PHP Twitter替換鏈接和與真實鏈接的hashtag

Question

我正在從Twitter API循環JSON響應。每個API響應給了我類似的tweet：

大家好，我是@約翰，我愛#soccer，請訪問我

我試圖取代@john，並插入<a href=http://twitter.com/john>@john</a>但逗號（,）@john後，是問題所在。

如何在標籤之前和之後替換點，逗號等？

Answer 1

$str = preg_replace("/@(\w+)/i", "<a href=\"http://twitter.com/$1\">$0</a>", $str); 

Answer 2

preg_replace("/@(\w+)/", "<a href=http://twitter.com/$1>@$1</a>", $string)" 

Answer 3

要做到主題標籤做它轉換主題標籤這個

$item_content = preg_replace("/#([a-z_0-9]+)/i", "<a href=\"http://twitter.com/search/$1\">$0</a>", $item_content); 

Answer 4

繼承人的功能，用戶提及和網址鏈接，使用來自Twitter的API鳴叫「實體」的數據。

<?php 

function tweet_html_text(array $tweet) { 
    $text = $tweet['text']; 

    // hastags 
    $linkified = array(); 
    foreach ($tweet['entities']['hashtags'] as $hashtag) { 
     $hash = $hashtag['text']; 

     if (in_array($hash, $linkified)) { 
      continue; // do not process same hash twice or more 
     } 
     $linkified[] = $hash; 

     // replace single words only, so looking for #Google we wont linkify >#Google<Reader 
     $text = preg_replace('/#\b' . $hash . '\b/', sprintf('<a href="https://twitter.com/search?q=%%23%2$s&src=hash">#%1$s</a>', $hash, urlencode($hash)), $text); 
    } 

    // user_mentions 
    $linkified = array(); 
    foreach ($tweet['entities']['user_mentions'] as $userMention) { 
     $name = $userMention['name']; 
     $screenName = $userMention['screen_name']; 

     if (in_array($screenName, $linkified)) { 
      continue; // do not process same user mention twice or more 
     } 
     $linkified[] = $screenName; 

     // replace single words only, so looking for @John we wont linkify >@John<Snow 
     $text = preg_replace('/@\b' . $screenName . '\b/', sprintf('<a href="https://www.twitter.com/%1$s" title="%2$s">@%1$s</a>', $screenName, $name), $text); 
    } 

    // urls 
    $linkified = array(); 
    foreach ($tweet['entities']['urls'] as $url) { 
     $url = $url['url']; 

     if (in_array($url, $linkified)) { 
      continue; // do not process same url twice or more 
     } 
     $linkified[] = $url; 

     $text = str_replace($url, sprintf('<a href="%1$s">%1$s</a>', $url), $text); 
    } 

    return $text; 
}