Question

我正在尝试做如下预备陈述：

    $stmt = $mysqli->prepare("SELECT COUNT(*) FROM table WHERE from = '?' && to = '?' LIMIT 1");
    $stmt->bind_param('ii', $fromId, $toId);
    $stmt->execute();
    $stmt->bind_result($db_offeredAlready);
    $stmt->fetch();
    $stmt->close();

我收到以下错误：

 Fatal error: Call to a member function bind_param() on a non-object in /path/to/script.php on line 97

添加`和'给我

    $stmt = $mysqli->prepare("SELECT COUNT(*) FROM `table` WHERE `from` = '?' && `to` = '?' LIMIT 1");

以及

 Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of variables doesn't match number of parameters in prepared statement in /path/to/script.php on line 97

显然BIND变量的数量与？的数量相匹配？分数。我已尝试添加和删除“和”的所有组合无济于事。这份准备好的陈述有什么问题？我可以用一个条件做一个SELECT就好了，但它有多个抱怨。

Answer 1

不需要引用占位符（?）。

$stmt = $mysqli->prepare("SELECT COUNT(*) FROM `table` WHERE `from` = ? && `to` = ? LIMIT 1");

MySQLi的多重条件，有什么不对

1 个答案: