試圖對數據庫運行,但結果的兩個MySQLi的查詢總是爲失敗的消息。
代碼(PHP):
// Checks whether submit button has been pressed
if (isset($_POST['submit']))
{
// Gets values from form fields
$email = mysql_real_escape_string(stripslashes($_POST['email']));
$password = mysql_real_escape_string(stripslashes($_POST['password']));
// Selects e-mail in database
$query = "SELECT * FROM userlogin WHERE email = '{$email}'";
$result = mysqli_query($link, $query) or die('Error [' . mysqli_error($link) . ']');
// Checks whether e-mail exist
if (mysqli_num_rows($result) != 0)
{
// Register new user
$query = "INSERT INTO
userlogin
(username, password, email, regdate)
VALUES
('James','{$password}', '{$email}', NOW())
";
// Submit query to database
$result = mysqli_query($link, $query) or die('Error [' . mysqli_error($link) . ']');
// Return success message
header('Location: index.php?notification=success');
}
else
{
// Return failure message
header('Location: index.php?notification=fail');
}
}
我缺少什麼?任何建議表示讚賞。
「這裏是代碼,爲我修復它」並不是Stack Overflow熱烈歡迎的一類問題。 –
你爲什麼要用'SELECT * FROM userlogin WHERE email ='{$ email}''?去'SELECT * FROM userlogin WHERE email ='$ email'' –
@YourCommonSense對不起。在你看來,你會如何推薦它? – kexxcream