我面對這一個問題一個存儲過程:PHP的SQL Server,請在致電參數
我犯了一個存儲過程如下:
CREATE PROCEDURE verify_user_connection
@username nvarchar(50),
@pass nvarchar(50)
AS
BEGIN
SELECT users.username, users.pass
FROM dbo.users
WHERE users.username = '@username'
AND users.pass = '@pass'
RETURN '@username'
END
,並從PHP代碼我叫他這個樣子:
<?php
/* Handle form buttons. */
if (isset($_POST['login'])) {
if (!empty($_POST['username']) || !empty($_POST['pass'])) {
$username = $_POST['username'];
$password = $_POST['pass'];
$sql = "{ CALL dbo.verify_user_connection (@username = ?, @pass = ?) }";
$param = array($username, $password);
$result = sqlsrv_query($conn, $sql, $param);
$row = sqlsrv_fetch_object($result);
echo $row;
if (!$row) {
die(print_r(sqlsrv_errors()));
}
else {
echo "record found";
}
}
else {
echo "something went wrong";
}
}
?>
好我有點新在微軟的SQLSRV驅動程序,我知道我的語法是不準確這裏 可能有人INLIGHT我的問題,並給我解釋我在做什麼WR翁?
在此先感謝
我根本不知道php,但在你的sp上,你應該刪除你的變量的''':users.username = @username AND users.pass = @pass RETURN @ username' – Lamak