mysqli prepared语句失败,无法执行

dgenwo3n  于 2021-06-18  发布在  Mysql
关注(0)|答案(0)|浏览(251)

我的sql语句有问题,它在第一个if语句中给出了错误,sql语句失败了。对于准备好的陈述,我使用了mmtuts和w3schools,但我不知道哪里出了问题。
这是我的密码:

function login($conn) {
  if (isset($_POST['submitLogin'])) {
    $username = $conn->escape_string($_POST['emailorusername']);
    $password = $conn->escape_string($_POST['password']);
    $sql = "SELECT * FROM users WHERE username = ?";
    $stmt = mysqli_stmt_init($conn);
    if (!mysqli_stmt_prepare($stmt, $sql)) {
      header("Location: ?error=sqlstatementfailed");
      exit();
    } else {
      mysqli_stmt_prepare($stmt, $sql);
      mysqli_stmt_bind_param($stmt, "s", $username);
      mysqli_stmt_execute($stmt);
      $result = mysqli_stmt_get_result($stmt);
      $row = mysqli_fetch_assoc($result);
      if (mysqli_num_rows($result) == 0) {
        header("Location: ?username=notfound");
        exit();
      } else {
        if (mysqli_num_rows($result) > 1) {
          header("Location: ?error=toomuchresults");
          exit();
        } else {
          header("Location: ?username=ok");
          if ($password !== $row['password']) {
            header("Location: ?password=false");
            exit();
          } else {
            header("Location: ?password=ok");
            $_SESSION['username'] = $row['username'];
            $_SESSION['email'] = $row['email'];
            $_SESSION['firstname'] = $row['firstname'];
            $_SESSION['lastname'] = $row['lastname'];
            header("Location: Index");
            exit();
          }
        }
      }
    }
  }
}

服务器已连接到数据库,因此这不是问题所在。
有人能帮忙吗?谢谢!
解决方案:
检查mysqli错误后,你准备,它会告诉你为什么失败。

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