我想做的是......我有一個(gè)名為 char 的數(shù)組和一個(gè)字符串$char = array("-","_","|","+","~");
$MyString = "this is test string";我想做的是用數(shù)組中的一個(gè)字符替換 $MyString 中的每個(gè)空格但我希望結(jié)果是這樣的$MyString = "this is test string";$ExplodedString = explode(" ",$MyString);foreach ($char as $sinchar) { $text1 = preg_replace("/ /", $sinchar, $MyString ,1); foreach ($char as $sinchar) { $text2 = preg_replace("/ /", $sinchar, $text1 ,1); //echo $text2."\n"; foreach ($char as $sinchar) { $text3 = preg_replace("/ /", $sinchar, $text2 ,1); echo $text3."\n"; } }}這段代碼對我來說工作得很好，但它是一個(gè)靜態(tài)代碼，我正在尋找更動態(tài)的東西，所以如果我添加一個(gè)具有更多空格的文本，我不必添加另一個(gè)循環(huán)。那么大家有什么解決辦法嗎？
1 回答
倚天杖
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超3個(gè)贊
您可以使用遞歸函數(shù)，將第一個(gè)單詞的排列與其余單詞的所有可能排列組合起來：
function computeAllPermutations(array $words, array $separators): array
{
switch (count($words)) {
case 0:
return [];
case 1:
return [$words[0]];
default:
$permutations = [];
foreach (computeAllPermutations(array_slice($words, 1), $separators) as $subPermutation) {
foreach ($separators as $separator) {
$permutations[] = $words[0] . $separator . $subPermutation;
}
}
return $permutations;
}
}
用法：
$words = preg_split('/\h+/', $MyString);
$permutations = computeAllPermutations($words, $char);
print_r($permutations);
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