Question

所以我在phpmyadmin和我的php代码中运行相同的查询。

查询是：

SELECT comments.roomID,comments.message, comments.dateTimeSent, sender.fname ,sender.lname,receiver.fname, receiver.lname
FROM comments
INNER JOIN users as sender ON comments.senderID = sender.id
INNER JOIN users as receiver ON comments.receiverID = receiver.id
INNER JOIN chatRooms ON comments.roomID = chatRooms.id WHERE comments.roomID = 8;

当我直接从phpmyadmin面板运行时，我得到的结果就是这个

但是当我在我的PHP代码中运行时，我得到了这个结果：

Array
(
    [roomID] => 8
    [message] => Hello from mysql database
    [dateTimeSent] => 2018-03-16 11:04:03
    [id] => 23
    [fname] => pavlos
    [lname] => elpidorou
)
Array
(
    [roomID] => 8
    [message] => asdasd;asda
    [dateTimeSent] => 2018-03-16 11:21:30
    [id] => 25
    [fname] => Antreas
    [lname] => antoniou
)

数组中缺少sender.fname，sender.lname，receiver.fname，receiver.lname

我用来执行查询的代码，得到的结果如下

foreach ($chatRoomArray as &$room) {
            $roomID = $room['id'];

            $query = "SELECT comments.roomID,comments.message, comments.dateTimeSent,sender.id, sender.fname ,sender.lname,receiver.fname, receiver.lname
                      FROM comments
                      INNER JOIN users as sender ON comments.senderID = sender.id
                      INNER JOIN users as receiver ON comments.receiverID = receiver.id
                      INNER JOIN chatRooms ON comments.roomID = chatRooms.id WHERE comments.roomID = ".$roomID;


            $stmt = $this->conn->prepare($query);
            $result = $stmt->execute();

            $commentArray = array();

            if ($result) {
                $num = $stmt->rowCount();
                if ($num > 0) {
                    while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
                        extract($row);
                        print_r($row);
                        $comment = array(
                            "commentID" => $row['id'],
                            "message"=>$row['message'],
                            "dateTimeSent"=>$row['dateTimeSent'],
                            "senderFname"=>$row['sender.fname'],
                            "senderLname" => $row['sender.lname'],
                            "receiverFname" => $row['receiver.fname'],
                            "receiverLname" => $row['receiver.lname']
                        );

                    }
                }

            }
        }

Answer 1

PHP + MySQL不会在字段名称中包含表名。

当您运行查询时，请在SELECT部分为其命名。

SELECT `sender`.`fname` AS `sender_fname`, ...

然后它将作为sender_fname位于您的数组中。我会避免在你的名字中使用圆点。

Answer 2

您必须添加唯一的别名

sender.fname, sender.lname, receiver.fname,receiver.lname

例如：

SELECT comments.roomID,comments.message, comments.dateTimeSent, sender.fname as 'fname1' ,sender.lname as'lname1',receiver.fname as 'fname2', receiver.lname as 'lname2'
FROM comments
INNER JOIN users as sender ON comments.senderID = sender.id
INNER JOIN users as receiver ON comments.receiverID = receiver.id
INNER JOIN chatRooms ON comments.roomID = chatRooms.id WHERE comments.roomID = 8;

然后进入php

"senderFname"=>$row['fname1'],
                            "senderLname" => $row['lname1'],
                            "receiverFname" => $row['fname2'],
                            "receiverLname" => $row['lname2']

MySQL查询不返回PHP中的值

2 个答案: