PHP mysqli->准备什么都不返回

Question

我正在尝试使用 mysqli 连接从我的数据库中检索行，但是无论我尝试什么，我都会继续收到 500 内部服务器错误。

$getUserQuery = "SELECT * FROM members WHERE name = ? AND member_id > 0";
$getUserStatement = $mysqli_conn->prepare($getUserQuery);
$getUserStatement->bind_param("s", $name);

$mysqli_conn->ping() 的值是 true，所以我知道数据库连接没有问题。 var_dump($getUserStatement) 导致 bool(false) 所以准备有一些问题。完整代码：

$user_dirty = $_GET['u'];
$pass_dirty = $_GET['p'];

$getUserQuery = "SELECT * FROM members WHERE name = ? AND member_id > 0";
$getUserStatement = $mysqli_conn->prepare($getUserQuery);

if ($getUserStatement) {
    echo($getUserStatement);
} else {
    echo("not good");
}

$getUserStatement->bind_param("s", $user_dirty);
$getUserStatement->execute();
$getUserResult = $getUserStatement->get_result();

以及我如何创建数据库连接：

$mysql_host = "host";
$mysql_user = "user";
$mysql_password = "pass";
$mysql_database = "db";

$mysqli_conn = new mysqli($mysql_host, $mysql_user, $mysql_password, $mysql_database);

Answer 1

您收到 500 内部错误，因为语句可能是错误的（https://stackoverflow.com/a/845025/2962442 打开显示错误），而不是应有的资源，这可能是由于连接丢失造成的，请在准备之前尝试 ping 或重新连接。

优秀代码示例：

$getUserQuery = "SELECT * FROM members WHERE name = ? AND member_id > 0";
$getUserStatement = $mysqli_conn->prepare($getUserQuery);
if ($getUserStatement) {
    $getUserStatement->bind_param("s", $name);
} else {
    //statement is false, not good
}

以及连接部分的好例子

$mysql_host = "host";
$mysql_user = "user";
$mysql_password = "pass";
$mysql_database = "db";

$mysqli_conn = new mysqli($mysql_host, $mysql_user, $mysql_password, $mysql_database);    
if (!$mysqli_conn) {
    //not connected...
}