Question

我有这两张桌子。连接“密钥”是conID，它包含在两个表中。所以现在我想写一些select语句，它会给我这样的东西：

John Smith dairy22 Texas 4000 smth4

迈克情况glenn32球衣1000 smth1 ...

＆GT;表“人”：

NUM   Name lastName  address    conID
-----------------------------------------
  1    John  Smith     dairy22   Texas
  2    Mike  Situation glenn32   Jersey
  3    Duke  Nukem     haris48   NYork
  4    Queen Lisa      court84   London

＆GT;表“国家”

conID   postNum   region
-------------------------
Jersey  1000      smth1
NYork   2000      smth2
London  3000      smth3
Texas   4000      smth4

！ - ＆GT; NUM是AUTO INCREMENT主键，如果可能，我不希望输出。

提前感谢您的帮助：））

Answer 1

这应该根据num列返回您想要的行：

SELECT Name, lastName, address, people.conID, postNum, region 
FROM people 
JOIN countries 
ON people.conID = countries.conID 
WHERE num=1

Answer 2

您可以使用以下MySQL查询：

SELECT
    Name,
    lastName,
    address,
    people.conID,
    postNum,
    region
FROM
    people
JOIN
    countries
        ON people.conID = countries.conID
WHERE
    num = 1

Answer 3

这可能会起到作用：

SELECT name, lastName, address, people.conID, postNum, region
FROM people
 JOIN countries ON people.conID = countries.conID;

Answer 4

您的SQL查询应如下所示：

SELECT p.Name, p.lastName, p.address, p.conID, c.postNum, c.region
FROM people p 
LEFT JOIN countries c ON p.conID = c.conID

Answer 5

Good ole USING条款：

SELECT Name, lastName, address, p.conID, postNum, region FROM people p
INNER JOIN countries USING (conID);

Answer 6

SELECT Name, lastName, address, people.conID, postnum, region
FROM people JOIN countries ON people.conID = countries.conID;

在SELECT（MYSQL / PHP）中连接2个表

6 个答案: