Php 从mysql读取列时尝试获取非对象的属性

我试图用php和mysqli阅读mysql中的一个专栏，但是我得到了这个错误<代码>注意：尝试获取非对象的属性

PHP

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说实话，我遗漏了什么？我没有很好地理解对象一章。

$db->query

返回一个无法直接迭代的结果对象。试着这样做：

$subscribers = $db->query('SELECT * FROM subscribers');
while ($subscriber = $subscribers->fetch_assoc()) {
    $name = $subscriber['name']; // name is the name of the column
    echo $name;
}

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因为您应该获取结果：

<?php
      $db = new mysqli(HOST, USER, PASSWORD, DATABASE);
      global $db;
      $subscribers = $db->query('SELECT * FROM subscribers');

      while ($subscriber = $subscribers->fetch_object()) {
          $name = $subscriber->name; // name is the name of the column
          echo $name;
      }

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@AbhikChakraborty，但是在foreach中还是有这样做的方法吗？您需要使用循环作为对象获取记录，因为光标指针将设置为记录集中的第一条记录。您最好学习如何遵循手动操作，而不是像这样毫无意义的“预感”
$subscribers = $db->query('SELECT * FROM subscribers');
while ($subscriber = $subscribers->fetch_assoc()) {
    $name = $subscriber['name']; // name is the name of the column
    echo $name;
}

<?php
      $db = new mysqli(HOST, USER, PASSWORD, DATABASE);
      global $db;
      $subscribers = $db->query('SELECT * FROM subscribers');

      while ($subscriber = $subscribers->fetch_object()) {
          $name = $subscriber->name; // name is the name of the column
          echo $name;
      }