点击一个按钮后更新数据库（它看起来很简单，但它不能工作请阅读）

Question

所以我有这个网页，每当我点击一个按钮添加行时，它会为我添加一个额外的行。之后，将会出现一个完成按钮，用户点击它时会更新数据库。这是我的错误进入的地方。每当我更新时，他们都找不到我的$ _Post ['id']。这些是我的页面代码，您可以在每次单击按钮时添加行。

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 

    <div id="div"></div> 
    <button id='add' value="Add Row">Add Row</button> 
    <form action = newExpertise.php method="post"> 

     <button id="finish" name = "finish" value ="Finish" onclick="clicktoindex()">Finish</button>    
     <input type='hidden' name='store' id='store'> 



    <script> 
     var x=1; 
     var count=0; 

     $('body').on('click','#add',function() 
     { 
      if(count <= 10) 
      { 
       $('#div').append("<form action = newExpertise.php method='post'><div class='line'><input type='text' name = 'txta"+x+"' id='txta"+ x +"'><span class =wordtab></span><input type='text' name = 'txtb"+x+"' id='txtb"+ x +"'><span class =wordtab></span><button class='delete' value ='Delete Row'>Delete Row</button></div><br></form>"); 
       count++; 
       x++; 
      } 
      else 
      alert("Maximum 10 Skills"); 
     }); 

     $('body').on('click','.delete',function() 
     { 
     $(this).closest('.line').remove(); 

     count--; 
     }); 

     function clicktoindex(){ 
      document.getElementById('store').value=count; 
     } 
    </script> 

我认为这是因为当我完成按钮点击，它刷新页面，因此，我的文本框都将消失，所以他们不能够检测到文本框的ID（$ _ POST [” txta1']）。我想用ajax来解决这个问题，但我希望我的页面在点击完成按钮后进入下一页。这些是我的代码，当我点击按钮完成。

if(isset($_POST['finish'])) 
    { 
      for($x=1; $x <= $_POST['store']; $x++) 
      { 
        $finishquery = "INSERT INTO particulars_expertise (Particulars_ID, Expertise, Years_experience) 
        VALUES('".$_POST['storeid']."','".$_POST['txta1']."', '".$_POST['txtb1']."')"; 

      if ($connect->query($finishquery) === TRUE) 
      { 

      } 

      else 
      { 
       echo "Error updating record: " . $connect->error; 
      } 
      } 
    } 

Answer 1

更改您的查询这样，你忘了添加)

$finishquery = "INSERT INTO `particulars_expertise`(`Particulars_ID`, `Expertise`, `Years_experience`) 
       VALUES('$_POST[storeid]','$_POST[txta1]', '$_POST[txtb1]')"; 

Answer 2

尝试下面的代码，你在上面的代码中的许多错误！

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 


<form action ="newExpertise.php" method="post"> 
    <div id="skills"> 
    <div class="skill"> 
     <input type="text" name="skills[]" /> 
     <button type="button" class="delete">Delete</button> 
    </div> 
    </div> 
    <button id="add" value="Add Row">Add Row</button> 

    <button type="submit" id="finish" name ="finish" value="Finish">Finish</button>    
    <input type='hidden' name='store' id='store'> 
</form> 



<script> 
    var x=1; 
    var count=0; 

    $('body').on('click','#add',function() 
    { 
     if(count <= 10) 
     { 
      $('#skills').append('<div class="skill"><input type="text name="skills[]" /><button type="button" class="delete">Delete</button></div>'); 
      count++; 
      x++; 
     } 
     else 
     alert("Maximum 10 Skills"); 
    }); 

    $('body').on('click','.delete',function() 
    { 
    $(this).closest('.skill').remove(); 

    count--; 
    }); 

    function clicktoindex(){ 
     document.getElementById('store').value=count; 
    }