Question

我有一个像这样的数组

array:32 [▼
  "ID" => "7917"
  "ProvinceCode" => "MB"
  "Create" => "2016-05-18 18:16:26.790"
  "DayOfTheWeek" => "4"
  "Giai1" => "28192"
  "Giai2" => "83509"
  "Giai3" => "51911-02858"
  "Giai4" => "14102-97270-96025-08465-89047-45904"
  "Giai5" => "7892-9140-4069-8499"
  "Giai6" => "6117-7471-5541-9119-4855-0566"
  "Giai7" => "843-860-023"
  "Giai8" => "71-13-55-89"
  "Giai9" => ""
  "Status" => "1"
]

我有一个int变量$position = 59，我的工作是通过计算来自Giai1 to Giai9的{{1}}来自59 times的字符来查找值，并获得此位置的值不包含字符0，如果-，那么位置$position = 59的获取值将会返回。

例如，在58位置找到值，返回值为20

我已经编写了这段代码来执行此操作

1 at 14102 in Giai4 (actually 19 count from 0)

预期结果将返回带有getted number标记的数组。例如，代码在位置$position = 59; $count = 0; foreach ($data['result'][0] as $key => $item) { if(preg_match('@Giai@s', $key)) { $_item = str_replace('-', '', $item); $count = $count + strlen($_item); $chars = str_split($item); $chars_sp = array_count_values($chars); $countChar = count($chars); if($count > $position) { //this block contains needed position $math = $count - $position; $secmath = strlen($_item) - $math; for($i=$secmath;$i>=0;$i--){ if($chars[$i] == '-'){ $splash_last++; } } $secmath = $secmath + $splash_last; if($chars[$secmath] == '-'){ echo "+1 - "; $secmath = $secmath + 1; } echo "Count: $count Match: $math Secmatch: $secmath Splash_last: $splash_last"; $chars[$secmath] = 'x' . $chars[$secmath] . 'y'; $edited = implode('', $chars); $data['result'][0][$key] = $edited; break; } } } dd($data['result'][0]); }找到了数字，并在59 (58 from 0)处首先签名，在x处签署。您可以在y

中看到此信息

Giai5

从1到50它工作正常，但在位置50之后，我得到的位置值总是错误的

有什么想法吗？

Answer 1

您可以通过一个简单的array_grep()来获取所有＆＃34; Giai＆＃34;键和implode将值连接到一个大字符串，然后选择该字符串的位置。

$giai = array_flip(preg_grep("/^Giai\d+$/", array_flip($a)));
echo implode("",$giai)[$pos];

https://eval.in/573746

Answer 2

如果我这次理解正确，你可以这样做：

$array = [
    "ID" => "7917",
    "ProvinceCode" => "MB",
    "Create" => "2016-05-18 18:16:26.790",
    "DayOfTheWeek" => "4",
    "Giai1" => "28192",
    "Giai2" => "83509",
    "Giai3" => "51911-02858",
    "Giai4" => "14102-97270-96025-08465-89047-45904",
    "Giai5" => "7892-9140-4069-8499",
    "Giai6" => "6117-7471-5541-9119-4855-0566",
    "Giai7" => "843-860-023",
    "Giai8" => "71-13-55-89",
    "Giai9" => "",
    "Status" => "1"
];

$position = 59;

$start = 0;
$end = 0;

foreach ($array as $key => &$value) {
    if (!preg_match('/Giai/', $key)) {
        continue;    
    }

    $start = $end + 1;
    $end = $start + strlen(str_replace('-', '', $value)) - 1;

    if (($start <= $position) && ($position <= $end)) {
        $counter = $start;
        $value = str_split($value);

        foreach ($value as &$char) {
            if ($char === '-') {
                continue;
            }

            if ($counter === $position) {
                $char = "x{$char}y";
                break;
            }

            $counter++;
        }

        $value = join($value);
    }
}

var_dump($array);

这是amalgamation。

代码有点冗长，但这是因为当你观察位置时你会跳过破折号（-），但是当你需要标记角色时你需要将它们带入帐户。从这段代码中你可以理解算法，然后按照你想要的方式重构代码。我建议从嵌套循环中逃脱，因为它们很难阅读。您可以通过将代码分解为函数或使用可用的数组函数来完成此操作。

PHP查找字符串值并按数组

2 个答案: