如何編寫嵌套查詢MySQL PHP

Question

我正在開發一個頁面以編輯董事會會議，並且我希望將未參加特定會議的所有董事會成員顯示為下面的checkox，如果要添加更多內容，請作為編輯參加會議，因此我這樣做：

我的代碼：

$q = "SELECT * FROM `boardteam`";
    $r = mysql_query($q);
    while ($dbfield = mysql_fetch_assoc($r))
    {
        $member_id =$dbfield['nationalID'];
        $query = "SELECT `attendance` FROM `meetingattendance` WHERE `meetingID` = '$mid' AND `attendance`!= '$member_id'";
    $res = mysql_query($query);
    if ($res)
        {
        $tname ="";
        switch ($dbfield['titleName'])
        {
            case "Dr":
            $tname .= "د.";
            break;
            case "Ms":
            $tname .= "السيدة.";
            break;
            case "Mr":
            $tname .= "السيد.";
            break;
        }
        $At .= "<input type='checkbox' name='moreAttendence[]' dir='rtl' value=".$dbfield['nationalID']."><div class='styled-checkbox'>".$tname." ".$dbfield['fName']." ".$dbfield['sName']." ".$dbfield['lName']."</div><br>";
        }

    }

D B：

CREATE TABLE `boardteam` (
`nationalID` int(10) NOT NULL,
`titleName` char(2) NOT NULL,
`fName` char(20) NOT NULL,
`sName` char(20) NOT NULL,
`lName` char(20) NOT NULL,
`gender` char(1) NOT NULL,
`birthDate` date DEFAULT NULL,
`materialStatus` char(15) DEFAULT NULL,
`jobTitle` varchar(100) NOT NULL,
`jobLocation` varchar(20) DEFAULT NULL,
`employer` varchar(100) DEFAULT NULL,
`email` varchar(100) NOT NULL,
`photo` varchar(255) DEFAULT NULL,
`academicGrade` char(15) DEFAULT NULL,
`employmentStartDate` date NOT NULL,
`employmentEndDate` date NOT NULL,
`employmentType` char(20) DEFAULT NULL,
`employmentStatus` char(15) DEFAULT NULL,
`jobStartDate` date DEFAULT NULL,
`jobNumber` int(10) DEFAULT NULL,
`cv` varchar(255) DEFAULT NULL,
PRIMARY KEY (`nationalID`)
)


CREATE TABLE `meetingattendance` (
 `meetingID` int(11) NOT NULL,
 `attendance` int(10) DEFAULT NULL,
 `absence` int(10) DEFAULT NULL,
 `reason` varchar(255) DEFAULT NULL,
 `additionalAttendance` varchar(255) DEFAULT NULL,
 KEY `absence` (`absence`),
 KEY `meeingID` (`meetingID`),
 KEY `attendance` (`attendance`),
 CONSTRAINT `meetingattendane_ibfk_1` FOREIGN KEY (`meetingID`) REFERENCES `boardmeetings` (`meetingID`),
 CONSTRAINT `meetingattendane_ibfk_2` FOREIGN KEY (`attendance`) REFERENCES `boardteam` (`nationalID`),
 CONSTRAINT `meetingattendane_ibfk_3` FOREIGN KEY (`absence`) REFERENCES `boardteam` (`nationalID`)
)

有了我的代碼，我讓所有董事會成員，包括參加會議的人，如何解決？

Answer 1

您需要使用LEFT JOIN才能在boardTeam中找到不在特定會議中的人員。 例如：

SELECT b.*, m.attendance
  FROM boardTeam b
    LEFT JOIN meetingattendance m
      ON b.nationalID = m.attendance AND m.meetingID = $mid
    WHERE m.meetingID IS NULL

如果要獲取所有董事會成員，然后在PHP中確定他們是否出席會議，只需刪除m.attendance IS NULL子句，如下所示：

SELECT b.*, m.attendance as attendance
  FROM boardTeam b
    LEFT JOIN meetingattendance m
      ON b.nationalID = m.attendance AND m.meetingID = $mid

現在，當您遍歷php中的響應行時，您可以進行這樣的測試（假設您將一行一行地放入$ row變量中）：

if($row['attendance'] != null)
{
  // attended meeting
}
else
{
  // did not attend meeting
}

另外，如注釋中所述，使用mysqli或pdo代替純mysql_函數

示例小提琴在這里： http ://sqlfiddle.com/#!9/ ba7d4/6