Question

我的代码在下面

<form action="POST">
<td class="tg-yw4l" ><?PHP echo $_CONFIG['item_Id']?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['item_Name']?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['itemPicture']?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['market_Value']?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['bptf_Value']?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['opskins_Value']?></td>
<td class="tg-yw4l" ><input name="sell" placeholder="<?PHP echo "$".$Sell_Price?>" value="<?PHP echo $Sell_Price?>"></td>
<td class="tg-yw4l" ><input name="keys" placeholder="<?PHP echo $Keys_Price."K"?>" value="<?PHP echo $Keys_Price?>"></td>
<td class="tg-yw4l" ><input name="refs" placeholder="<?PHP echo $Refs_Price."R"?>" value="<?PHP echo $Refs_Price?>"></td>
</form>

它在表中，输入名称sell可以通过我的PHP后端代码保存到SQL，并且不确定为什么名称键和名称引用不起作用；我的意思是它确实有文本框，但没有将其保存到SQL。

这是它的后端PHP代码：

if (isset($_POST['submit'])) {
  if($s_Bot_Id == $f_Bot_Id ){
    if(isset($_POST['sell'])){
      $content = $_POST['sell'];
      $sql_table = "bot_items_db";
      $sql = "UPDATE $sql_table SET Sell_Price=? WHERE Bot_Id=?";
      $id = $s_Bot_Id;
      $stmt = $conn->prepare($sql);
      $stmt->bind_param('ss', $content, $id);
      if ($stmt->execute()){
        return $stmt->affected_rows;
        $stmt->close(); 
      }else {
        echo "save not working";
      }
    }
    if(isset($_POST['keys'])){
      $content = $_POST['keys'];
      $sql_table = "bot_items_db";
      $sql = "UPDATE $sql_table SET `Keys_Price` = ? WHERE Bot_Id = ?;";
      $id = $s_Bot_Id;
      $stmt = $conn->prepare($sql);
      $stmt->bind_param('ss', $content, $id);
      if ($stmt->execute()){
        return $stmt->affected_rows;
        $stmt->close(); 
      }else {
        echo "save not working";
      }
    }
    if(isset($_POST['refs'])){
      $content = $_POST['refs'];
      $sql_table = "bot_items_db";
      $sql = "UPDATE $sql_table SET Refined_Price=? WHERE Bot_Id=?";
      $id = $s_Bot_Id;
      $stmt = $conn->prepare($sql);
      $stmt->bind_param('ss', $content, $id);
      if ($stmt->execute()){
        return $stmt->affected_rows;
        $stmt->close(); 
      }else {
        echo "save not working";
      }
    }

Answer 1

您还应该在html上为它们添加一个ID，以获取您无法通过名称获取数据的数据

<form action="POST">
<td class="tg-yw4l" ><?PHP echo $_CONFIG['item_Id'];?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['item_Name'];?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['itemPicture'];?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['market_Value'];?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['bptf_Value'];?></td>
<td class="tg-yw4l" ><?PHP echo $_CONFIG['opskins_Value'];?></td>
<td class="tg-yw4l" ><input id="sell" name="sell" placeholder="<?PHP echo "$".$Sell_Price;?>" value="<?PHP echo $Sell_Price;?>"></td>
<td class="tg-yw4l" ><input id="keys" name="keys" placeholder="<?PHP echo $Keys_Price."K";?>" value="<?PHP echo $Keys_Price;?>"></td>
<td class="tg-yw4l" ><input id="refs" name="refs" placeholder="<?PHP echo $Refs_Price."R";?>" value="<?PHP echo $Refs_Price;?>"></td>
</form>

如何在PHP中制作可编辑的表值

1 个答案: