不確定爲什麼我要獲取此PHP警告消息。看起來準備好的語句中有四個參數,並且bind_param()中還有四個變量。謝謝你的幫助!PHP警告:mysqli_stmt :: bind_param():變量數量與預準備語句中的參數數量不匹配
if($stmt = $mysqli -> prepare("SELECT url, month, year, cover_image FROM back_issues ORDER BY year DESC, month DESC")) {
$stmt -> bind_param("ssis", $url, $month, $year, $cover_image);
$stmt -> execute();
$stmt -> bind_result($url, $month, $year, $cover_image);
$stmt -> fetch();
while ($stmt->fetch()) {
echo "<li class='item'><a href='$url'><img src='$cover_image' alt='$cover_image' width='' height='' /></a><br /><span class='monthIssue'>$month $year</span></li>";
}
$stmt -> close();
$mysqli -> close();
}
你在'bind_param結合四個變量的查詢( )',但沒有佔位符 – andrewsi