問題描述有三個數組,如下:// 數字數組$numbers = ['17', '38', '12', '40', '20', '47', '45', '22', '02', '19', '29', '10', '27', '41', '26', '47', '18', '24', '22', '27', '33', '47', '25', '09', '16', '32', '01', '28', '18', '38'];// 單雙數組$oddEven = ['1', '2', '2', '2', '2', '1', '1', '2', '2', '1', '1', '2', '1', '1', '2', '1', '2', '2', '2', '1', '1', '1', '1', '1', '2', '2', '1', '2', '2', '2'];// 匹配數組$matchArr = ['1', '2', '2', '2'];現在我想用$matchArr匹配$oddEven,請問下怎么才能把$oddEven中的3個['1', '2', '2', '2']給匹配出來?同時能對應上$numbers中的數字?我希望的結果是:$result[0] = ['17', '38', '12', '40'];
$result[1] = ['47', '18', '24', '22'];
$result[2] = ['01', '28', '18', '38'];或者合并在一起也可以,可以有重復的值:$result = ['17', '38', '12', '40', '47', '18', '24', '22', '01', '28', '18', '38'];我試過把數組轉成字符串使用preg_match_all()進行匹配,但是這樣就無法與$numbers對應了。實在想不到什么辦法了,請大家幫幫我,謝謝了。
1 回答

手掌心
TA貢獻1942條經驗 獲得超3個贊
思路:
確認matchArr在oddEven的位置,比如1,10,20
根據上面獲取的位置,取出number里的數據,最后拼接
實現:
$numbers = ['17', '38', '12', '40', '20', '47', '45', '22', '02', '19', '29', '10', '27', '41', '26', '47', '18', '24', '22', '27', '33', '47', '25', '09', '16', '32', '01', '28', '18', '38'];// 單雙數組$oddEven = ['1', '2', '2', '2', '2', '1', '1', '2', '2', '1', '1', '2', '1', '1', '2', '1', '2', '2', '2', '1', '1', '1', '1', '1', '2', '2', '1', '2', '2', '2']; $oddEvenStrs = implode("", $oddEven);// 匹配數組$matchArr = ['1', '2', '2', '2']; $matchArrStrs = implode("", $matchArr); var_dump($oddEvenStrs); var_dump($matchArrStrs); var_dump(count($numbers));function findIt($from, $find, $numbers, $pos) { $result = ''; $pos = strpos($from, $find, $pos); if ($pos !== false) { $result .= implode(",", array_slice($numbers, $pos, 4)); $pos += 4; } if (count($numbers) > $pos) { return $result . "," . findIt($from, $find, $numbers, $pos); } return $result; } $result = findIt($oddEvenStrs, $matchArrStrs, $numbers, 0);echo $result; var_dump(explode(",", $result));
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