php - 无法访问其他文件中的 mysqli 连接对象？

Question

这个问题已经有答案了:

"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP (29 个回答)

已关闭 5 年前。

dbConnection.php

<?php
include_once('config.php');
$dbConnection = new mysqli(DB_HOST,DB_USER,DB_PASS,DB_NAME);
  if($dbConnection->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
  }
  mysqli_set_charset($dbConnection, 'utf8');
?>

新闻.php

<?php
  require_once('dbConnection.php');
  function getNews($request){
      $sql = "select * from news";
      if (!$result = $dbConnection->query($sql)) {
          die('There was an error running the query [' . $dbConnection->error . ']');
      }

      $news = array();
      while ($row = $result->fetch_assoc() ){
        $news[]=$row;
      }

      $result->free();
      $dbConnection->close();
      return $news;
  }

  $latestNews = getNews($_REQUEST);
  echo json_encode($latestNews);
 ?>

我收到错误 undefined variable :dbConnection on line xx。有人可以帮我解决这个问题吗？

Answer 1

您收到此错误是因为您尝试在函数内访问此变量。

您应该将变量作为参数传递，如下所示:

function getNews($request, $dbConnection) {...}
$latestNews = getNews($_REQUEST, $dbConnection);

或者使用全局:

function getNews($request)
{
    global $dbConnection;
    ...
}