Question

我不知道这段代码有什么问题。数据没有插入数据库。

       $connection = mysqli_connect('localhost', 'root', '', 'login-system');
if ($connection) {
    echo "everything works fine";
} else {
    echo "there is something wrong";
}
if (isset($POST['submit'])) {
    $username = $POST['username'];
    $password = $POST['password'];
    $query = "INSERT INTO users(username, password)";
    $query .= 'VALUES("$username", "$password")';
    $result = mysqli_query($connection,$query);
    if (!$result) {
    die('Query Failed'. mysqli_error());    
    }
}

您对此有何建议？请记住，服务器已连接到数据库，但未插入数据。

谢谢

Answer 1

问题是您正在使用$POST['submit']，这是错误的，您需要使用

$_POST['submit']

因为php没有任何名为$POST的变量。您的完整代码应如下所示。

include_once 'db.php';

if (isset($_POST['submit'])) {
    $username = $_POST['username'];
    $password = $_POST['password'];
    $query = "INSERT INTO users(username, password) VALUES('$username', '$password')";
    $result = mysqli_query($connection,$query);

    if (!$result) {
        die('Query Failed'. mysqli_error($connection));    
    }
}

注意：：您的代码可以进行SQL注入攻击您需要使用准备好的查询。

Answer 2

尝试不同的方式：）

 $connection = mysqli_connect('localhost', 'root', '', 'login-system');
  if ($connection) {
    echo "everything works fine";
  } else {
    echo "there is something wrong";
  }
 if ($_SERVER['REQUEST_METHOD']=='POST') {
  $username = $_POST['username'];
  $password = $_POST['password'];
  $query = sprint_f("INSERT INTO users set username='%s', password='%s'" ,$username,$password);
 $result = mysqli_query($connection,$query);
  if (!$result) {
   die('Query Failed'. mysqli_error());    
 }

}

Answer 3

您的$ query字符串更改为此查询宽度。操作者附加2个字符串

include_once 'db.php';

if (isset($POST['submit'])) {
    $username = $POST['username'];
    $password = $POST['password'];
    $query = "INSERT INTO users(username, password)";
    $query .= 'VALUES("'.$username.'", "'.$password.'")';
    $result = mysqli_query($connection,$query);

    if (!$result) {
        die('Query Failed'. mysqli_error());    
    }
}

Answer 4

在查询串联中的VALUES之前添加一些空格

 $connection = mysqli_connect('localhost', 'root', '', 'login-system');
if ($connection) {
    echo "everything works fine";
} else {
    echo "there is something wrong";
}
if (isset($_POST['submit'])) {
    $username = $_POST['username'];
    $password = $_POST['password'];
    $query = "INSERT INTO users(username, password)";
    $query .= ' VALUES("$username", "$password")';// Here add space before VALUES
    $result = mysqli_query($connection,$query);
    if (!$result) {
    die('Query Failed'. mysqli_error());    
    }
}

Answer 5

我在我这边尝试过。我发现 $ _ POST 变量没有 submit ，因此它不会进入if块。如果条件允许，将其移除。新代码如下所示。

if ((is_array($_POST)) && (count($_POST) > 1)) {
  $username = $_POST['username'];
  $password = $_POST['password'];
  $query = "INSERT INTO users(username, password)";
  $query .= " VALUES('$username', '$password')";
  $result = mysqli_query($connection,$query);
  if (!$result) {
    die('Query Failed'. mysqli_error());    
  }
}

希望它会对您有所帮助。让我知道是否有任何问题。编码愉快：）

为什么数据没有插入数据库

5 个答案: