phpmyadmin中coulmns中缺少数据_Php_Mysql_Phpmyadmin

phpmyadmin中coulmns中缺少数据

php mysql phpmyadmin

phpmyadmin中coulmns中缺少数据,php,mysql,phpmyadmin,Php,Mysql,Phpmyadmin,您好，我在数据库中输入了4条记录，一切正常，我通过表单输入的记录已成功添加到数据库中，但当我浏览表时，有4条记录可用，但记录内容缺失，请帮助。谢谢代码 <body> <?php $servername = "localhost"; $username = "root"; $password = ""; $dbname = "test"; $conn = new mysqli($servername, $username,

您好，我在数据库中输入了4条记录，一切正常，我通过表单输入的记录已成功添加到数据库中，但当我浏览表时，有4条记录可用，但记录内容缺失，请帮助。谢谢

代码

  <body> 
    <?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "test";
    $conn = new mysqli($servername, $username, $password, $dbname);
    if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
    }
    $sql = "INSERT INTO names (firstname, secondname) VALUES ('', '')";
    if ($conn->query($sql) === TRUE) {
      echo "New record created successfully";
    } else {
      echo "Error: " . $sql . "<br>" . $conn->error;
    }
    $conn->close();
    ?> 
    <?php echo $_POST["firstname"]; ?><br /> 
    <?php echo $_POST["secondname"]; ?> 
  </body> 
</html>

<body> 
    <?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "test";
    $conn = new mysqli($servername, $username, $password, $dbname);
    if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
    }
    $stmt = $conn->prepare("INSERT INTO names (firstname, secondname) VALUES  (?, ?)");
    $stmt->bind_param("ss", $_POST["firstname"], $_POST["secondname"]);
    if ($stmt->execute()) {
      echo "New record created successfully";
    } else {
      echo "Error: " . $sql . "<br>" . $conn->error;
    }
    $conn->close();
    ?> 
    <?php echo $_POST["firstname"]; ?><br /> 
    <?php echo $_POST["secondname"]; ?> 
  </body> 
</html>


您没有向列传递任何值
改变
$sql = "INSERT INTO names (firstname, secondname) VALUES ('', '')"; 

到
但是，使用
准备好的语句是用于执行相同语句（或
类似）SQL语句重复执行，效率高。准备好
语句对于SQL注入非常有用，因为参数
随后使用不同协议传输的值需要
不能正确地逃脱。如果原始语句模板不可用
从外部输入派生，SQL注入无法发生
更新代码
  <body> 
    <?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "test";
    $conn = new mysqli($servername, $username, $password, $dbname);
    if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
    }
    $sql = "INSERT INTO names (firstname, secondname) VALUES ('', '')";
    if ($conn->query($sql) === TRUE) {
      echo "New record created successfully";
    } else {
      echo "Error: " . $sql . "<br>" . $conn->error;
    }
    $conn->close();
    ?> 
    <?php echo $_POST["firstname"]; ?><br /> 
    <?php echo $_POST["secondname"]; ?> 
  </body> 
</html>

<body> 
    <?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "test";
    $conn = new mysqli($servername, $username, $password, $dbname);
    if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
    }
    $stmt = $conn->prepare("INSERT INTO names (firstname, secondname) VALUES  (?, ?)");
    $stmt->bind_param("ss", $_POST["firstname"], $_POST["secondname"]);
    if ($stmt->execute()) {
      echo "New record created successfully";
    } else {
      echo "Error: " . $sql . "<br>" . $conn->error;
    }
    $conn->close();
    ?> 
    <?php echo $_POST["firstname"]; ?><br /> 
    <?php echo $_POST["secondname"]; ?> 
  </body> 
</html>




您能详细描述一下吗？您需要向我们展示一些代码，以便我们能够帮助您发布您的代码（试用过的代码）非常感谢@Nana Partykar:-）很乐意提供帮助@塔希尔汗·拜。但是，请再看一遍我的答案。它将帮助您避免SQL注入。