Question

我有两张表complain和repair。如果ass_to_per中不存在repair complain，我希望从id或complain获取repair [最新一个]。< / p>

说明：

我想从ass_to_per表中获取complain，也从repair表中获取repair。但是它们之间存在关系，在com_id中有一个名为ass_to_per的字段，它是一个外键。所以我想从complain获取repair，并检查ass_to_per是否有外键，如果有的话，请检查修复的sqlfiddle。如果有，那么得到它作为结果。

我有CREATE TABLE `complain` ( `id` int(11) NOT NULL, `ass_to_per` varchar(50) NOT NULL ); INSERT INTO `complain` (`id`, `ass_to_per`) VALUES (1, 'frayne'), (2, 'murad'), (4, 'frayne'), (5, 'murad'), (6, 'frayne'), (7, 'frayne');，对于在线测试：sqlfiddle.com

下面给出的表格和数据。

投诉

CREATE TABLE `repair` (
  `id` int(11) NOT NULL,
  `com_id` int(11) NOT NULL,
  `ass_to_per` varchar(50) NOT NULL
);
INSERT INTO `repair` (`id`, `com_id`, `ass_to_per`) VALUES
(1, 1, 'frayne'),
(2, 1, 'murad'),
(3, 4, 'frayne'),
(4, 6, 'murad'),
(5, 2, 'murad'),
(6, 5, 'frayne');

修复

SELECT * FROM `complain` 
WHERE `id` IN (SELECT DISTINCT(`com_id`) FROM `repair` WHERE `ass_to_per` = 'frayne') OR `ass_to_per`='frayne'

我的查询：

id |    ass_to_per
--------------
1  |    frayne
4  |    frayne
5  |    murad
6  |    frayne
7  |    frayne

查询结果

id | ass_to_per[complain] | ass_to_per[repair]
--------------
1  |    murad             |    frayne
2  |    murad             |    murad
4  |    frayne            |    frayne          //need this one
5  |    murad             |    frayne          //need this one
6  |    frayne            |    murad
7  |    frayne            |                    //need this one

分析

id |    ass_to_per
--------------
4  |    frayne        //ass_to_per from repair
5  |    frayne        //ass_to_per from repair
7  |    frayne        //ass_to_per from complain

预期结果：

<?php
$servername = "localhost";
$username = "username";
$password = "password";


$conn = mysqli_connect($servername, $username, $password);

if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";
?>

if ($result = mysqli_query($link, "SELECT * FROM City", MYSQLI_USE_RESULT)) {

    if (!mysqli_query($link, "SET @a:='this will not work'")) {
        printf("Error: %s\n", mysqli_error($link));
    }
    mysqli_free_result($result);
}

mysqli_close($link);
?>

Answer 1

COALESCE()将输出它找到的第一个非null参数，因此使用连接可能会从抱怨或修复中获取ass_to_per，具体取决于存在哪个：

SELECT 
    complain.id, 
    COALESCE(repair3.ass_to_per, complain.ass_to_per) as ass_to_per
FROM complain
LEFT JOIN 
    (SELECT max(id) as maxid, com_id FROM repair GROUP BY com_id) as repair2 
    ON complain.id = repair2.com_id
LEFT JOIN repair as repair3
    ON repair2.maxid = repair3.id
GROUP BY complain.id

如果你想在计算出的ass_to_per上进一步过滤（比如'frayne'中的例子），只需将此选项嵌入子查询：

SELECT * 
FROM (
  SELECT 
    complain.id, 
    COALESCE(repair3.ass_to_per, complain.ass_to_per) as ass_to_per
  FROM complain
  LEFT JOIN (SELECT max(id) as maxid, com_id FROM repair GROUP BY com_id) AS repair2 
    ON complain.id = repair2.com_id
  LEFT JOIN repair as repair3
    ON repair2.maxid = repair3.id
  GROUP BY complain.id
) AS mydata 
WHERE mydata.ass_to_per = 'frayne'
ORDER BY mydata.id;

在这里小提琴：http://sqlfiddle.com/#!9/33433/49

Answer 2

SELECT c.id
  FROM complain c 
  LEFT 
  JOIN 
     ( SELECT x.*
         FROM repair x
         JOIN 
            ( SELECT com_id,MAX(id) id FROM repair GROUP BY com_id ) y
           ON y.com_id = x.com_id
          AND y.id = x.id
     ) r
    ON r.com_id = c.id
 WHERE COALESCE(r.ass_to_per,c.ass_to_per) = 'frayne';

MySql - 从表2中获取字段，如果不是，则从表1中获取

2 个答案: