Question

我正在查询产生所需结果的两个表。我需要从这个结果数组中有条件地插入元素到另一个表中。请原谅我缺乏理解，但我对php和mysql相当新。

结果字段为：bus_id, age, race, and sex。我想有条件地将这些字段的元素插入到另一个包含字段的表中：bus_id, no_people, males, females, and ethnicity%。

结果数组中的bus_id确实=插入表格中的bus_id。

例如，我想计算分配给特定bus_id的人数，并确定人数，男性，女性和种族百分比。

我的思维过程可能看似令人费解，但请原谅我是一个新手。如果你能把我放在正确的方向，我们将不胜感激。

好的，这是我对前两张表的查询结果表：

bus_name bus_id cus_name age race sex DHL 10 tom selleck 32 asian male Fedex 17 jennifer lopez 34 black female UPS 1 jet li 26 black male UPS 1 hugh douglas 40 black male UPS 1 ray lewis 40 white male UPS 1 michael vick 38 black male UPS 1 ed reed 44 white male UPS 1 joe flacco 17 white male UPS 1 bob dole 26 latin male UPS 1 deion sanders 15 white male UPS 1 bill clinton 15 latin male UPS 1 lebron james 41 white male

我想在另一个名为“comp_struct”的表中有条件地插入这些元素，其中包含以下字段：bus_id，males，female，ethnicity％（可能会将其更改为多样性百分比）忘记现在的种族％。

所以我希望它在完成后看起来像这样。

bus_id   no_people     males     females    

  1        10            10        0
  17        1             0        1
  10        1             1        0

这是我用我的代码创建第一个表结果的程度。

$query1 = $db->query("SELECT bus.bus_name, bus.bus_id, bus_users.purch, bus_users.cus_name,  bus_users.age, bus_users.race, bus_users.sex FROM bus, bus_users WHERE bus_users.purch = bus.bus_id ORDER BY bus.bus_name") or die($db->error);


while($result = mysqli_fetch_assoc($query1))
{
(   "SELECT bus_id, 
        COUNT(*), 
        SUM(IF(sex='male', 1, 0)) AS males,
        SUM(IF(sex='female', 1, 0)) AS females
FROM @query1
GROUP BY bus_id



INSERT INTO comp_struct (bus_id, no_people, males, females)
SELECT bus_id, 
        COUNT(*) AS no_people, 
        SUM(IF(sex='male', 1, 0)) AS males,
        SUM(IF(sex='female', 1, 0)) AS females
FROM @query1
GROUP BY bus_id");

毋庸置疑，这并没有产生预期的效果。

Answer 1

暂时忽略种族字段，您可以获得所需的值： -

SELECT bus_id, 
        COUNT(*), 
        SUM(IF(sex='male', 1, 0)) AS males,
        SUM(IF(sex='female', 1, 0)) AS females
FROM table_one
GROUP BY bus_id

并插入： -

INSERT INTO table_two(bus_id, no_people, males, females)
SELECT bus_id, 
        COUNT(*) AS no_people, 
        SUM(IF(sex='male', 1, 0)) AS males,
        SUM(IF(sex='female', 1, 0)) AS females
FROM table_one
GROUP BY bus_id

编辑 - 基于您现在给出的详细信息，单个语句将按如下方式执行插入（无需先单独选择行）

INSERT INTO table_two(bus_id, no_people, males, females)
SELECT bus.bus_id , 
        COUNT(*) AS no_people, 
        SUM(IF(bus_users.sex ='male', 1, 0)) AS males,
        SUM(IF(bus_users.sex ='female', 1, 0)) AS females
FROM bus
INNER JOIN bus_users 
ON bus_users.purch = bus.bus_id 
GROUP BY bus.bus_id

编辑 - 这也可以让你获得种族： -

SELECT bus_id, no_people, males, females, GROUP_CONCAT(CONCAT(race, race_percentage, '%'))
FROM
(
    SELECT Sub1.bus_id, Sub2.no_people, Sub2.males, Sub2.females, Sub1.race, (Sub1.race_count / Sub2.no_people) * 100 AS race_percentage
    FROM
    (
        SELECT bus.bus_id , 
                bus_users.race, 
                COUNT(*) AS race_count
        FROM bus
        INNER JOIN bus_users 
        ON bus_users.purch = bus.bus_id 
        GROUP BY bus.bus_id, race
    ) Sub1
    INNER JOIN
    (
        SELECT bus.bus_id , 
                COUNT(*) AS no_people, 
                SUM(IF(bus_users.sex ='male', 1, 0)) AS males,
                SUM(IF(bus_users.sex ='female', 1, 0)) AS females
        FROM bus
        INNER JOIN bus_users 
        ON bus_users.purch = bus.bus_id 
        GROUP BY bus.bus_id
    ) Sub2
    ON Sub1.bus_id = Sub2.bus_id
) Sub0
GROUP BY bus_id

SQL在这里小提琴： -

http://www.sqlfiddle.com/#!2/c8c9c/1

有条件地将查询中的元素插入另一个数据库表

1 个答案: