如何比較來自不同查詢的2個單詞？

Question

請幫我，如何比較2名稱從不同的查詢？ 我想比較subject1與subject2，如果它的相同單詞將打印'是'（如圖像）。

所以這是我的代碼，我的代碼剛剛讀取了subject1的最後一個查詢。 那麼如何比較來自subject1和所有subject2的所有查詢？

<table class="table table-bordered"> 
       <tr> 
        <td class="table-bordered">No</td> 
        <td class="table-bordered">Date</td> 
        <td class="table-bordered">Subject Name</td> 
        <td class="table-bordered">Subject No</td> 
        <td class="table-bordered">Revision No</td> 
        <td class="table-bordered">Trainer</td> 
        <td class="table-bordered">Institution</td> 
       </tr> 
       <?php 
        $query = mysql_query("SELECT * FROM training_detail AS s JOIN materi AS t JOIN training_history AS u JOIN employee AS v WHERE v.employee_id = $id1 AND s.nik LIKE v.nik AND u.id_materi LIKE t.id_materi AND s.id_riwayat_training LIKE u.id_riwayat_training "); 
        $i=1; 
        while($row1 = mysql_fetch_array($query)) 
        { 
         $date = $row1['date']; 
         $subject1 = $row1['subject_name']; 
       ?> 
       <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['subject_no']; ?></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>  
       <?php 
        $i++; 
        } 
       ?> 
      </table> 
      </br> 
      <table class="table table-bordered"> 
       <tr> 
        Training yang belum diikuti 
        <td class="table-bordered">No</td> 
        <td class="table-bordered"></td> 
        <td class="table-bordered">subject id</td> 
        <td class="table-bordered">subject namei</td> 
        <td class="table-bordered">subject no</td> 
       </tr> 
       <?php 
        $query = mysql_query("SELECT * FROM job_name AS r JOIN subject AS q JOIN employee AS p WHERE q.subject_id LIKE r.header_id AND r.job_id LIKE p.id_jabatan AND p.id_karyawan = $id1 ORDER BY q.id_materi ASC "); 
        $i=1; 
        while($row2 = mysql_fetch_array($query)) 
        { 
         $subject2 = $row2['subject_name']; 
         if (strcasecmp($subject1, $subject2) == 0) 
         { 
       ?> 
       <tr> 
        <td class="table-bordered"><?php echo $i; ?></td> 
        <td class="table-bordered"><?php echo "YES" ?></td> 
        <td class="table-bordered"><?php echo $row2['subject_id'] ?></td> 
        <td class="table-bordered"><?php echo $subject2; ?></td> 
        <td class="table-bordered"><?php echo $row2['subject_no']; ?></td> 
       </tr>  
       <?php 
         } 
         else 
         { 
       ?>   
          <tr> 
           <td class="table-bordered"><?php echo $i; ?></td> 
           <td class="table-bordered"><?php echo "NO" ?></td> 
           <td class="table-bordered"><?php echo $row2['subject_id'] ?></td> 
           <td class="table-bordered"><?php echo $subject2; ?></td> 
           <td class="table-bordered"><?php echo $row2['subject_no']; ?></td> 
          </tr> 
       <?php   
         } 
        $i++;  
        } 

       ?> 
      </table> 

Answer 1

你需要使用嵌套循環我猜。在第一個裏面使用第二個。

順便說一句http://php.net/manual/en/control-structures.alternative-syntax.php，它可能會讓你的代碼看起來更好。