计算PHP字符串中某个特定字符的所有出现次数的最有效方法是什么？

Question

计算PHP字符串中某个特定字符的所有出现次数的最有效方法是什么？

Answer 1

用这个：

echo substr_count("abca", "a"); // will echo 2

Answer 2

您不能将字符提供给preg_match_all吗？

Answer 3

Not sure what kind of a response you're looking for, but here's a function that might do it:

function findChar($c, $str) {
    indexes = array();
    for($i=0; $i<strlen($str); $i++) {
        if ($str{$i}==$c) $indexes[] = $i;
    }
    return $indexes;
}

Pass it the character you're looking for and the string you want to look:

$mystring = "She shells out C# code on the sea shore";
$mychar = "s";
$myindexes = $findChar($mychar, $mystring);
print_r($myindexes);

It should give you something like

Array (
    [0] => 0
    [1] => 4
    [2] => 9
    [3] => 31
    [4] => 35
)

or something...

Answer 4

If you are going to be repeatedly checking the same string, it'd be smart to have some sort of trie or even assoc array for it otherwise, the straightforward way to do it is...

for($i = 0; $i < strlen($s); $i++)
  if($s[i] == $c)
    echo "{$s[i]} at position $i";

Answer 5

This was working for me. Please try below code :

$strone = 'Sourabh Bhutani';
$strtwo = 'a';
echo parsestr($strone, $strtwo);

function parsestr ($strone, $strtwo) {
$len = 0;
while ($strtwo{$len} != '') {
    $len++;
}

$nr = 0;

while ($strone{$nr} != '')
{
    if($strone{$nr} != ' ')
    {
        $data[$nr] = $strone{$nr};
    }
    $nr++;
}

$newdata = $data;

if($len > 1)
{
    $newdata = array();
    $j = 0;
    foreach($data as $val)
    {
        $str .= $val;
        if($j == ($len -1))
        {
            $newdata[] = $str;
            $str = '';
            $j = 0;
        }
        else
            $j++;
    }
}
$i = 0;

foreach ($newdata as $val) {
    if($val == $strtwo)
    {
        $i++;
    }
}
return $i;
}