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這是我網站的代碼。它旨在註冊新用戶,只有登錄管理員才能執行此操作。該帳戶已成功創建,但似乎沒有爲mysqli_num_rows返回1。我打印出num_rows返回的值,並且什麼都沒有出來,甚至沒有出現0。我已經在這方面工作了好幾天了,我對PHP編碼很新,所以這可能是一個明顯的錯誤,但我看不到它。任何幫助將不勝感激。我搜索谷歌,但找不到任何東西。爲什麼mysqli_num_rows不返回值?
if($_SERVER['REQUEST_METHOD']=="POST"){
$name = mysqli_real_escape_string($_POST['fname']);
$email = mysqli_real_escape_string($_POST['email']);
$user = mysqli_real_escape_string($_POST['user']);
$pass = mysqli_real_escape_string(md5($_POST['password']));
$authority = mysqli_real_escape_string($_POST['use']);
$queryCheck = "SELECT `user` FROM `user` WHERE `user`='$user';";
$resultCheck = mysqli_query($link, $queryCheck);
$rowCheck = mysqli_num_rows($resultCheck);
if($rowCheck != 1){
$query = "INSERT INTO `user` (`id`, `name`, `email`,`user`,`pass`) VALUES ($authority,'$name','$email','$user','$pass');";
//echo $query;
$result = mysqli_query($link, $query);
$row = mysqli_affected_rows($result);
if($row == 1){
echo "User Created Successfully!<br/>";
echo '<br/><br/><br/><a href="admin.php">Return to Administration Page</a>';
}else{
echo "User was not created, Please try again.";
}
}else{
echo 'User already exists.';
}
}
什麼是'var_dump($ link)' –
和'var_dump($ rowCheck)'返回的是什麼?真/假還是0/1? –
運行上述程序後,您的程序是否能夠註冊數據庫的任何條目 –