无法将数据保存到数据库中PHP

Question

我在这里遇到这个问题，当我按下Add按钮时，它应该将我选择的选项保存到我的数据库表中。

但是当我按下它时，我的数据库表中没有收到任何数据。

我的代码有问题吗？我需要找到一种将数据保存到指定表格的正确方法。

任何帮助将不胜感激。谢谢

<?php 
    include "..\subjects\connect3.php"; 
    //echo "Connection successs"; 

    $query = "SELECT * FROM programmes_list"; 
    $result = mysqli_query($link, $query); 
?> 

<form name = "form1" action="dropdownindex.php" method="post"> 
    <table> 
     <tr> 
     <td>Select Pragramme</td> 
     <td> 
      <select id="programmedd" onChange="change_programme()"> 
      <option>select</option> 
      <?php while($row=mysqli_fetch_array($result)) { ?> 
      <option value="<?php echo $row["ID"]; ?>"><?php echo $row["programme_name"]; ?></option> 
      <?php } ?> 
      </select> 
     </td> 
     </tr> 

     <tr> 
     <td>Select intake</td> 
     <td> 
      <div id="intake"> 
      <select> 
       <option>Select</option> 
      </select> 
      </div> 
     </td> 
     </tr> 

     <tr> 
     <td>Select Subjects</td> 
     <td> 
      <div id="subject"> 
      <select > 
       <option>Select</option> 
      </select> 
      </div> 
     </td> 
     </tr> 

     <input type="submit" value="Add" name="send"> 

    </table> 
</form> 


<?php 
    if(isset($_POST['Add'])) { 

     //print_r($_POST); 
     $course1 = implode(',',$_POST['programmedd']); 
     $course2 = implode(',',$_POST['intake']); 
     $course3 = implode(',',$_POST['subject']); 

     $db->query("INSERT INTO programmes(programme_registered, intake_registered, subjects_registered) 
       VALUES (' ".$course1." ',' ".$course2." ', ' ".$course3." ')"); 

     echo $db->affected_rows; 
    } 
?> 

<script type="text/javascript"> 
    function change_programme() 
    { 
     var xmlhttp=new XMLHttpRequest(); 
     xmlhttp.open("GET","ajax.php?programme="+document.getElementById("programmedd").value,false); 
     xmlhttp.send(null); 

     document.getElementById("intake").innerHTML=xmlhttp.responseText; 

     if(document.getElementById("programmedd").value=="Select"){ 

      document.getElementById("subject").innerHTML="<select><option>Select</option></select>";  
     } 
    } 


    function change_intake() 
    { 
     var xmlhttp=new XMLHttpRequest(); 
     xmlhttp.open("GET","ajax.php?intake="+document.getElementById("intakedd").value,false); 
     xmlhttp.send(null); 

     document.getElementById("subject").innerHTML=xmlhttp.responseText; 
    } 
</script> 


//ajax.php 
<?php 
    $dbhost = 'localhost' ; 
    $username = 'root' ; 
    $password = '' ; 
    $db = 'programmes' ; 

    $link = mysqli_connect("$dbhost", "$username", "$password"); 

    mysqli_select_db($link, $db); 

    if (isset($_GET["programme"])) { 
     $programme = $_GET["programme"]; 
    } else { 
     $programme = ""; 
    } 

    if (isset($_GET["intake"])) { 
     $intake = $_GET["intake"]; 
    } else { 
     $intake = ""; 
    } 

    if ($programme!="") { 
     $res=mysqli_query($link, "select * from intakes where intake_no = $programme"); 
     echo "<select id='intakedd' onChange='change_intake()'>"; 
     echo "<option>" ; echo "Select" ; echo "</option>"; 
     while($value = mysqli_fetch_assoc($res)) { 

     echo "<option value=".$value['ID'].">"; 
     echo $value["intake_list"]; 
     echo "</option>"; 
     } 
     echo "</select>"; 
    } 

    if ($intake!="") { 
     $res=mysqli_query($link, "select * from subject_list where subject_no = $intake"); 
     echo "<select>"; 
     echo "<option>" ; echo "Select" ; echo "</option>"; 
     while($value = mysqli_fetch_assoc($res)) { 

     echo "<option value=".$value['ID'].">"; 
     echo $value["subjects"]; 
     echo "</option>"; 
     } 
     echo "</select>"; 
    } 

?> 

Answer 1

您的错误是您检查按钮单击的位置。

更改这个

如果（isset（$ _ POST [ '发送']））

对于未来的目的和引用，在做上述情况，包括您的按钮的name属性，而不是值属性