[英]Remove '%' from string using preg_replace()
How do I remove % from a string using preg_replace function? 如何使用preg_replace函数从字符串中删除% ?
Here is what I currently have, but the % sign is blocking the twitter share button from processing: 这是我目前拥有的东西,但是%符号阻止了twitter共享按钮的处理:
<div class="twitter">
<a href="http://twitter.com/share?text=Just%20Sorc\'d:%20%20'. preg_replace('/((?:http|https|ftp):\/\/(?:[A-Z0-9][A-Z0-9_-]*(?:\.[A-Z0-9][A-Z0-9_-]*)+):?(\d+)?\/?[^\s\"\']+)/i','<a href="$1" rel="nofollow" target="blank">$1</a>',$post).'%20--%20via:%20%20http://www.sorcd.com/'.$session_user_id.'%20%20at%20%20" target="_blank" data-count="none"><img src="https://abs.twimg.com/a/1378977615/images/resources/twitter-bird-white-on-blue.png" width="16px" alt="Tweet"></a>
</div>
Firstly, Your string/URL has URI encoded values like %20 which means they are spaces. 首先,您的字符串/ URL具有URI编码的值,例如%20,这意味着它们是空格。 Removing % will not solve the problem.
删除%不会解决问题。
Secondly, this is a possible duplicate of how-to-remove-sign-in-php-string . 其次,这可能是如何删除php-string的副本。
You can use urldecode() to decode it. 您可以使用urldecode()对其进行解码。 '%20' means space
'%20'表示空格
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