如何从给定字符串中随机选择5个字符?他们可以重复。
说我的字符串是这样的:
static $chars = "123456789bcdfghjkmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ";
我只想从该变量中取出5个随机字符。感谢任何可以帮助我的人!
答案 0 :(得分:3)
function gen_code() {
$charset = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
return substr(str_shuffle($charset), 0, 5);
}
答案 1 :(得分:2)
正在运作请尝试一下,
<?php
$length = 5;
$randomString = substr(str_shuffle("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"), 0, $length);
echo $randomString;
?>
答案 2 :(得分:1)
只需创建5个随机索引并从字符串中抓取字符:
$chars = "123456789bcdfghjkmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ";
$length = strlen($chars) - 1;
$randchars = '';
for ($i = 0; $i < 5; $i++) {
$position = mt_rand(0, $length);
$randchars .= $chars[$position];
}
echo $randchars;
如果你只是想获得一个5个字符长的随机字符串,那么有更好的方法。从操作系统获取随机数据,然后对其进行编码,将是理想的方式:
function random_string($length) {
$raw = (int) ($length * 3 / 4 + 1);
$bytes = mcrypt_create_iv($raw, MCRYPT_DEV_URANDOM);
$rand = str_replace('+', '.', base64_encode($bytes));
return substr($rand, 0, $length);
}
echo random_string(5);
答案 3 :(得分:0)
试试这个
static $chars = "123456789bcdfghjkmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ";
$chars=str_shuffle($chars);
$finalString=substr($chars, 0,5);
答案 4 :(得分:0)
只需将$chars
传递给此功能,它就会返回5个随机字符。
// to generate random string
function rand_str($length = 5, $chars = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ')
{
// Length of character list
$chars_length = (strlen($chars) - 1);
// Start our string
$string = $chars{rand(0, $chars_length)};
// Generate random string
for ($i = 1; $i < $length; $i = strlen($string))
{
// Grab a random character from our list
$r = $chars{rand(0, $chars_length)};
// Make sure the same two characters don't appear next to each other
if ($r != $string{$i - 1}) $string .= $r;
}
// Return the string
return $string;
}
// function ends here
答案 5 :(得分:0)
$var = explode(',' , '1,2,3,4,5,6,7,8,9,a,b,c,d,f,g,h,j,k,m,n,p,q,r,s,t,v,w,x,y,z,B,C,D,F,G,H,J,K,L,M,N,P,Q,R,S,T,V,W,X,Y,Z');
$string ='';
for($i=0; $i <= 4; $i++)
{
$string .= $var[rand(0, count($var)-1)];
}
echo $string;
编辑*有一个错误,“count($ var)”应为“count($ var)-1”
答案 6 :(得分:0)
//define a function
function generateRandomString($length = 5){
$chars = "123456789bcdfghjkmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ";
return substr(str_shuffle($chars),0,$length);
}
//usage
echo generateRandomString(5); //random string legth: 5
echo generateRandomString(6); //random string legth: 6
echo generateRandomString(7); //random string legth: 7