How to write nested query MySQL PHP - php

Question

I'm developing a page to edit board meetings and I want to display all board members who did not attend specific meeting as a checkox located below who attend as an edit in case of user want to add more so I did this:
My code:
$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`)
)
With my code I got all board members including who attend, How to fix that ??

Answer 1

You need to use a LEFT JOIN in order to find people in the boardTeam who were not in a specific meeting. eg:
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
If you want to get ALL board members, and then determine within PHP if they attended the meeting or not, simply remove the m.attendance IS NULL clause, as such:
SELECT b.*, m.attendance as attendance
FROM boardTeam b
LEFT JOIN meetingattendance m
ON b.nationalID = m.attendance AND m.meetingID = $mid
and now when you loop through the response rows in php, you can test as such (assuming you fetch your rows one by one into a $row variable):
if($row['attendance'] != null)
{
// attended meeting
}
else
{
// did not attend meeting
}
Also, as mentioned in the comments, use mysqli, or pdo instead of pure mysql_ functions
Example fiddle here: http://sqlfiddle.com/#!9/ba7d4/6