錯誤PHP Twitter分享

Question

我有這個代碼的問題：

<?php echo '<a href="https://twitter.com/share" class="twitter-share-button" data-via="User" data-size="large">Tweet</a> 
<script>!function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)?'http':'https';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+'://platform.twitter.com/widgets.js';fjs.parentNode.insertBefore(js,fjs);}}(document, 'script', 'twitter-wjs');</script>'; ?> 

我得到的錯誤：

解析錯誤：解析錯誤，期待','' or ';'」

Answer 1

您可能需要以逃避該字符串中的所有單引號或不使用PHP輸出該代碼（推薦）：

<?php 
//some php code 
?> 
<a href="https://twitter.com/share" class="twitter-share-button" data-via="User" data-size="large">Tweet</a> 
<script>!function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)?''http':'https';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+'://platform.twitter.com/widgets.js';fjs.parentNode.insertBefore(js,fjs);}}(document, 'script', 'twitter-wjs');</script> 
<?php 
// more PHP code 
?> 

轉義引號：

<?php echo '<a href="https://twitter.com/share" class="twitter-share-button" data-via="User" data-size="large">Tweet</a> 
<script>!function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)?\'http\':\'https\';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+\'://platform.twitter.com/widgets.js\';fjs.parentNode.insertBefore(js,fjs);}}(document, \'script\', \'twitter-wjs\');</script>'; ?>