Question

我有一个CROSS JOIN查询，用于查看哪种物料数量组合产生最佳输出。

DECLARE @last_found DECIMAL(10, 2) = 0
DECLARE @calculated DECIMAL(10, 2)
DECLARE @n_count INT
DECLARE @tbl1n INT
DECLARE @tbl2n INT
DECLARE @tbl3n INT


DROP TABLE IF EXISTS #tbl1
DROP TABLE IF EXISTS #tbl2
DROP TABLE IF EXISTS #tbl3

;WITH numbers AS (
    SELECT ROW_NUMBER() OVER (ORDER BY [value]) AS n
    FROM string_split('1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20', ',')
)
SELECT n, (n * 10000 * (1 + IIF(n > 1, (0.50/19.00) * (n - 1), 0))) AS price
INTO #tbl1 FROM numbers

;WITH numbers AS (
    SELECT ROW_NUMBER() OVER (ORDER BY [value]) AS n
    FROM string_split('1,2,3,4,5,6,7,8,9,10,11,12', ',')
)
SELECT n, (n * 15000 * (1 + IIF(n > 1, (0.50/11.00) * (n - 1), 0))) AS price
INTO #tbl2 FROM numbers

;WITH numbers AS (
    SELECT ROW_NUMBER() OVER (ORDER BY [value]) AS n
    FROM string_split('1,2,3,4,5,6', ',')
)
SELECT n, (n * 20000 * (1 + IIF(n > 1, (0.50/5.00) * (n - 1), 0))) AS price
INTO #tbl3 FROM numbers


SELECT
    @n_count = (tbl1.n + tbl2.n + tbl3.n),
    @calculated = IIF(@n_count = 10, (tbl1.price + tbl2.price + tbl3.price), 0),
    @tbl1n = IIF(@calculated > @last_found, tbl1.n, @tbl1n),
    @tbl2n = IIF(@calculated > @last_found, tbl2.n, @tbl2n),
    @tbl3n = IIF(@calculated > @last_found, tbl3.n, @tbl3n),
    @last_found = IIF(@calculated > @last_found, @calculated, @last_found)
FROM #tbl1 tbl1
CROSS JOIN #tbl2 tbl2
CROSS JOIN #tbl3 tbl3


SELECT @last_found AS highest_value, @tbl1n AS tbl1n, @tbl2n AS tbl2n, @tbl3n AS tbl3n,
    t1.price AS tbl1_price, t2.price AS tbl2_price, t3.price AS tbl3_price
FROM #tbl1 t1
INNER JOIN #tbl2 t2 ON t1.n = @tbl1n AND t2.n = @tbl2n
INNER JOIN #tbl3 t3 ON t3.n = @tbl3n

可以看出，如果查询找到的值大于先前找到的最高值，则它将使用多个@itemN = IIF(@calculated > @last_found, tbl.n, @itemN)实例存储组合。

是否可以一次性分配所有@tblXn变量？我可以使用CONCAT，但是我认为它可能会减慢查询速度，因为它是字符串操作。

仅供参考-'n'是介于0和20之间的值。

Answer 1

您可以使用const puppeteer = require('puppeteer'); const express = require('express'); const fs = require('fs'); var port = process.env.PORT || 80; const app = express(); global.browser = null; const setup = async function(){ console.log("Initializing Puppeteer...") browser = await puppeteer.launch({ args: [ '--no-sandbox', '--disable-setuid-sandbox', '--disable-dev-shm-usage=true', '--disable-accelerated-2d-canvas=true', '--disable-gpu', ] }); browser.on('disconnected', setup); } app.get('/', (req, res) => { setup(); console.log(`Started Puppeteer with pid ${browser.process().pid}`); res.status(200).send('Hello, world!').end(); }); // Start the server const PORT = process.env.PORT || 8080; app.listen(PORT, () => { setup(); console.log(`App listening on port ${PORT}`); console.log('Press Ctrl+C to quit.'); }); // [END gae_node_request_example] module.exports = app;：

apply

注意：您可以进一步将值分配给变量。

Answer 2

您根本不应该为此使用变量。

这将更简单地编写如下（无需依赖跨行分配变量的无证/毫无保证的行为）

SELECT TOP 1 CAST(combined_price AS DECIMAL(10, 2)) AS highest_value,
             tbl1.n                                 AS tbl1n,
             tbl2.n                                 AS tbl2n,
             tbl3.n                                 AS tbl3n,
             tbl1.price                             AS tbl1_price,
             tbl2.price                             AS tbl2_price,
             tbl3.price                             AS tbl3_price
FROM   #tbl1 tbl1
       CROSS JOIN #tbl2 tbl2
       CROSS JOIN #tbl3 tbl3
       CROSS APPLY (VALUES (tbl1.price + tbl2.price + tbl3.price, 
                            tbl1.n + tbl2.n + tbl3.n)) CA(combined_price, combined_n)
WHERE  combined_n = 10
ORDER  BY combined_price DESC

在SELECT语句的一个IIF（）中设置多个变量的值

2 个答案: