<p>这是我的代码片段：</p> <pre class="brush:php;toolbar:false;">$stmt = $mysqli-&gt;prepare("SELECT DISTINCT model FROM vehicle_types WHERE year = ? AND make = '?' ORDER by model"); $stmt-&gt;bind_param('is', $year, $make); $stmt-&gt;execute();</pre> <p>当我输出$year和$make的值时，我能看到这些值，但是当我运行这个脚本时，我得到一个空值，并且我的日志文件中出现以下警告：</p> <blockquote> <p>PHP Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement</p> </blockquote> <p>在这种情况下，year在数据库中的类型是int(10)，我已经尝试将其转换为int类型进行传递，而make是一个带有utf8_unicode_ci编码的varchar(20)。我有什么遗漏吗？</p>

$stmt = $mysqli->prepare("
    SELECT DISTINCT model FROM vehicle_types WHERE year = ? AND make = ? ORDER by model
");
$stmt->bind_param('is', $year, $make);
$stmt->execute();