<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
include("../script/connection.php");
$link = mysqli_connect($SQLhost, $SQLusername, $SQLpassword, $SQLdatabase);
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$key = '4dc5f0ea67ca3791614b28c310cd1c79';
$username = 'jean8mathieu';
$query = "SELECT * FROM Account WHERE accountKey=? AND accountUsername=?";
if ($stmt = mysqli_prepare($link, $query)) {
mysqli_stmt_bind_param($stmt, 'ss', $key, $username);
mysqli_stmt_execute($stmt);
mysqli_stmt_store_result($stmt);
printf("Number of rows: %d.\n", mysqli_stmt_num_rows($stmt));
mysqli_stmt_close($stmt);
} else {
printf('errno: %d, error: %s', $mysqli->errno, $mysqli->error);
exit;
}
mysqli_close($link);
那麼,什麼是'?'?還要注意你正在使用'準備'。 – 2014-10-19 00:34:37
查詢很好。我沒有得到任何錯誤。它只是返回0我試圖只把accountKey或accountUsername和他們都沒有改變任何東西... – 2014-10-19 00:38:42
也許我應該更直接。說什麼是'$ key'和'$ username'。它是否存在於'Account'中? – 2014-10-19 00:38:57