scalar-Aggregate Comparison查询技术的一种变体，与自动加入汇总结果相比，该查询技术在性能更好的查询中将实现您想要的目标。正如您在下面看到的那样，此操作仅执行一次连接，然后执行3个级联的聚合层以实现最终输出。
SELECT
yr,
SUBSTRING(MAX(CONCAT(LPAD(total_sales, 16, '0'), products)), 17) AS best_seller,
MAX(total_sales) AS sales_amount
FROM (
SELECT
yr,
total_sales,
GROUP_CONCAT(name order by name) AS products
FROM (
SELECT
YEAR(s.created_at) AS yr,
p.id,
p.name,
SUM(s.units * p.unit_price) AS total_sales
FROM Product AS p
INNER JOIN Sale AS s on s.product_id = p.id
GROUP BY YEAR(s.created_at), p.id
) AS pys
GROUP BY yr, total_sales
) AS py
GROUP BY yr
ORDER BY yr DESC;
这是一个DB Fiddle演示：https://dbfiddle.uk/?rdbms=mysql_5.6&fiddle=e14ed7f396738ce064f1998bdb4310ac
答案 3 :(得分：0)
尝试一下：
var gl;
var canvas;
var img = new Image();
// img.onload = run;
img.crossOrigin = 'anonymous';
img.src = 'https://threejsfundamentals.org/threejs/resources/images/heightmap-96x64.png';
var gridWidth;
var gridDepth;
var gridPoints = [];
var gridIndices = [];
var rowOff = 0;
var rowStride = gridWidth + 1;
var numVertices = (gridWidth * 2 * (gridDepth + 1)) + (gridDepth * 2 * (gridWidth + 1));
//creating plane
function generateHeightPoints() {
var ctx = document.createElement("canvas").getContext("2d"); //using 2d canvas to read image
ctx.canvas.width = img.width;
ctx.canvas.height = img.height;
ctx.drawImage(img, 0, 0);
var imgData = ctx.getImageData(0, 0, ctx.canvas.width, ctx.canvas.height);
gridWidth = imgData.width - 1;
gridDepth = imgData.height - 1;
for (var z = 0; z <= gridDepth; ++z) {
for (var x = 0; x <= gridWidth; ++x) {
var offset = (z * imgData.width + x) * 4;
var height = imgData.data[offset] * 10 / 255;
gridPoints.push(x, height, z);
}
}
}
function generateIndices() {
for (var z = 0; z<=gridDepth; ++z) {
rowOff = z*rowStride;
for(var x = 0; x<gridWidth; ++x) {
gridIndices.push(rowOff+x,rowOff+x+1);
}
}
for (var x = 0; x<=gridWidth; ++x) {
for(var z = 0; z<gridDepth; ++z) {
rowOff = z * rowStride;
gridIndices.push(rowOff+x,rowOff+x+rowStride);
}
}
}
//init
//program init
window.onload = function init()
{
canvas = document.getElementById( "gl-canvas" );
gl = WebGLUtils.setupWebGL( canvas );
if ( !gl ) { alert( "WebGL isn't available" ); }
gl.viewport(0, 0, gl.canvas.width, gl.canvas.height);
var program = initShaders( gl, "vertex-shader", "fragment-shader" );
gl.useProgram( program );
generateHeightPoints();
generateIndices();
var positionBuffer = gl.createBuffer();
gl.bindBuffer(gl.ARRAY_BUFFER, positionBuffer);
gl.bufferData(gl.ARRAY_BUFFER, new Float32Array(gridPoints),
gl.STATIC_DRAW);
var indexBuffer = gl.createBuffer();
gl.bindBuffer(gl.ELEMENT_ARRAY_BUFFER, indexBuffer);
gl.bufferData(gl.ELEMENT_ARRAY_BUFFER, new Uint16Array(gridIndices),
gl.STATIC_DRAW);
var vPosition = gl.getAttribLocation( program, "vPosition" );
gl.vertexAttribPointer( vPosition, 3, gl.FLOAT, false, 0, 0 );
gl.enableVertexAttribArray( vPosition );
var matrixLoc = gl.getUniformLocation(program, 'matrix');
var m4 = twgl.m4;
var projection = m4.perspective(60 * Math.PI / 180, gl.canvas.clientWidth /
gl.canvas.clientHeight, 0.1, 100);
var cameraPosition = [-18, 15, -10];
var target = [gridWidth / 2, -10, gridDepth / 2];
var up = [0, 1, 0];
var camera = m4.lookAt(cameraPosition, target, up);
var view = m4.inverse(camera);
var mat = m4.multiply(projection, view);
gl.uniformMatrix4fv(matrixLoc, false, mat);
render();
}
function render() {
gl.drawElements(gl.LINES, numVertices, gl.UNSIGNED_SHORT, 0);
gl.drawElements(gl.LINES,numVertices,gl.UNSIGNED_SHORT,0);
requestAnimFrame(render);
}
答案 4 :(得分：0)
如果要获得每年销量最高的产品，请使用rank()和汇总查询：
select ps.*
from (select p.name, year(s.created_at) as year, sum(p.unit_price*s.units) as sales
rank() over (partition by year(s.created_at) order by sum(p.unit_price*s.units) desc as seqnum
from Product p inner join
sale s
on p.id = s.product_id
group by name, year
) ps
where seqnum <= 1; -- You can change "1" for more rows