Home > php教程 > php手册 > php中截取字符串支持utf-8

php中截取字符串支持utf-8

WBOY
Release: 2016-06-13 12:33:48
Original
992 people have browsed it

截取字符串
$string="2006年4月我又长大了一岁!";
echo substr($string,1)."..."; 
 //截取字符串 
function SubstrGB($in,$num)
{
 $pos=0;
 $out="";
 while($pos {
  $c=substr($in,$pos,1);
  if($c=="\n") break;
  if(ord($c)>128)
  {
   $out.=$c;
   $pos++;
   $c=substr($in,$pos,1);
   $out.=$c;
  }
  else
  {
   $out.=$c;
  }
  $pos++;
  if($pos>=$num) break;
 }
        return $out;

  echo SubstrGB($string,8) ;
 ?> 
/***************************************************************************
 *            cut_string.php
 *        ------------------------------
 *    Date        : Jul 16, 2005
 *    Copyright    : none
 *    Mail        : 
 *
 *    作用:截取中文字符.
 *
 *
 ***************************************************************************/
function cut_str($string, $sublen, $start = 0, $code = 'UTF-8')
{
    if($code == 'UTF-8')
    {
        $pa = "/[x01-x7f]|[xc2-xdf][x80-xbf]|xe0[xa0-xbf][x80-xbf]|[xe1-xef][x80-xbf][x80-xbf]|xf0[x90-xbf][x80-xbf][x80-xbf]|[xf1-xf7][x80-xbf][x80-xbf][x80-xbf]/";
        preg_match_all($pa, $string, $t_string);
        if(count($t_string[0]) - $start > $sublen) return join('', array_slice($t_string[0], $start, $sublen))."...";
        return join('', array_slice($t_string[0], $start, $sublen));
    }
    else
    {
        $start  = $start*2;
        $sublen = $sublen*2;
        $strlen = strlen($string);
        $tmpstr = '';
        for($i=0; $i        {
            if($i>=$start && $i            {
            if(ord(substr($string, $i, 1))>129) $tmpstr.= substr($string, $i, 2);
            else $tmpstr.= substr($string, $i, 1);
            } 
            if(ord(substr($string, $i, 1))>129) $i++;
        }
        if(strlen($tmpstr)        return $tmpstr;
    }
}
    echo "
".cut_str($string,8,$start=0,$code='sdf') ;
?> 

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Recommendations
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template