检查字符串是否为不带正则表达式的字母数字

Question

I want to check if a string is alpha-numeric and don't intend to use Regex . I have am getting the correct answer but somehow the program is throwing an error stating undefined offset . I have checked the array keys and seemingly they are fine.

$str="hello";
$arr=str_split($str);//convert a string to an array
$a=0;
$d=0;

for($i=0;$i<=count($arr);$i++)
{
    if($arr[$i]>='a' && $arr[$i]<='z' || $arr[$i]>='A' && $arr[$i]<='Z')
    {
        $a=1;
    }
    elseif($arr[$i]>='0' && $arr[$i]<='9')
    {
        $d=1;
    }

}
if($a==1&&$d==1)
{
    echo "Alphanumeric";
}
else
{
    echo "Not alphanumeric";
}

Answer 1

数组从索引零开始，所以结束ist i<count

for($i=0;$i<count($arr);$i++)

Answer 2

You should check out the ctype_alnum ( string $text ) ; function.

ctype_alnum

$str="hello";

if(ctype_alnum($str))
{
    echo "Alphanumeric";
}
else
{
    echo "Not alphanumeric";
}

Answer 3

您应该使用$i < count($arr)或$i <= count($arr) -1因为数组从0开始，只有$i <= count($arr)会导致undefined offset错误消息。