Question

嗨，我是PHP的新手，我正在尝试编写用于登录的准备语句，但是如下面的代码所示，我的查询没有获得任何行，而是获得field_count。

如果我把那个

SELECT * FROM users WHERE UserName = 'DemoUser'

直接数据库中的

我得到一行，但$stmt->num_rows为零。我错过了什么。

// in connections.php
$dbh = new mysqli($host, $username, $password, $database);

include('connections.php');
include('config.php');
include('comman_functions.php');

if(isset($_POST['password']) && $_POST['password']!=''
    && isset($_POST['username']) && $_POST['username']!=''
    && isset($_POST['location']) && $_POST['location']!='') {

    $username = "DemoUser";
    $stmt = $dbh->prepare("SELECT * FROM users WHERE UserName =? ");
    $stmt->bind_param("s", $username);
    $stmt->execute();  
    error_log($stmt->num_rows); 
    if($stmt->num_rows>0){
    $row = $stmt->fetch();
    error_log("logged in!!!");

    $stmt->close();

    } else {
        $_SESSION['error'] = 'Invalid Username or Password';
        header('Location:index.php');
    }
} else {
    $_SESSION['error'] = 'Please provide Username and Password';
    header('Location:index.php');
}

Answer 1

试试这个！

您需要在$stmt->bind_result();

之后添加$stmt->execute();

$dbh = new mysqli($host, $username, $password, $database);// in connections.php



<?php
  include('connections.php');
  include('config.php');
  include('comman_functions.php');

  if(isset($_POST['password']) && $_POST['password']!='' && isset($_POST['username']) && $_POST['username']!='' && isset($_POST['location']) && $_POST['location']!=''){

    $username = "DemoUser";
    $stmt = $dbh->prepare("SELECT * FROM users WHERE UserName =? ");
      $stmt->bind_param("s", $username);
      $stmt->execute();  
      $stmt->bind_result();
      error_log($stmt->num_rows); 
    if($stmt->num_rows>0){
      $row = $stmt->fetch();
        error_log("logged in!!!");

      $stmt->close();

    }else{
      $_SESSION['error'] = 'Invalid Username or Password';
      header('Location:index.php');
    }
  }else{
    $_SESSION['error'] = 'Please provide Username and Password';
    header('Location:index.php');
  }

?>

Answer 2

如果您在使用num_rows时遇到问题，则必须先声明->store_result()。

<?php
$mysqli = new mysqli("localhost","root", "", "tables");

$query = $mysqli->prepare("SELECT * FROM table1");
$query->execute();
$query->store_result();

$rows = $query->num_rows;

echo $rows;

通过准备语句mysqli php获得不受欢迎的结果

2 个答案: