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消息:试图获取非对象Codeigniter的属性

[英]Message: Trying to get property of non-object Codeigniter

I'm trying to create a simple piece of PHP code to display tweets using the twitter API, that just displays tweets about EVE Online - as a test. 我正在尝试创建一个简单的PHP代码,以使用twitter API显示推文,该推文仅显示有关EVE Online的推文-作为测试。 The code below does work however, I get the following error PHP notice after it: 下面的代码可以正常工作,但是我收到以下错误PHP通知:

A PHP Error was encountered 遇到PHP错误

Severity: Notice 严重程度:注意

Message: Trying to get property of non-object 消息:试图获取非对象的属性

Filename: pages/about.php 文件名:pages / about.php

Line Number: 17 行号:17

Here is my code: 这是我的代码:

<?php include('twitteroauth.php');?>
<?php include('get_tweet.php');?>
<div id="tweets">
<?php
    $tweets = $twitter->get('https://api.twitter.com/1.1/search/tweets.jsonq=EveOnline&
    result_type=recent&count=10'    );
foreach($tweets as $tweet)
{
    foreach($tweet as $t)
    {
    echo '<div id="tweetwrap"><img src="'.$t->user->profile_image_url.'"/>'
        .$t->text.'</div>';
    }
}
?>
</div>

I'm not using CodeIgniter's controllers or methods to do this. 我没有使用CodeIgniter的控制器或方法来执行此操作。 I'm trying to do it outside of CodeIgniter whilst using CodeIgniter for everything else. 我正在尝试在CodeIgniter之外执行此操作,同时将CodeIgniter用于其他所有操作。 I'm sure there's a way, however I don't understand why CodeIgniter displays errors when I'm not using any of it's controllers or methods. 我敢肯定有办法,但是我不明白为什么在不使用任何控制器或方法时CodeIgniter会显示错误。

I've also tried this: 我也尝试过这个:

    echo '<div id="tweetwrap"><img src="'.$t->user['profile_image_url'].'"/>'
        .$t['text'].'</div>';

But this also does not work. 但这也不起作用。 Please help. 请帮忙。

got it, basically i replaced the foreach loop with this: 知道了,基本上我用以下代码替换了foreach循环:

<?php include('twitteroauth.php');?>
<?php include('get_tweet.php');?>
<div id="tweets">
<?php
foreach($tweets->statuses as $status)
{
    echo '<div id="tweetwrap"><img src="'.$status->user-
>profile_image_url.'"/>';
    echo "User: " .$status->user->screen_name."</br>";
    echo "<p>Tweet: " .$status->text."</p></br>";
    echo "</br></div>";
}
?>

If its an array use this: 如果是数组,请使用以下命令:

echo '<div id="tweetwrap"><img src="'.$t['user']['profile_image_url'].'"/>'
    .$t['text'].'</div>';

instead of: 代替:

echo '<div id="tweetwrap"><img src="'.$t->user['profile_image_url'].'"/>'
    .$t['text'].'</div>';

EDIT 编辑

See this tutorial 请参阅本教程

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