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而不是将Twitter前缀用于推文,Twitter现在在转推的推文上提供转发的标志。我想这显示在我的twitter推特上的推特鸣叫...即原始用户信息是嵌入在饲料twitter.com/userTwitter机器人未在推特上转发推文
这是我到目前为止的代码(使用API 1.1 ):
$twitter = new TwitterOAuth(CONSUMER_KEY, CONSUMER_SECRET, ACCESS_TOKEN, ACCESS_TOKEN_SECRET);
$twitter->host = "http://search.twitter.com/";
$search = $twitter->get('search', array('q' => '-escort -RT -ADRTBot #abudhabi', 'count' => 5));
$twitter->host = "https://api.twitter.com/1.1/";
foreach($search->results as $tweet) {
$status = $tweet->text;
if(strlen($status) > 140) $status = substr($status, 0, 139);
$twitter->post('statuses/retweet/$tweet->id', array('status' => $status));
print "STATUS: $tweet->id $status<br>";
}
任何想法将受到感谢!
上述不起作用...我仍然在努力使用新的api 1.1转推。
下面是修改后的代码我到目前为止:
<?php
require_once('twitteroauth/twitteroauth.php');
define('CONSUMER_KEY', 'xxxxx');
define('CONSUMER_SECRET', 'xxxxx');
define('ACCESS_TOKEN', 'xxxxx');
define('ACCESS_TOKEN_SECRET', 'xxxxx');
$twitter = new TwitterOAuth(CONSUMER_KEY, CONSUMER_SECRET, ACCESS_TOKEN, ACCESS_TOKEN_SECRET);
$tweets = $twitter->get("https://api.twitter.com/1.1/search/tweets.json?q=-escort%20-RT%20-ADRTBot%20abudhabi&count=5");
$twitter->host = "https://api.twitter.com/1.1/";
foreach($tweets as $tweet) {
foreach($tweet as $chirp) {
$id = $chirp->id_str;
//testing that data coming through... and it is
echo "<br>THIS IS THE ID: $id<br>";
echo "statuses/retweet/$id.json<br>";
echo "$chirp->text<br>";
$twitter->post('https://api.twitter.com/1.1/statuses/retweet/$id.json');
}
}
echo json_encode($tweets);
?>
但是,不张贴到Twitter ......我失去了什么?
非常感谢,
[R