動態地從MySQL

Question

生成HTML表我有三個MySQL表科目，考試，examinfo

表 - 科目

1. subjectid
2. 主旨名稱
3. subjectExamid

表 - EXAMINFO

1. examid
2. 考試

TABLE EXAMINATIONS

1. FNAME
2. L-NAME
3. studentid
4. 得分

5. subjectid

   ON subject.subjectExamid = exam.examid

   ON examination.subjectid = subject.subjectid

現在我想生成HTML表格，顯示學生成績每得到紙對主題

結構表輸出針對每個主題 學生細節得分

編輯的代碼示例

<?php 
    $examinid = 3; 
    $subjects = mysqli_query(
     $con," 
      SELECT * FROM subjects 
      WHERE examid = '$examinid' 
      ORDER BY shortname ASC 

    "); 
    $content = mysqli_query(
     $con," 
      SELECT DISTINCT exam.idcandidate, exam.sex, exam.fname, exam.lname 
      FROM examinations 
      AS exam 
      INNER JOIN examinfo 
      AS info 
      ON exam.id_subject = info.idsubject 
      WHERE info.idexam = '$examinid'   
    "); 
?> 
<div id="table_1"> 
    <table cellpadding="0" cellspacing="0" border="0"> 
     <tr> 
      <td class="table1tr">#</td> 
      <td class="table1tr">Candidate</td> 
      <td class="table1tr">ID</td> 
      <td class="table1tr">Sex</td> 
      <?php 
       // output subjects 
       while($subRow = mysqli_fetch_array($subjects)){ 
        $arbv = strtoupper($subRow['shortname']); 
        $subjectname = ucwords(strtolower($subRow['subjectname']." - ".$subRow['subjectid']."")); 
      ?> 
      <td class="table1tr" title="<?php echo $subjectname; ?>"> 
       <?php echo $arbv; ?> 
      </td> 
      <?php 
       } 
      ?>     
      <td class="table1tr">Exam</td> 
     </tr> 
     <?php 
      while($stdnt = mysqli_fetch_array($content)){ 
       $fullname = ucwords(strtolower("$stdnt[lname] $stdnt[fname]")); 
       $studentid = str_replace(array('/', 'M', 'W', 'S', 'F', '-'), "",$stdnt['idcandidate']); 
       if($sex = $stdnt['sex'] == Male){ 
        $sex = M; 
       }else{ $sex = F; } 
       $id_subject = $stdnt['id_subject']; 
       $x++; 
       $zebra_1 = ($x%2)? 'TableZebra_1': 'TableZebra_2'; 
     ?>   
     <tr> 
      <td class="<?Php echo $zebra_1; ?>"><?php echo $count++; ?></td>  
      <td class="<?Php echo $zebra_1; ?>"><?Php echo $fullname; ?></td> 
      <td class="<?Php echo $zebra_1; ?>"><?php echo $studentid; ?></td>  
      <td class="<?Php echo $zebra_1; ?>"><?php echo $sex; ?></td> 
      <td class="<?Php echo $zebra_1; ?>"> 
      <!-- Problem is here how to output the subject grades $grade --> 
      <!-- 
       My first unsuccessful approach 

       SELECT score 
       FROM examinations AS test 
       INNER JOIN examinfo AS testinfo ON testinfo.idsubject = test.id_subject 
       WHERE testinfo.idexam 
       IN (

       SELECT idexam 
       FROM examinfo 
       WHERE idexam = $examinid 
       ) 
       AND test.id_subject = $id_subject AND test.idcandidate = '$studentid' 

       Then output results - But this falls it shows one student subjects in one cell 
      --> 
      </td> 
      <td class="<?Php echo $zebra_1; ?>">Exam</td> 
     </tr>   
      <?php 
       } // loop content 
      ?>  

    </table> 
</div> 

Answer 1

如果您的解決方案是不CONCAT（）;你可能首先

按照簡單的步驟

1 loop $contents // to get info such as studentid 
2 inside the loop of $contents loop $subjects // to get all subjects including subjectids 
3 inside $subject loop, loop examinations table where studentid = '$studentid' AND subjectid = '$subjectid' 
if step three return null echo empty cell otherwise echo cell with score 

我沒有時間來測試這一點，但你可以遵循的步驟，它會工作，否則嘗試谷歌搜索