php-pdo代码比较两个表中的数据并返回结果

我有两个表

表1

书籍收据优惠券

100 701 501502

100 702 503504

100 703 505506

表2

书籍1收据1耦合1

100 701 501502

100 702 503504

我正在根据bookno从表1中搜索数据。

现在我需要检查我是否输入bookno=100并点击搜索

然后它检查表2的所有收据和优惠券…

并且仅返回表2中不存在的那些值。

在上述示例中

下一行不在表2中，但在表1中存在

100 703 505506

所以查询只返回此行

请帮我做这个。。。

<?php   
$receipt = "";
$db=new PDO('mysql:host=localhost;dbname=circulation_scheme_prepaid','root','');
if($_POST && isset($_POST['search']))
{       
    $result=$db->prepare('SELECT DISTINCT receipt_no FROM scheme_master WHERE book_no2=:book_no2');
    $result->bindParam(':book_no2',$_POST['book_no']);
    $result->execute(); 
    $data1 = $result->fetchAll();
    $coupons = array(); 
        foreach($data1 as $row)
        {    
            $receipt[] = $row['receipt_no']; 
        }   
}
?>

    SELECT DISTINCT master.receipt_no 
    FROM scheme_master AS master 
        JOIN book_issue_return  AS bir  ON master.bookno2 = bir.book_no 
 WHERE master.book_no2=:book_no2  AND (
    bir.receiptno  NOT IN 
        (SELECT master.receipt_no FROM scheme_master AS master ) AND 
    bir.couponno NOT IN (SELECT master.receiptno FROM scheme_master AS master ))

$data1 = $result->fetchAll();
if (count($data1)) {
    // return all the data
} else {
    // return message
}

SELECT DISTINCT SM.receipt_no FROM scheme_master AS SM
JOIN book_issue_return AS bir ON SM.book_no2 = bir.book_no
WHERE SM.book_no2 = :book_no2 AND (SM.receipt_no NOT IN (SELECT bir.receiptno FROM book_issue_return AS bir))