在PHP

Question

我需要更換這個HTML代碼的鏈接替換詞？

在控制面板中我看不到演員表我只看到這一點：[xfvalue_actors]

，我試圖取代，因爲這：

<a href="http://myfilms.ga/actor/<?php $str = '[xfvalue_actors]'; echo preg_replace('/, /', '"></a><a href="http://myfilms.ga/actor/', $str, 20); ?>/">[xfvalue_actors]</a><br /> 

，但我得到它：

<a href="http://myfilms.ga/actor/Mamoru Miyano"></a><a href="http://myfilms.ga/actor/Kappei Yamaguchi"></a><a href="http://myfilms.ga/actor/Aya Hirano">Mamoru Miyano, Kappei Yamaguchi, Aya Hirano</a><br /> 

Answer 1

你能試試這個，告訴我是怎麼回事加工？

<?php 
$str = '[xfvalue_actors]'; 
$arrActors = explode(',', $str); 
$out = ''; 
foreach ($arrActors as $actor) { 
    $out .= "<a href='http://myfilms.ga/actor/{$actor}'>{$actor}</a><br />,"; 
} 
echo "Actors: " . substr($out, 0, -1); 
?> 

結果我得到：

Actors: <a href='http://myfilms.ga/actor/Mamoru Miyano'>Mamoru Miyano</a><br />,<a href='http://myfilms.ga/actor/ Kappei Yamaguchi'> Kappei Yamaguchi</a><br />,<a href='http://myfilms.ga/actor/ Aya Hirano'> Aya Hirano</a><br />

希望這是你所需要的！ :)

Answer 2

您可以使用str_replace如果要替換字符串的一部分

$old = "Actors: Mamoru Miyano, Kappei Yamaguchi, Aya Hirano"; 
$new = "Actors: <a href="http://example.com/actor/Mamoru Miyano>Mamoru Miyano</a>, <a href="http://example.com/actor/Kappei Yamaguchi>Kappei Yamaguchi</a>, <a href="http://example.com/actor/Aya Hirano>Aya Hirano</a>"; 


str_replace($old,$new,$your_string) 

但在你的情況你指的是HTML。所以可能是你可以使用jQuery

var replaced = $("body").html().replace('Actors: Mamoru Miyano, Kappei Yamaguchi, Aya Hirano','The new string'); 
$("body").html(replaced); 

Answer 3

如果你試圖基於某種condition..try改變，如果循環（PHP）

<?php if(condition){ ?> Actors: <a href="http://example.com/actor/Mamoru Miyano>Mamoru Miyano</a>, <a href="http://example.com/actor/Kappei Yamaguchi>Kappei Yamaguchi</a>, <a href="http://example.com/actor/Aya Hirano>Aya Hirano</a><?php }else { ?>Actors: Mamoru Miyano, Kappei Yamaguchi, Aya Hirano <?php } ?>