Home > Backend Development > PHP Tutorial > php截取字符串函数分享_PHP

php截取字符串函数分享_PHP

WBOY
Release: 2016-05-31 13:17:33
Original
802 people have browsed it

经常看到有新手问PHP有没有类似asp的left函数或right函数,实现截取某字符串左边或右边开始N个字符的函数。答案当然是有的。PHP中的substr函数就可以做的到,只不过PHP把二个函数合二为一了,这里再给大家分享一个更加优秀的截取字符串的函数。

代码如下:


/**
     * 方法库-截取字符串-【该函数作者未知】
     * @param string  $string 字符串 
     * @param int     $length 字符长度
     * @param string  $dot    截取后是否添加...
     * @param string  $charset编码
     * @return string
     */
    public function cutstr($string, $length, $dot = ' ...', $charset = 'utf-8') {
        if (strlen($string)             return $string;
        }
        $string = str_replace(array('&', '"', ''), array('&', '"', ''), $string);
        $strcut = '';
        if (strtolower($charset) == 'utf-8') {
            $n = $tn = $noc = 0;
            while ($n                 $t = ord($string[$n]);                //ASCIIֵ
                if($t == 9 || $t == 10 || (32                     $tn = 1; $n++; $noc++;
                } elseif (194                     $tn = 2; $n += 2; $noc += 2;
                } elseif (224                     $tn = 3; $n += 3; $noc += 2;
                } elseif (240                     $tn = 4; $n += 4; $noc += 2;
                } elseif (248                     $tn = 5; $n += 5; $noc += 2;
                } elseif ($t == 252 || $t == 253) {
                    $tn = 6; $n += 6; $noc += 2;
                } else {
                    $n++;
                }
                if($noc >= $length) {
                    break;
                }
            }
            if ($noc > $length) {
                $n -= $tn;
            }
            $strcut = substr($string, 0, $n);
        } else {
            for ($i = 0; $i                 $strcut .= ord($string[$i]) > 127 ? $string[$i].$string[++$i] : $string[$i];
            }
        }
        $strcut = str_replace(array('&', '"', ''), array('&', '"', ''), $strcut);
        return $strcut.$dot;
    }

以上就是本文的全部内容,希望大家能够喜欢。

Related labels:
php
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 Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template