我正在创建一个注册页面。表格包含电子邮件,密码,确认密码,学生证,名字,第二名,课程,性别和DoB。我希望页面收集该信息并将其存储在数据库中。
在monent,我收到此错误:mysqli_num_rows()期望参数1为mysqli_result,第22行给出布尔值。
<?php
require_once 'connect.php';
if (isset($_POST['reg'])){
$dob = $_POST['date'];
$Student_ID = $_POST['Student_ID'];
$gender = $_POST['gender'];
$course = $_POST['Course'];
$email = $_POST['inputEmail'];
$password = $_POST['inputPassword'];
$FN = $_POST['FirstName'];
$SN = $_POST['SecondName'];
$sql = "INSERT INTO tblaccounts (Email, Password, Student_ID, FirstName, SecondName, Course, Gender, DoB) VALUES ('".$email."','".$password."','".$Student_ID."','".$FN."','".$SN."','".$course."','".$gender."','".$dob."')";
$result = mysqli_query($connection, $sql) or die("Database Connection Failed" . mysqli_error($connection));
$count = mysqli_num_rows($result);
if ($count == 1){
echo "Account created!";
header('Location: profile.php');
//session_start();
//$id = session_id();
$cookie_name = "user";
$cookie_value = $username;
setcookie($cookie_name, $cookie_value);
//$result2 = mysqli_query($connection, "SELECT Student_ID FROM `tblaccounts` WHERE Email='$username'");
$row2 = mysqli_fetch_assoc($userid);
$cookie_name2 = "userID";
$cookie_value2 = $userid['Student_ID'];
setcookie($cookie_name2, $cookie_value2);
} else {
echo "Login Failed!:";#
?><br/><a href ="login.php">Go back to the login screen.</a><?php
}
}
?>
答案 0 :(得分:0)
$ count = mysqli_num_rows($ result);
上述命令不适用于Insert查询,它通常用于检索查询。
而不是你使用mysqli_affected_rows()
mysqli_affected_rows() - 获取先前MySQL操作中受影响的行数