[英]find and replace all occurrences of string [php shortcodes]
我正在使用此代碼用包含圖像的鏈接替換 CMS 中的短代碼,但它僅替換第一個短代碼
$string = $row['Content'];
if(stristr($string,'[gal=')){
$startTag = "[gal=";
$endTag = "]";
$pos1 = strpos($string, $startTag) + strlen($startTag);
$pos2 = strpos($string, $endTag);
$gal = substr($string, $pos1, $pos2-$pos1);
$q=$db->prepare("select * from images where Gal_ID = :gal");
$q->execute(["gal"=>$gal]);
$imgs='';
while($r=$q->fetch(PDO::FETCH_ASSOC)){
$images[] = $r['Image'];
}
foreach($images as $val){
$imgs .= "<a href='gallery/large/$val' class='fancybox-thumbs' rel='gallery'><img src='gallery/thumb/$val'></a>";
}
$result = substr_replace($string, $imgs, $pos1, $pos2-$pos1);
$result = str_replace($startTag,'',$result);
$result = str_replace($endTag,'',$result);
echo $result;
}
else{
echo $string;
}
字符串包含一些段落和 2 個短代碼
[gal=36] and [gal=37]
結果是僅用鏈接和圖像替換第一個短代碼,但第二個短代碼顯示如下:“37”只是數字。 那么如何遍歷所有短代碼以將它們替換為鏈接而不是第一個短代碼
這是我上面描述的完整示例。
//get matches
if(preg_match_all('/\[gal=(\d+)\]/i', $string, $matches) > 0){
//query for all images. You could/should bind this, but since the expression
//matches only numbers, it is technically not possible to inject anything.
//However best practices are going to be "always bind".
$q=$db->prepare("select Gal_ID, Image from images where Gal_ID in (".implode(',', $matches[1]).")");
$q->execute();
//format the images into an array
$images = array();
while($r=$q->fetch(PDO::FETCH_ASSOC)){
$images[$r['Gal_ID']][] = "<a href='gallery/large/{$r['Image']}' class='fancybox-thumbs' rel='gallery'><img src='gallery/thumb/{$r['Image']}'></a>";
}
//replace shortcode with images
$result = preg_replace_callback('/\[gal=(\d+)\]/i', function($match) use ($images){
if(isset($images[$match[1]])){
return implode('', $images[$match[1]]);
} else {
return $match[0];
}
}, $string);
echo $result;
}
我盡可能多地測試了它,但我沒有 PDO 和/或你的桌子。 這應該可以替代你上面的內容。
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