PHP 提取图片img标记中的任意属性的简单实例

复制代码 代码如下:

$str = '

//1、取整个图片代码
preg_match('/]*?srcs*=s*('|")(.*?)\1[^>]*?/?s*>/i',$str,$match);
echo $match[0];

//2、取width属性
preg_match('//i',$str,$match);
echo $match[1];

//3、取height属性
preg_match('//i',$str,$match);
echo $match[1];

//4、取src
preg_match('//i',$str,$match);
echo $match[1];

//1、将src="/uploads/images/20100516000.jpg"替换为src="/uploads/uc/images/20100516000.jpg")
print preg_replace('/()/i',"${1}uc/images/${3}",$str);
echo "

//2、将src="/uploads/images/20100516000.jpg"替换为src="/uploads/uc/images/20100516000.jpg",并省去宽和高
print preg_replace('/(/i',"${1} ${2}uc/images/${3}>",$str);

?>