SQL QUERY 没有达到我的预期

SQL QUERY Not resulting what i expect

我对这段代码感到困惑,我在 mysql 上创建了存储过程 这是我的存储过程代码

CREATE DEFINER=`root`@`localhost` PROCEDURE `getChats`(
    IN `idUser` VARCHAR(50)

)
LANGUAGE SQL
NOT DETERMINISTIC
READS SQL DATA
SQL SECURITY DEFINER
COMMENT ''
BEGIN

SET @sqlQuery = 'SELECT user.id AS owner_id, user.name, user.image, chats.sender, chats.recipient, chats.date_sent, chats.content FROM chats 
INNER JOIN (
    SELECT MAX( id_chat ) AS id_chat
    FROM chats AS alt 
    WHERE alt.recipient = ?
    OR alt.sender = ?
    GROUP BY LEAST(recipient, sender), greatest(recipient, sender)
) AS chat ON chats.id_chat = chat.id_chat
INNER JOIN user ON user.id = CASE WHEN chats.sender = ?
    THEN chats.recipient
    ELSE chats.sender
END';

PREPARE sqlQuery FROM @sqlQuery;
execute sqlQuery using @idUser, @idUser, @idUser;


END

这是CALL getChats('USER15703082085d99007096571')

的结果

https://puu.sh/EpRhZ.png Result from Heidi SQL

https://puu.sh/EpRmz.png Result from phpmyadmin

https://puu.sh/EpRub.png Result from php pdo

我的php代码

$conn = new PDO(sprintf("mysql:host=%s;dbname=%s", "localhost",  "kosan_finder"), "root", "");
$sql = "CALL getChats(?)";
$stmt = $conn->prepare($sql);
$stmt->bindValue(1, $PostData->sender);

$stmt->execute();
echo "Result stmt: ";
print_r($stmt->fetchAll(PDO::FETCH_ASSOC));
echo "\nError info: ";
print_r($stmt->errorInfo());

我需要的是我可以在 php

上获取数据

无需在存储过程中使用动态SQL。此外,过程参数可以按原样使用(而不是@param)。

正确使用参数的代码如下所示:

CREATE DEFINER=`root`@`localhost` PROCEDURE `getChats`(
idUser VARCHAR(50)
)
BEGIN

SELECT user.id AS owner_id, user.name, user.image, chats.sender, chats.recipient, chats.date_sent, chats.content 
FROM chats INNER JOIN (
    SELECT MAX( id_chat ) AS id_chat
    FROM chats AS alt 
    WHERE alt.recipient = idUser OR alt.sender = idUser
    GROUP BY LEAST(recipient, sender), greatest(recipient, sender)
) AS chat ON chats.id_chat = chat.id_chat
INNER JOIN user ON user.id = 
    CASE WHEN chats.sender = idUser
    THEN chats.recipient
    ELSE chats.sender
END;

END

在查询中使用 GROUP BY 看起来很奇怪。不确定您要用它做什么。