准备更新我看了PHP手册了一个多小时,并试图谷歌的答案,但无济于事:(什么是错我的代码 - 在PHP
$stmt = $mysqli->prepare("UPDATE players SET energy=:energy, confidence=:confidence, morale=:morale WHERE playerID=:playerID ");
$stmt->bindParam(':energy', $energy);
$stmt->bindParam(':confidence', $confidence);
$stmt->bindParam(':morale', $morale);
$stmt->bindParam(':playerID', $playerID);
$playerID=1;
$energy = 1000;
$confidence = 1100;
$morale = 1200;
$stmt->execute();
但是当我尝试运行它我收到以下错误
Fatal error: Call to a member function prepare() on a non-object in /home/www/websites/www.cricket.cliftonbazaar.com/gm/rungame/rungame.php on line 136
注意线136是准备线路
编辑:所有的变量和表名是正确的,他们一直在三重检查
检查是否启用PDO – Gowri
请将您的代码发布到/创建$ mysqli – breiti
您应该以'var_dump'ing'$ mysqli'开头,PHP认为它不是一个对象。 – soulmerge