我试图查看“uid'针对student_uid"学生"如果已经有uid匹配,则阻止注册的表。但是它没有工作,我得到以下错误:PHP警告:mysqli_num_rows()期望参数1是mysqli_result,在第31行给出的布尔值。我对php很新,所以我'我很困惑。
这是我的代码:
<?php
if (isset($_POST['submit'])) {
include_once 'dbh.inc.php';
$first = mysqli_real_escape_string($conn, $_POST['first']);
$last = mysqli_real_escape_string($conn, $_POST['last']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$uid = mysqli_real_escape_string($conn, $_POST['uid']);
$pwd = mysqli_real_escape_string($conn, $_POST['pwd']);
//Error handlers
//Check for empty fields
if (empty($first) || empty($last) || empty($email) || empty($uid) || empty($pwd)) {
header("Location: ../signup.php?signup=empty");
exit();
} else {
//Check if input characters are valid
if (!preg_match("/^[a-zA-Z]*$/", $first) || !preg_match("/^[a-zA-Z]*$/", $last)) {
header("Location: ../signup.php?signup=invalid");
exit();
} else {
//Check if email is valid
if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
header("Location: ../signup.php?signup=email");
exit();
} else {
$sql = "SELECT * FROM students WHERE student_uid='uid'";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
if ($resultCheck > 0) {
header("Location: ../signup.php?signup=usertaken");
exit();
} else {
//Hashing the password
$hashedPwd = password_hash($pwd, PASSWORD_DEFAULT);
//Insert the student into the database
$sql = "INSERT INTO students (student_first, student_last, student_email, student_uid, student_pwd) VALUES ('$first', '$last', '$email', '$uid', '$hashedPwd');";
mysqli_query($conn, $sql);
header("Location: ../signup.php?signup=success");
exit();
}
}
}
}
} else {
header("Location: ../signup.php");
exit();
}