Question

我正在开发一个用于编辑董事会会议的页面，我想要显示所有 没有参加 特定会议的董事会成员作为位于下方的检查员编辑如果用户想要添加更多，所以我这样做：

我的代码：

$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>";
        }

    }

DB：

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`)
)

通过我的代码，我得到了所有董事会成员，包括谁参加，如何解决？

Answer 1

您需要使用LEFT JOIN才能找到未参加特定会议的BoardTeam中的人员。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

如果您想获得所有董事会成员，然后在PHP中确定他们是否参加了会议，只需删除m.attendance IS NULL条款，如下：

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

现在当你遍历php中的响应行时，你可以这样测试（假设你逐行将你的行提取到$ row变量中）：

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

另外，正如评论中所提到的，使用mysqli或pdo而不是纯mysql_函数

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

如何编写嵌套查询MySQL PHP

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