予取一些数据从相同的表名,但具有不同的条件,余查询它的3倍,其结果是:如何从不同条件的相同表中求和数据?
101=>1
301=>1
501=>2
502=>4
---------------
101=>2
501=>1
---------------
101=>1
501=>1
其中第一列是教室和第2栏是值。什么是同一个教室,总结这些价值的最佳方式,因此结果将是:
101 = 4
301 = 1
501 = 4
502 = 4
我的查询命令:
$query = $db->prepare("SELECT COUNT(std_id) AS total, std_class FROM attendance WHERE att_mode IN('mode-01','mode-04') AND att_attend='0' AND att_date=CURDATE() GROUP BY std_class ORDER BY std_class");
$query->execute();
while ($char = $query->fetch(PDO::FETCH_OBJ)) {
$labels2[] = $char->std_class.'-'.$char->total;
$link[] = $char->std_class;
}
$query = $db->prepare("SELECT COUNT(std_id) AS total_sick,std_class FROM attendance WHERE att_mode ='mode-01' AND att_attend='2' AND att_date=CURDATE() GROUP BY std_class ORDER BY std_class");
$query->execute();
while($char = $query->fetch(PDO::FETCH_OBJ)){
$sickCount[] = $char->std_class.'-'.$char->total_sick;
}
$query = $db->prepare("SELECT COUNT(std_id) AS total_leave,std_class FROM attendance WHERE att_mode ='mode-01' AND att_attend='3' AND att_date=CURDATE() GROUP BY std_class ORDER BY std_class");
$query->execute();
while($char = $query->fetch(PDO::FETCH_OBJ)){
$leaveCount[] = $char->std_class.'-'.$char->total_leave;
}
我得到这个错误:SQLSTATE [21000]:基数冲突:1242子查询返回多于1行 – Wasinha