PHP - MySQLi 准备好的语句

Question

$name = $_GET['user'];
if(isset($_GET['user']) && strlen($_GET['user'])>0) {
    $mysqli = new mysqli($dbhost, $dbuser, $dbpass, $db);
    $stmt = $mysqli->prepare("SELECT username FROM users WHERE username=?");
    $stmt->bind_param('s', $name);
    $stmt->execute();
    while($stmt->fetch()) {
        if($stmt->num_rows == 0) {
            header("Location: home?errormsg=notfound");
            exit();
        }
    }
    $stmt->store_result();
    $stmt->close();
}
$mysqli->close();

因此，上面的代码检查数据库中是否存在 $_GET['name'] ，如果不存在，则重定向到 home?errormsg=notfound 但它将数据库中存在的用户名重定向到链接'home?errormsg=notfound' 也是如此。您能提出解决这个问题的方法吗？

Answer 1

您必须在 $stmt->num_rows 之前调用 $stmt->store_result()。

并且您的 $stmt->fetch() 不是必需的，因为您不使用所选数据。

如果您在 num_rows 之后调用 store_result() ，它将不起作用。

来自manual page的部分评论:

If you do not use mysqli_stmt_store_result( ), and immediatley call this function after executing a prepared statement, this function will usually return 0 as it has no way to know how many rows are in the result set as the result set is not saved in memory yet.

因此您的代码应如下所示:

$name = $_GET['user'];
if(isset($_GET['user']) && strlen($_GET['user'])>0) {
    $mysqli = new mysqli($dbhost, $dbuser, $dbpass, $db);
    $stmt = $mysqli->prepare("SELECT username FROM users WHERE username=?");
    $stmt->bind_param('s', $name);
    $stmt->execute();
    $stmt->store_result();
    if($stmt->num_rows == 0) {
        header("Location: home?errormsg=notfound");
        exit();
    }
    $stmt->close();
}
$mysqli->close();