Question

我正在构建一个带有php的Twitter客户端，并在我尝试运行该程序时收到此错误消息。当我在本地运行它时，它似乎与服务器有关？我有php instlled。我使用Aptana作为我的IDE。有什么问题？

$url = "https://api.twitter.com/1.1/statuses/user_timeline.json";
$requestMethod = "GET";
if (isset($_GET['user'])) {$user = $_GET['user'];} else {$user = "hello";}
if (isset($_GET['count'])) {$count = $_GET['count'];} else {$count = 20;}    
$getfield = '?screen_name=hello&count=20';
$twitter = new TwitterAPIExchange($settings);
$string = json_decode(
    $twitter->setGetfield($getfield)->buildOauth($url, $requestMethod)->performRequest(), $assoc = TRUE
);

if($string["errors"][0]["message"] != "") {
    echo "<h3>Sorry, there was a problem.</h3><p>Twitter returned the following error message:</p><p><em>" . $string[errors][0]["message"] . "</em></p>";
    exit();
}

foreach($string as $items) {
    echo "Time and Date of Tweet: ".$items['created_at']."<br />";
    echo "Tweet: ". $items['text']."<br />";
    echo "Tweeted by: ". $items['user']['name']."<br />";
    echo "Screen name: ". $items['user']['screen_name']."<br />";
    echo "Followers: ". $items['user']['followers_count']."<br />";
    echo "Friends: ". $items['user']['friends_count']."<br />";
    echo "Listed: ". $items['user']['listed_count']."<br /><hr />";
}
?>

错误消息：

Sorry, there was a problem.

Twitter returned the following error message:

".$string[errors][0]["message"]. "
";exit();} foreach($string as $items) { echo "Time and Date of Tweet: ".$items['created_at']."
"; echo "Tweet: ". $items['text']."
"; echo "Tweeted by: ". $items['user']['name']."
"; echo "Screen name: ". $items['user']['screen_name']."
"; echo "Followers: ". $items['user']['followers_count']."
"; echo "Friends: ". $items['user']['friends_count']."
"; echo "Listed: ". $items['user']['listed_count']."

视图源格式的

错误消息：

<?php
require_once('TwitterAPIExchange.php");
$settings = array(
'oauth_access...    
);
$url = "https://api.twitter.com/1.1/statuses/user_timeline.json";
$requestMethod = "GET";
if (isset($_GET['user'])) {$user = $_GET['user'];} else {$user = "iagdotme";}
if (isset($_GET['count'])) {$count = $_GET['count'];} else {$count = 20;}
$getfield = '?screen_name=iagdotme&count=20';
$twitter = new TwitterAPIExchange($settings);
$string = json_decode($twitter->setGetfield($getfield)
->buildOauth($url, $requestMethod)
->performRequest(),$assoc = TRUE);
if($string["errors"][0]["message"] != "") {echo "<h3>Sorry, there was a problem.</h3>
<p>Twitter returned the following error message:</p><p><em>".$string[errors][0]["message"].
"</em></p>";exit();}
foreach($string as $items)
    {
        echo "Time and Date of Tweet: ".$items['created_at']."<br />";
        echo "Tweet: ". $items['text']."<br />";
        echo "Tweeted by: ". $items['user']['name']."<br />";
        echo "Screen name: ". $items['user']['screen_name']."<br />";
        echo "Followers: ". $items['user']['followers_count']."<br />";
        echo "Friends: ". $items['user']['friends_count']."<br />";
        echo "Listed: ". $items['user']['listed_count']."<br /><hr />";
    }
?>

有一条说明：Saw＆＃34; <?＆＃34;。可能的原因：尝试在HTML中使用XML处理指令。（HTML不支持XML处理说明。）

这是我正在关注的教程：https://iag.me/socialmedia/build-your-first-twitter-app-using-php-in-8-easy-steps/

Answer 1

根据您的错误消息，您的PHP代码根本不会执行。这是PHP代码的纯HTML输出。

您的问题可能是配置您的Web服务器（apache，nginx，...）以解释PHP，或者在任何PHP代码之前添加缺少的开始标记（<?php）。

Answer 2

我使用不同的方法来访问Twitter的API，但在我看来，您可能只是请求用户ID而不是用户时间线。更改URL，使其如下所示：

$url = 'https://api.twitter.com/1.1/statuses/user_timeline.json';  
$getfield = '?screen_name=J7mbo';  
$requestMethod = 'GET';  

$twitter = new TwitterAPIExchange($settings);  
echo $twitter->setGetfield($getfield)  
    ->buildOauth($url, $requestMethod)  
    ->performRequest();

Answer 3

首先将$string[errors][0]["message"]更改为$string["errors"][0]["message"]。

然后您必须确保您的计算机上安装了PHP（供Web服务器使用）。（Tutorials for PHP, MySQL and Apache | Tutorials for PHP, MySQL and nginx）

使用<?php起始标记代替<?。

您的浏览器似乎将您的PHP文件视为HTML文件。

Answer 4

PHP可以运行，我能够在我有权访问的服务器上运行它。为此，我需要的只是TwitterAPIExchange.php和您提交的PHP。话虽如此，我建议你改变

$string[errors][0]["message"]

到

$string["errors"][0]["message"]

以便它可以正确返回错误并且还要改变

$getfield = '?screen_name=iagdotme&count=20';

to（注意下面的双引号，这些很重要）

$getfield = "?screen_name=$user&count=$count";

以便您实际使用前面的变量。

问题可能在于在PC上运行文件;我建议找一个网络托管服务提供商。

使用Twitter API的php循环错误消息

4 个答案: