php - PHP脚本。检查用户是否已经在MySQL数据库中

Question

我正在为Facebook创建应用程序，所有使用该应用程序的用户都已写入数据库。所以我需要经常检查如果用户已经在数据库中，如何正确执行呢？

所以我有变量$name = $user_profile['name'];它获取用户名

这是我将用户插入数据库的功能。

    $mysqli = new mysqli("asd","asd","pw","asd");
if ($stmt = $mysqli->prepare("INSERT into myTable (score, userName) VALUE (?,?) ")) {
if (!$mysqli->set_charset("utf8")) {
    printf("Error loading character set utf8: %s\n", $mysqli->error);
} else {

}
$stmt->bind_param('ss', $score, $name);
   $stmt->execute();
   if ($stmt->error != '') {
       echo ' error:'.$stmt->error;
   } else {
       echo 'success';
   }
   $stmt->close();
} else {
   echo 'error:'.$mysqli->error;
}
$mysqli->close();

现在，我需要 If函数来检查用户是否已经在数据库中。据我了解，应该像插入数据库一样，只是 INSERT into我需要使用 SELECT from才能成功完成它？谢谢。

我读过这篇文章： Best way to check for existing user in mySQL database?但它不会帮助我。

更新

现在我的代码如下：

   $mysqli = new mysqli("host","asd","pw","asdf");
   echo "1";
 $sql = "SELECT COUNT(*) AS num FROM myTable WHERE userName = ?";
    echo "2";
 if ($stmt = $mysqli->prepare("SELECT COUNT(*) AS num FROM under WHERE userName = ? ")) {
    echo "3";
    $stmt->bind_param('s', $name);
       echo "4";
    $stmt->execute();
       echo "5";
    $results = $stmt->get_result();
       echo "6";
    $data = mysqli_fetch_assoc($results);
       echo "7";
    }

它打印1 2 3 4 5.表示这行 $results = $stmt->get_result();不正确，因为在它之后不打印值。这里有什么问题？

Answer 1

这个查询很简单：

SELECT
    COUNT(*)
FROM
    myTable
WHERE
    userName = :user_name

如果结果不为0，则该用户存在。

PHP中的mysqli实现

$sql = "SELECT
    COUNT(*) AS num
FROM
    myTable
WHERE
    userName = ?";

if ($stmt = $mysqli->prepare($sql)) {
    $stmt->bind_param('s', $name);
    $stmt->execute();
    $results = $stmt->get_result();
    $data = mysqli_fetch_assoc($results);
    if($data['num'] != 0)
    {
        print "user already exists\n";
    }
}