我已经为此绞尽脑汁好几天了，我似乎被困住了。我是PHP新手，所以请原谅我的粗心和错误。
给定一个模式，例如电子邮件地址"ab@?b?"我是否需要替换'的任何实例?'a-e'， '@'和'.'中的所有可能的排列
这是我现在的记录:
function permute($str, $ch, $i, $n, &$output) {
if ($i == $n) {
array_push($output, $str);
} else {
for ($x = $i; $x < $n; $x++) {
for ($y = 0; $y < count($ch); $y++) {
$str[$x] = $ch[$y];
permute($str, $ch, $i + 1, $n, $output);
}
}
}
}
# each ? in the pattern to be replaced by all possible permutations of characters in chars array
# a through e as well as @ and .
$output = array();
$chars = range('a', 'e');
array_push($chars, '@');
array_push($chars, '.');
# the pattern to be checked
$pattern = "ab@?b?.ca";
permute($pattern, $chars, 0, strlen($pattern), $output);
…这很接近我想要的，但不完全正确。该函数对字符串的每个字符进行操作，但它应该只对'?'进行操作。我还能做些什么我还没做的事吗?如果我弄明白了，我会在评论中回复并编辑!
这是我的工作解决方案:
function permute($str, $arr, $i, $n, &$result) {
$nLen = strlen($n);
// cycle through every position of needle
while (($i = strpos($str, $n, $i)) !== false) {
$i = $i + $nLen;
// cycle through each replacement value
foreach ($arr as $value) {
$modified = substr_replace($str, $value, $i - $nLen, $nLen);
// if there are no needles left, save it
if (stristr($modified, $n) === false) {
$result[] = $modified;
}
permute($modified, $arr, $i, $n, $result);
}
}
}
# each ? in the pattern to be replaced by all possible permutations of characters in chars array
# a through e as well as @ and .
$chars = range('a', 'e');
array_push($chars, '@');
array_push($chars, '.');
# the pattern to be checked
$pattern = "ab@?b?.ca";
$result = array();
$needle = '?';
$index = 0;
permute($pattern, $chars, $index, $needle, $result);
var_dump($result);
这假定您只想保存没有指针的值。例如，而不是:
array(63) {
[0]=>
string(9) "ab@ab?.ca"
[1]=>
string(9) "ab@aba.ca"
[2]=>
string(9) "ab@abb.ca"
// etc...
输出:
array(49) {
[0]=>
string(9) "ab@aba.ca"
[1]=>
string(9) "ab@abb.ca"
[2]=>
string(9) "ab@abc.ca"
// etc...
如果您确实想要第一个结果，那么只需删除stristr($modified, $n) === false条件。
无需写出算法，就可以完成相同的操作:
$var = "ab@?b?.co";
$var = str_replace("?","",$var);
print $var;
//ab@b.co