我知道还有其他人收到了这个错误消息。但这并不是我需要的答案。它不断地给我带来错误。。我真的需要解决这个问题。这是我的代码:
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST' && isset($_POST['submit']))
{
$DBConnect = mysqli_connect("localhost", "root", "");
$email = mysqli_real_escape_string($DBConnect, $_POST['email']);
$wachtwoord = mysqli_real_escape_string($DBConnect, $_POST['wachtwoord']);
if(!empty($email) && !empty($wachtwoord))
{
$selectGebruiker = mysqli_query($DBConnect, "SELECT lidID, wachtwoord FROM geregistreerden WHERE email='$email'");
while ($row = $selectGebruiker->fetch_assoc())
{
$email = $row['email'];
$wachtwoordHash = $row['wachtwoord'];
}
$checkGebruiker = mysqli_num_rows($selectGebruiker);
$wachtwoordCheck = password_verify ( $wachtwoord , $wachtwoordHash );
if( $checkGebruiker && $wachtwoordCheck )
{
$_SESSION['email'] = $email;
echo "gelukt!!";
}
else
{
$loginError = 'Email / wachtwoord is onjuist';
echo "Onjuist";
}
}
else
{
$loginError = 'Email / wachtwoord is onjuist.';
}
}
?>
检查您使用mysqli_query运行的查询。
成功查询时返回结果对象,失败时返回布尔值FALSE,这就是为什么会出现错误。
您可以在文档中找到更多信息:http://php.net/manual/en/mysqli.query.php