Question

在我的表情符号代码中，它只替换了我的代码中的第一个，只表示:)可以显示图像。

我的代码中有什么问题。

我的代码

function smileys($text){
// Smiley to image
$icons = array(
    ':)' => '<img src="smilies/smile.gif" border="0" alt="" />',
    ':-)' => '<img src="smilies/smile.gif" border="0" alt="" />',
    ':D' => '<img src="smilies/smile.gif" border="0" alt="" />',
    ':-(' => '<img src="smilies/angry.gif" border="0" alt="" />',
    'angel' => '<img src="smilies/angel.gif" border="0" alt="" />',
    'at' => '<img src="smilies/at.gif" border="0" alt="" />',
    ':-D' => '<img src="smilies/biggrin.gif" border="0" alt="" />',
    'lol' => '<img src="smilies/biggrin.gif" border="0" alt="" />',
    ':-]' => '<img src="smilies/blush.gif" border="0" alt="" />',
    ':-?' => '<img src="smilies/confused.gif" border="0" alt="" />',
    'B-)' => '<img src="smilies/cool.gif" border="0" alt="" />',
    'B)' => '<img src="smilies/cool.gif" border="0" alt="" />',
    ';)' => '<img src="smilies/dodgy.gif" border="0" alt="" />',
    ':(' => '<img src="smilies/sad.gif" border="0" alt="" />',
    ':-(' => '<img src="smilies/sad.gif" border="0" alt="" />',
    'shy' => '<img src="smilies/shy.gif" border="0" alt="" />',
    '|-)' => '<img src="smilies/sleepy.gif" border="0" alt="" />',
    ':-P' => '<img src="smilies/tongue.gif" border="0" alt="" />',
    ':-|' => '<img src="smilies/undecided.gif" border="0" alt="" />',
    ';-)' => '<img src="smilies/wink.gif" border="0" alt="" />',
);

// Now you need find and replace
foreach($icons as $search => $replace){
     $text = preg_replace("#(?<=\s|^)" . preg_quote($search) . "#", $replace, $text);
return $text;
    }
}

echo''.smileys($description).'';

Answer 1

这是因为您在第一次替换后返回循环。将返回移动到循环外部。

    foreach ($icons as $search => $replace) {
        $text = preg_replace("#(?<=\s|^)" . preg_quote($search) . "#", $replace, $text);

    }
    return $text;

仅数组第一个表情符号替换了我的代码

1 个答案: