Question

我的数据库中有3个表（standards，courses，students）。我想在Fname表格中显示gender和students，这些表格已在选定的＆＃39;标准＆＃39;和＆＃39;当然＆＃39;从下拉列表 - 但是＆＃34;提交＆＃34;按钮似乎不起作用。代码如下所示。我确信它与数据库相关联，作为“课程”的下拉菜单。和＆＃39;标准＆＃39;工作：

 <html>
  <head>
    <title>Courses</title>
    <link rel="stylesheet" type="text/css" href="css/style.css">
   </head>
    <body>
     <div id="container">
        <div id="wrapper">
           <h1> Students</h1>   
              <div id="data">
            <form action="index.php" method="POST">
         <select name="standards">
            <option>Standard</option>
           <?php
            include 'includes/dbconnect.php';
             $query1 = "SELECT * FROM standards";
            $result1 = mysql_query($query1);
          while($rows1 = mysql_fetch_array($result1)){
              $standardID = $rows1['id'];
              $rowsData1 = $rows1['standardName'];

          ?>
           <option value="<?php echo $standardID; ?>">
        <?php
          echo $rowsData1; ?></option>
       <?php
         }
    ?>
    </select>
    </div>
    <div id="data2">
    <select name="courses">
    <option>Courses</option>
       <?php
         $query2 = "SELECT * FROM courses";
         $result2 = mysql_query($query2);
         while($rows2 = mysql_fetch_array($result2)){
              $coursesID = $rows2['id'];
              $rowsData2 = $rows2['courseName'];
          ?>
          <option value="<?php echo $courseID;?>">
       <?php echo $rowsData2; ?></option> <?php }?>
     </select>
     </div>

     <div id="submit">
     <input type="submit" name="submit" id="submit" value="submit"/>

     <table border="1" id="table1">
       <tr>
           <th>Student Name</th>
           <th>Gender</th>
       </tr>  

       <?php
         if(isset($_POST['submit'])){
         $standardName = $_POST['standards'];
         $courseName = $_POST['courses'];
         $query3 = "SELECT students.Fname, students.gender
            FROM students
            WHERE students.standardID = '$standardName'
            AND students.courseID = '$courseName'";
        $result3 = mysql_query($query3);
         while($rows3 = mysql_fetch_array($result3)){
              //$dataID = $rows3['id'];
              $studentName = $rows3['FName'];
              $gender = $rows3['gender'];
       ?>    
       <tr>
          <td><?php echo $studentName; ?></td>
          <td><?php echo $gender; ?></td>
          <tr>
          <?php
          }
          }
          ?>
          </table>

          </div> 
         </div>
        </body>
       </html>

Answer 1

使用post作为方法在表单标记内标记输入元素。提交类型的输入元素用于提交表单。

Answer 2

如果该数据包含submit字段，则您只对提交的表单数据执行任何操作。

if(isset($_POST['submit'])){

但是你有：

<input type="submit" name="submit" id="submit"/>

您的提交按钮没有值，因此表格数据中不会包含任何内容。

无法显示数据库

2 个答案: