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我正在创建简单的注册脚本，该脚本具有检查用户名和电子邮件是否已被占用的功能。我导入了 php 文件，该文件创建了与 DB 的 mysqli 连接并将其保存到$mysqli变量中。现在的问题 - 为什么我在我的函数中看不到$mysqli变量？我必须将它作为参数传递，以便我可以使用它。这是代码

<?php
require_once "./incl/db.php";
require_once "./Logger.class.php";
$logger = new Logger("register.php.log");

function isTaken($username, $email, $mysqli){
  $ret = 0;
  if($stmt = $mysqli->prepare("SELECT email,nick FROM users WHERE email = ? OR nick = ? LIMIT 1")){
    $stmt->bind_param("ss", $email, $username);
    if($stmt->execute()){
      $stmt->store_result();
      if($stmt->num_rows === 0){
        $stmt->free_result();
        return $ret;
      }else{
        $stmt->bind_result($dbEmail, $dbNick);
         while($stmt->fetch()){
          if(strcmp(strtolower($dbEmail), strtolower($email)) == 0)
            $ret+= 1;
          if(strcmp(strtolower($dbNick), strtolower($username)) == 0)
            $ret+= 2;
          return $ret;
         }
      }
    }else{
      $logger->logError("Error executing stmt(isTaken)! ".$mysqli->error.$logger->newLine.$stmt->error);
      die("stmt error");
    }
  }else{
    $logger->logError("Error preparing stmt(isTaken)! ".$mysqli->error.$logger->newLine.$stmt->error);
    die("Error preparing stmt!");
  }
}

session_start();
ob_start();

header ('HTTP/1.1 302 Found');
if(isset($_POST["registrovat"]) && !empty($_POST["mail"])
  && !empty($_POST["password"]) && !empty($_POST["username"])){
    $email = trim($_POST["mail"]);
    $username = trim($_POST["username"]);
    $password = trim($_POST["password"]);

    for($i = 0; $i < 10; $i++){
      $password = hash("sha256", $password);
    }

    if(filter_var($email, FILTER_VALIDATE_EMAIL)){
      //echo "E-mail valid";
      $isTaken = isTaken($username, $email, $mysqli);
      if($isTaken === 0){
        if($stmt = $mysqli->prepare("INSERT INTO users (id, nick, password, email, rights) VALUES(NULL, ?, ?, ?, 0)")){
          if($stmt->bind_param("sss", $username, $password, $email)){
            if($stmt->execute()){
              $id = $stmt->insert_id;
              header('Location: http://'.$_SERVER["SERVER_NAME"].'/profil/'.$id);
            }else{
              $logger->logError("Error executing stmt! ".$mysqli->error.$logger->newLine.$stmt->error);
              die("Error executing stmt");
            }
          }else{
            $logger->logError("Error binding params(reg): ".$mysqli->error.$logger->newLine.$stmt->error);
            die("error");
          }
          $stmt->close();
        }else{
          $logger->logError("Error preparing stmt(reg)! ".$mysqli->error.$logger->newLine.$stmt->error);
          die("error stmt!");
        }
      }else{
        $text;
        switch($isTaken){
          case 1: $text = "E-mail already exists"; break;   
          case 2: $text = "Username already exists"; break;         
          case 3: $text = "E-mail and username already exists"; break;
          default: $text = "default"; break;
        }
        die($text);
      }
    }else{
      $logger->logError("E-mail not valid: ".$email);
      $logger->logInfo(var_export($_POST, true));
      die("E-mail not valid");
    }
}else{
  $logger->logError("Fields empty.");
  $logger->logInfo(var_export($_POST, true));
  die("Fields empty!");

}
ob_end_flush();
?>
4

3 回答 3

2

如果要访问全局变量而不将其作为参数传递，则必须使用global声明：

function isTaken($username, $email){
    global $mysqli;
于 2013-07-16T09:46:43.283 回答
1

该变量超出了函数的范围。你可以这样做，或者像 Barmar 说的那样把它变成一个全局变量：

function isTaken($username, $email){
    require_once "./incl/db.php";
    ...
}
于 2013-07-16T09:48:10.060 回答
1

因为除非您导入它或将其声明为全局，否则它超出了函数的范围。

每个函数都是一个封装的代码块，只知道自己，你只能看到函数内设置的变量或超级全局变量，$_POST除非你将它们传递给函数或声明

global $mysqli;

在函数内部告诉它从外部使用变量

于 2013-07-16T09:48:40.637 回答