<?php
include('dbConnection.php');
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title></title>
</head>
<body>
<?php
//get the values from the form, using the POST method.
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$profession = $_POST['profession'];
$gender = $_POST['gender'];
$date_of_birth = $_POST['date_of_birth'];
$email = $_POST['first_name'];
$username = $_POST['username'];
$password = $_POST['password'];
$query = "SELECT username FROM user WHERE username = '$username' ";
$result = mysqli_query($link, $query) ;
if (mysqli_num_rows($result) >= 1) {
echo $message ="WARNING: Book already exist!<br/>";
}
else {
$queryInsert = "INSERT INTO user (first_name,last_name,profession,gender,date_of_birth,email,username,password)" .
"VALUES ('$first_name', '$last_name', '$profession', '$gender', '$date_of_birth', '$email', '$username', '$password')";
$queryInsert = mysqli_query($link,$resultInsert);
$resultInsert = mysqli_query($link,$queryInsert);
{
echo "<center><h2>It has been succesfully added!</h2><br>
Click <a href='manageBooks.php'>here</a> to manage book(s)<br></center>";
}}
?>
</body>
</html>
嗨,我遇到了3個主要問題,當我跑這組代碼。mysqli空查詢問題
說明:未定義變量:resultInsert在C:\ XAMPP \ htdocs中\ PhpProject1 \ doRegister.php在線38
警告:mysqli_query():用C空查詢:\ XAMPP \ htdocs中\ PhpProject1 \ doRegister.php在線38
警告:mysqli_query():用C空查詢:\ XAMPP \ htdocs中\ PhpProject1 \ doRegister.php第40行
任何人都能夠幫助引導我沿着什麼我需要相應地改變?
您需要了解您的代碼中至少發生了什麼。 –
嗯,它主要是空的查詢,我真的不明白 – marcus
編輯:你在一個*不正確的地方使用'$ resultInsert'。你的查詢應該是'$ status = mysqli_query($ link,$ queryInsert); '。並且您正在執行查詢兩次。 –