满足条件时执行SQLite查询_Sql_Sqlite_Case When

满足条件时执行SQLite查询

sql sqlite

满足条件时执行SQLite查询,sql,sqlite,case-when,Sql,Sqlite,Case When,我试图提出一个SQLite查询，它将在满足条件时检索查询中两个给定值A和B之间的所有行值如果给定值B大于表中B的最大值： -检索A和B之间的所有值样本表：库存 Prod_name | model | location | tesla | "5.6.1" | CA toyota | "4.7.1" | WA kia | "6.8.1" | MD tesla | "2.6.2" |

我试图提出一个SQLite查询，它将在满足条件时检索查询中两个给定值A和B之间的所有行值

如果给定值B大于表中B的最大值： -检索A和B之间的所有值

样本表：库存

Prod_name |    model     | location | 
tesla     |   "5.6.1"    |    CA
toyota    |   "4.7.1"    |     WA
kia       |   "6.8.1"    |     MD
tesla     |   "2.6.2"    |     CA
chev      |   "7.8.4"    |     AZ


Input given : model between ("5.0.0" to "8.2.0")
Output : (telsa,5.6.1,CA),(kia,6.8.1,MD) , (chev,7.8.4,AZ)

Input given : model between ("5.0.0" to "6.9.0")
Output: Query should not run as "7.8.4" > "6.9.0" 
i.e ( the max value in the table is greater than the upper limit of input query. 
Also to note is the model name is TEXT format. I need help to retrieving   

I have tried "CASE" statements of sqlite but was not able to retrieve 
multiple columns in the subquery.

我相信以下几点可以满足您的要求：-

/* Query using model in n.n.n format */
SELECT * FROM inventory 
WHERE 
    ((1000000 * substr(model,1,instr(model,'.')-1)) + 
    (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
    replace(model,'.','000') % 1000)
    BETWEEN
        (
            SELECT 1000000  * substr('5.0.0',1,instr('5.0.0','.') -1) 
                + (1000 * replace(substr('5.0.0',instr('5.0.0','.') + 1),'.','x'))
                + replace('5.0.0','.','000') % 1000
        )
        AND
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        )
    /* MAX COndition */
    AND 
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        ) 
            >  
        (
            SELECT MAX(((1000000 * substr(model,1,instr(model,'.')-1)) 
                + (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) 
                + replace(model,'.','000') % 1000)) 
            FROM inventory
        )
ORDER BY 
    (1000000 * substr(model,1,instr(model,'.')-1)) + 
            (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
            replace(model,'.','000') % 1000
;

我很想知道如何在当前的解决方案中使用这一点。或者如果你有其他的方法

我认为你使用了一个格式为n.n.n的模型，把事情复杂化了

如果将该模型转换为整数值，则可以大大简化问题

如果您确实希望将模型保持为n.n.n，那么可以修改表以添加一列，该列将模型存储为整数。e、 g.作为其中一员，您可以使用：-

ALTER TABLE inventory ADD COLUMN model_value INTEGER DEFAULT -1;

这将添加列model_值更改之后可以进行大规模更新，然后为现有行设置值，例如：-

UPDATE inventory SET model_value = 
    (1000000 * substr(model,1,instr(model,'.')-1)) + 
    (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
    replace(model,'.','000') % 1000;

为了避免需要更改插入和预先计算模型_值，您可以添加插入后触发器，例如：-

CREATE TRIGGER IF NOT EXISTS inventory_generate_modelvalue  AFTER INSERT ON inventory
BEGIN
    UPDATE inventory 
        SET model_value = (1000000 * substr(model,1,instr(model,'.')-1)) + 
            (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
            replace(model,'.','000') % 1000
    WHERE model_value < 0 OR model_value IS NULL
        ;
END;

如果要将模型值转换为n.n.n格式，可以基于以下内容使用：-

SELECT prod_name, 
    CAST (model_value / 1000000 AS TEXT)
        ||'.'
        || CAST((model_value % 1000000) / 1000 AS TEXT)
        ||'.'
        ||CAST(model_value % 1000 AS TEXT) 
        AS model, 
    location 
FROM inventory;

当然，如果您的程序中有一个函数，或者使用整数值而不是n.n.n，那么事情就更简单了

测试以下代码用于测试上述内容：-

DROP TABLE IF EXISTS inventory;
DROP TRIGGER IF EXISTS inventory_generate_modelvalue;
CREATE TABLE IF NOT EXISTS inventory (prod_name TEXT ,model TEXT,location TEXT);
INSERT INTO inventory VALUES ('tesla','5.6.1','CA'),('toyota','4.7.1','WA'),('kia','6.8.1','MD'),('tesla','2.6.2','CA'),('chev','7.8.4','AZ') ;

/* Add new column for model as an integer value */
ALTER TABLE inventory ADD COLUMN model_value INTEGER DEFAULT -1;

/* Update existing data for new column */
UPDATE inventory SET model_value = 
    (1000000 * substr(model,1,instr(model,'.')-1)) + 
    (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
    replace(model,'.','000') % 1000;

CREATE TRIGGER IF NOT EXISTS inventory_generate_modelvalue  AFTER INSERT ON inventory
BEGIN
    UPDATE inventory 
        SET model_value = (1000000 * substr(model,1,instr(model,'.')-1)) + 
            (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
            replace(model,'.','000') % 1000
    WHERE model_value < 0 OR model_value IS NULL
        ;
END;

-- INSERT INTO inventory VALUES('my new model','5.0.1','AA',null),('another','0.999.999','ZZ',-1);

SELECT * FROM inventory;

/* Query using model in n.n.n format */
SELECT * FROM inventory 
WHERE 
    ((1000000 * substr(model,1,instr(model,'.')-1)) + 
    (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
    replace(model,'.','000') % 1000)
    BETWEEN
        (
            SELECT 1000000  * substr('5.0.0',1,instr('5.0.0','.') -1) 
                + (1000 * replace(substr('5.0.0',instr('5.0.0','.') + 1),'.','x'))
                + replace('5.0.0','.','000') % 1000
        )
        AND
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        )
    /* MAX COndition */
    AND 
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        ) 
            >  
        (
            SELECT MAX(((1000000 * substr(model,1,instr(model,'.')-1)) 
                + (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) 
                + replace(model,'.','000') % 1000)) 
            FROM inventory
        )
ORDER BY 
    (1000000 * substr(model,1,instr(model,'.')-1)) + 
            (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
            replace(model,'.','000') % 1000
;

/* Query using model_value) */
SELECT * FROM inventory
WHERE model_value 
    BETWEEN 
        (
            SELECT 1000000  * substr('5.0.0',1,instr('5.0.0','.') -1) 
                + (1000 * replace(substr('5.0.0',instr('5.0.0','.') + 1),'.','x'))
                + replace('5.0.0','.','000') % 1000
        ) 
        AND
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        )
    AND 
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        ) 
            > 
        (SELECT MAX(model_value) FROM inventory)

ORDER BY model_value
;

SELECT prod_name, 
    CAST (model_value / 1000000 AS TEXT)
        ||'.'
        || CAST((model_value % 1000000) / 1000 AS TEXT)
        ||'.'
        ||CAST(model_value % 1000 AS TEXT) 
        AS model, 
    location 
FROM inventory;

这里最好的答案可能是将主要版本号和次要版本号存储在单独的列中。版本号是否总是介于0和9之间，或者也可能是两位数字？可能是@TimBiegeleisen的重复数字可以是两位或三位。@MikeT我很想知道如何在当前的解决方案中使用它。或者如果你有其他的方法？

/* Query using model_value) */
SELECT * FROM inventory
WHERE model_value 
    BETWEEN 
        (
            SELECT 1000000  * substr('5.0.0',1,instr('5.0.0','.') -1) 
                + (1000 * replace(substr('5.0.0',instr('5.0.0','.') + 1),'.','x'))
                + replace('5.0.0','.','000') % 1000
        ) 
        AND
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        )
    AND 
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        ) 
            > 
        (SELECT MAX(model_value) FROM inventory)

ORDER BY model_value
;

SELECT prod_name, 
    CAST (model_value / 1000000 AS TEXT)
        ||'.'
        || CAST((model_value % 1000000) / 1000 AS TEXT)
        ||'.'
        ||CAST(model_value % 1000 AS TEXT) 
        AS model, 
    location 
FROM inventory;

DROP TABLE IF EXISTS inventory;
DROP TRIGGER IF EXISTS inventory_generate_modelvalue;
CREATE TABLE IF NOT EXISTS inventory (prod_name TEXT ,model TEXT,location TEXT);
INSERT INTO inventory VALUES ('tesla','5.6.1','CA'),('toyota','4.7.1','WA'),('kia','6.8.1','MD'),('tesla','2.6.2','CA'),('chev','7.8.4','AZ') ;

/* Add new column for model as an integer value */
ALTER TABLE inventory ADD COLUMN model_value INTEGER DEFAULT -1;

/* Update existing data for new column */
UPDATE inventory SET model_value = 
    (1000000 * substr(model,1,instr(model,'.')-1)) + 
    (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
    replace(model,'.','000') % 1000;

CREATE TRIGGER IF NOT EXISTS inventory_generate_modelvalue  AFTER INSERT ON inventory
BEGIN
    UPDATE inventory 
        SET model_value = (1000000 * substr(model,1,instr(model,'.')-1)) + 
            (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
            replace(model,'.','000') % 1000
    WHERE model_value < 0 OR model_value IS NULL
        ;
END;

-- INSERT INTO inventory VALUES('my new model','5.0.1','AA',null),('another','0.999.999','ZZ',-1);

SELECT * FROM inventory;

/* Query using model in n.n.n format */
SELECT * FROM inventory 
WHERE 
    ((1000000 * substr(model,1,instr(model,'.')-1)) + 
    (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
    replace(model,'.','000') % 1000)
    BETWEEN
        (
            SELECT 1000000  * substr('5.0.0',1,instr('5.0.0','.') -1) 
                + (1000 * replace(substr('5.0.0',instr('5.0.0','.') + 1),'.','x'))
                + replace('5.0.0','.','000') % 1000
        )
        AND
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        )
    /* MAX COndition */
    AND 
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        ) 
            >  
        (
            SELECT MAX(((1000000 * substr(model,1,instr(model,'.')-1)) 
                + (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) 
                + replace(model,'.','000') % 1000)) 
            FROM inventory
        )
ORDER BY 
    (1000000 * substr(model,1,instr(model,'.')-1)) + 
            (1000 * replace(substr(model,instr(model,'.') + 1),'.','x')) +
            replace(model,'.','000') % 1000
;

/* Query using model_value) */
SELECT * FROM inventory
WHERE model_value 
    BETWEEN 
        (
            SELECT 1000000  * substr('5.0.0',1,instr('5.0.0','.') -1) 
                + (1000 * replace(substr('5.0.0',instr('5.0.0','.') + 1),'.','x'))
                + replace('5.0.0','.','000') % 1000
        ) 
        AND
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        )
    AND 
        (
            SELECT 1000000  * substr('8.5.0',1,instr('8.5.0','.') -1) 
                + (1000 * replace(substr('8.5.0',instr('8.5.0','.') + 1),'.','x'))
                + replace('8.5.0','.','000') % 1000
        ) 
            > 
        (SELECT MAX(model_value) FROM inventory)

ORDER BY model_value
;

SELECT prod_name, 
    CAST (model_value / 1000000 AS TEXT)
        ||'.'
        || CAST((model_value % 1000000) / 1000 AS TEXT)
        ||'.'
        ||CAST(model_value % 1000 AS TEXT) 
        AS model, 
    location 
FROM inventory;