mysqli 对象没有执行方法

Question

一直在阅读，但我找不到准确的答案。我只想使用准备好的语句查询数据库，但我得到了：

调用未定义的方法 mysqli::execute()

这是我目前拥有的：

$conn=mysqli_connect($dbhost, $dbuser, $dbpass, $db) or die(mysqli_error($conn));

$conn->prepare("SELECT AccomodationID, PUPoint, PUTime, Entity_ID, Comments FROM ACCOMMODATION ORDER BY PUPoint ASC");
$conn->execute();
$conn->bind_result($AccomodationID, $PUPoint, $PUTime, $Entity_ID, $Comments);
$conn->store_result();

编辑

已解决。非常感谢 TML 的帮助。 mysqli->prepare 返回一个mysqli_stmt 对象。

$stmt = $conn->prepare("SELECT AccomodationID, PUPoint, PUTime, Entity_ID, Comments FROM ACCOMMODATION ORDER BY PUPoint ASC");

然后我可以使用该对象的方法，如下所示：

$stmt->execute();
$stmt->bind_result($AccomodationID, $PUPoint, $PUTime, $Entity_ID, $Comments);
$stmt->store_result();

Answer 1

mysqli::prepare 返回 mysqli_stmt 对象。 execute() 是该对象上的方法，而不是原始 mysqli 对象上的方法。

$conn=mysqli_connect($dbhost, $dbuser, $dbpass, $db) or die(mysqli_error($conn));

$conn->prepare("SELECT AccomodationID, PUPoint, PUTime, Entity_ID, Comments FROM ACCOMMODATION ORDER BY PUPoint ASC");
$stmt = $conn->execute();
$stmt->execute();
$stmt->bind_result($AccomodationID, $PUPoint, $PUTime, $Entity_ID, $Comments);
$stmt->store_result();