Question

我做了这种在数据库中添加数据的形式。我的考试表包含exam_code（PK），exam_title和subject_code（FK）。这是设计

<div style="width:800px;height:auto;margin-left:auto;margin-right:auto;margin-top:50px;">
  <form action="" method="POST" class="form-horizontal" role="form">
    <div class="form-group">
      <div class="col-xs-6 col-sm-3 ">
        <input name="code" type="text" class="form-control" id="excode" placeholder="Enter Exam Code">
      </div>
      <div class="col-xs-6 col-sm-3 ">
        <input name="title" type="text" class="form-control" id="extitle" placeholder="Enter Exam Title">
      </div>
      <div class="col-xs-6 col-sm-3 ">
        <select name="subjcode" class="form-control">
            <option selected="selected">Choose subject</option>
            <option disabled="disabled">---------------------------------</option>
            <?php 
               include('db.php');
               $subj = $connect->query("SELECT subject_code FROM subject");
               while($row1 = mysqli_fetch_array($subj)){
                  echo "<option value = $row1[subject_code]>$row1[subject_code]</option>";
               } 
            ?>
        </select>
        </div>
        <div class="col-xs-6 col-sm-3 ">
        <input type="submit" name="add" class="btn btn-default" value="Add" />
      </div>
    </div>
  </form>
</div>

我的查询在这里是否正确？无论如何我都想不到插入数据。这里..

<?php 
    include('db.php');
    if(isset($_POST['add'])){
        $excode = $_POST['code'];
        $extitle = $_POST['title'];
        $subcode = $_POST['subjcode'];

        $examinsert = $connect->query("INSERT INTO exam (exam_code, exam_title, subject_code) VALUES ('$excode', '$extitle', '$subcode')");

        if(!$examinsert){
             die("<script>
                    alert('Error encountered, Reloading page');
                    window.location.href='teacher.php';
                 </script>");
        }else{
            die("<script>
                   alert('Your exam title has been added. You will see your titles in the Examination title section below!');
                   window.location.href='teacher.php';
                 </script>");
         }
     }

 ?>

Answer 1

更改您的PHP代码来自

<?php 
     include('db.php');
     $subj = $connect->query("SELECT subject_code FROM subject");
     while($row1 = mysqli_fetch_array($subj)){
       echo "<option value = $row1[subject_code]>$row1[subject_code]</option>";
     } 
?>

到此

<?php 
         include('db.php');
         $subj = $connect->query("SELECT subject_code FROM subject");
         while($row1 = mysqli_fetch_array($subj)){
           echo "<option value = ".$row1[subject_code].">".$row1[subject_code]."</option>";
         } 
    ?>

Answer 2

将值存储在Variable中，然后将其存储起来。

    <?php 
             include('db.php');
             $subj = $connect->query("SELECT subject_code FROM subject");
             while($row1 = mysqli_fetch_array($subj)){
               $subjectCode = $row1[subject_code];
               echo "<option value = $subjectCode>$subjectCode</option>";
             } 
        ?>

它的作品。

Answer 3

我通过添加禁用外键的查询来修复它。

$set = $connect->query('SET foreign_key_checks = 0');
/*insert query*/
$set1 = $connect->query('SET foreign_key_checks = 1');

如何从数据库中获取select-option标签的值

3 个答案: