如何比较2个表中每行的2个值？

Question

如何比较2表中的2个值？

我已经使用2 while（nested while），但表中检查一个一个，第二个表中的结果是从第一个表中的每一行。 如何使结果打印一次？

我想这样 result's image

，这里的结果是我目前的代码

<?php    
        $query1 = mysql_query("SELECT * FROM riwayat_training_detail AS s JOIN materi AS t JOIN riwayat_training AS u JOIN karyawn AS v WHERE v.id_karyawan = $id1 AND s.nik LIKE v.nik AND u.subject no LIKE t.subject no AND s.id_riwayat_training LIKE u.id_riwayat_training "); 
        $i=1; 
        while($row1 = mysql_fetch_array($query1)) 
        { 
         $date = $row1['date']; 
         $subject1 = $row1['subject name']; 
       ?> 
      <table class="table table-bordered"> 
       <tr> 
        Training yang diikuti 
        <td class="table-bordered">No</td> 
        <td class="table-bordered">date</td> 
        <td class="table-bordered">subject date</td> 
        <td class="table-bordered">No SOP</td> 
        <td class="table-bordered">revision no</td> 
        <td class="table-bordered">Trainer</td> 
        <td class="table-bordered">institution</td> 
       </tr> 

       <tr> 
        <td class="table-bordered"><?php echo $i; ?></td> 
        <td class="table-bordered"><?php echo date("j/F/Y", strtotime($date)); ?></td> 
        <td class="table-bordered"><?php echo $subject1; ?></td> 
        <td class="table-bordered"><?php echo $row1['no_sop']; ?></td> 
        <td class="table-bordered"><?php echo $row1['revision_no']; ?></td> 
        <td class="table-bordered"><?php echo $row1['trainer']; ?></td> 
        <td class="table-bordered"><?php echo $row1['institution']; ?></td> 
       </tr> 
      </br> 
      <table class="table table-bordered"> 
       <tr> 
        Training yang belum diikuti 
        <td class="table-bordered">No</td> 
        <td class="table-bordered">subject no</td> 
        <td class="table-bordered">subject date</td> 
        <td class="table-bordered">No SOP</td> 
       </tr> 
       <?php 
        $query2 = mysql_query("SELECT * FROM header_jabatan AS r JOIN materi AS q JOIN karyawn AS p WHERE q.subject no LIKE r.id_header AND r.id_jabatan LIKE p.id_jabatan AND p.id_karyawan = $id1 ORDER BY q.subject no ASC"); 
        $x=1; 
        while($row2 = mysql_fetch_array($query2)) 
        { 
         $subject2 = $row2['subject name']; 
         if (strcasecmp($materi1, $materi2) != 0) 
         { 
       ?> 
       <tr> 
        <td class="table-bordered"><?php echo $i; ?></td> 
        <td class="table-bordered"><?php echo $row2['subject no'] ?></td> 
        <td class="table-bordered"><?php echo $subject22; ?></td> 
        <td class="table-bordered"><?php echo $row2['no_sop']; ?></td> 
       </tr>  
       <?php 
         } 
        $x++;  
        } 
        $i++; 
       } 

       ?> 
      </table> 
     </table> 

请帮我我应该怎么办？

这是我用同一​​主题的第二个问题，我真的需要帮助，即时通讯坚持这个..

Answer 1

可以使用CASE声明检查此链接，详细介绍如何使用它。

check this link

Answer 2

这是不是沿着你想要的路线？

UPDATE table2 SET Status = "YES" WHERE SubjectName IN (SELECT SubjectName FROM table1) 

还是你想要它也使用修订？您的照片只会谈论SubjectName。

Answer 3

在你的情况对上述结果的图像时，最好使用LEFT JOIN或RIGHT JOIN

见下面的例子：

> CREATE TABLE table1 (subject VARCHAR(10), value INT); 
> CREATE TABLE table2 (id INT, subject VARCHAR(10)); 

> INSERT INTO table1 VALUES ('XXX', 23),('ZZZ', 24); 
> INSERT INTO table2 VALUES (1, 'XXX'),(2, 'YYY'), (3, 'ZZZ'); 

> SELECT * FROM table1; 
+---------+-------+ 
| subject | value | 
+---------+-------+ 
| XXX  | 23 | 
| ZZZ  | 24 | 
+---------+-------+ 

> SELECT * FROM table2; 
+------+---------+ 
| id | subject | 
+------+---------+ 
| 1 | XXX  | 
| 2 | YYY  | 
| 3 | ZZZ  | 
+------+---------+ 
> SELECT a.*, IF(b.value IS NULL, 'No', 'Yes') as status FROM table2 as a LEFT JOIN table1 as b ON b.subject = a.subject; 
+------+---------+--------+ 
| id | subject | status | 
+------+---------+--------+ 
| 1 | XXX  | Yes | 
| 3 | ZZZ  | Yes | 
| 2 | YYY  | No  | 
+------+---------+--------+