Question

这是我拥有的数据结构（它是为了更清晰的理解而简化）：

• USA
  • Alabama
    • Montgomery
    • Birmingham
  • Arizona
    • Phoenix
    • Mesa
    • Gilbert
• Germany
  • West Germany
    • Bonn
    • Cologne

我需要返回给定节点的所有路径 - 即：如果用户输入Arizona，我需要返回USA → Arizona。如果输入Birmingham，我需要返回USA → Alabama → Birmingham。

在PHP中是否有简单的方法来搜索这样的结构？

Answer 1

如果您没有庞大的数据结构，则可以使用XML解析。它众所周知且易于实施。它具有访问父元素的期望能力。

这是一个简单的例子：

$xml = <<<XML
<list>
  <state name="USA">
    <region name="Alabama">
      <city name="Montgomery" />
      <city name="Birmingham" />
    </region>
    <region name="Arizona">
      <city name="Phoenix" />
      <city name="Mesa" />
      <city name="Gilbert" />
    </region>
  </state>
  <state name="Germany">
    <region name="West Germany">
      <city name="Bonn" />
      <city name="Cologne" />
    </region>
  </state>
</list>
XML;


$doc = new \DOMDocument;
$doc->preserveWhiteSpace = false;
$doc->loadXML($xml);

$xpath = new \DOMXPath($doc);
// XPath query to match all elements with
// attribute name equals to your searched phrase
$locations = $xpath->query("//*[@name='Cologne']");

function parse($list) {

  $response  = [];

  foreach ($list as $node) {
      $response[] = $node->attributes->getNamedItem('name')->nodeValue;
      $parentNode = $node->parentNode;
      // traverse up to root element
      // root element has no attributes
      // feel free to use any other condition, such as checking to element's name
      while ($parentNode->hasAttributes()) {
          $response[] = $parentNode->attributes->getNamedItem('name')->nodeValue;
          $parentNode = $parentNode->parentNode;
      }
  }

  return $response;
}

$parsedLocations = array_reverse(parse($locations));

echo implode(' →  ', $parsedLocations), PHP_EOL;

Answer 2

这是一种可能的策略，逐个构建路径：从数组的第一级开始，检查searc项是否等于键。如果没有，则检查值，否则如果值是数组（is_array()），则使用前缀以递归方式重复搜索。

数据集：

$str = array(
    "USA" => array(
        "Alabama" => array(
            "Montgomery",
            "Birmingham"
        ),
        "Arizona" => array(
            "Phoenix",
            "",
            "Gilbert"
        ),
        "West Germany" => array(
            "Bonn",
            "",
            "Cologne"
        )
    ),
    "Germany" => array(
        "West Germany" => array(
            "Bonn",
            "Mesa",
            "Cologne"
        )
    )
);

功能：

function getPath($haystack, $needle, $prefix=""){
    $path = "";
    foreach($haystack as $key=>$value){

        if($path!="")break;

        if($key===$needle){
            return $prefix.$key;
            break;
        }
        elseif($value===$needle) {
            return $prefix.$value;
            break;
        }
        elseif(is_array($value)) {
            $path.=getPath($value,$needle,$prefix.$key."=>");   
        }   
    }
    return $path;
}

测试：

echo getPath($str,"Mesa");

如果有重复，您将获得第一个结果。如果找不到搜索词，则会得到一个空字符串。

Answer 3

因为＆＃34;数据结构＆＃34;是非常模糊的，你唯一的暗示是你使用PHP，我会假设你的数据结构＆＃34;表示以下内容：

[
    'USA' =>
    [
        'Alabama' =>
        [
            'Montgomery',
            'Birmingham'
        ],
        'Arizona' =>
        [
            'Phoenix',
            'Mesa',
            'Gilbert'
        ]
    ],
    'Germany' =>
    [
        'West Germany' =>
        [
            'Bonn',
            'Cologne'
        ]
    ]
]

我认为你希望你的结果在表格中

['USA', 'Alabama', 'Birmingham']

如果不是这种情况，请告知我们您的数据实际可用情况以及您希望如何获得结果。

在PHP中是否有简单的方法来搜索这样的结构？

这取决于你对＆＃34;简单＆＃34;的定义对我来说，适合单一功能的解决方案是简单的＆＃34; 但是，没有开箱即用的解决方案，您可以在单行中使用。

如果您只需要找到＆＃34; leafs＆＃34;，您可以RecursiveIteratorIterator使用foreach而不是RecursiveArrayIterator。{ 但是既然你需要找到中间密钥，那它就不是一个选择 this StackOverflow question也是如此。

您可以使用array_walk_recursive或ArrayIterator，但在此示例中，除了复杂的事情之外，他们无法真正做foreach循环所能做的任何事情。<登记/> 所以我只是选择function findMyThing($needle, $haystack) // Keep argument order from PHP array functions { // We need to set up a stack array + a while loop to avoid recursive functions for those are evil. // Recursive functions would also complicate things further in regard of returning. $stack = [ [ 'prefix' => [], 'value' => $haystack ] ]; // As long as there's still something there, don't stop while(count($stack) > 0) { // Copy the current stack and create a new, empty one $currentStack = $stack; $stack = []; // Work the stack for($i = 0; $i < count($currentStack); $i++) { // Iterate over the actual array foreach($currentStack[$i]['value'] as $key => $value) { // If the value is an array, then // 1. the key is a string (so we need to match against it) // 2. we might have to go deeper if(is_array($value)) { // We need to build the current prefix list regardless of what we're gonna do below $prefix = $currentStack[$i]['prefix']; $prefix[] = $key; // If the current key, is the one we're looking for, heureka! if($key == $needle) { return $prefix; } // Otherwise, push prefix & value onto the stack for the next loop to pick up else { $stack[] = [ 'prefix' => $prefix, 'value' => $value ]; } } // If the value is NOT an array, then // 1. the key is an integer, so we DO NOT want to match against it // 2. we need to match against the value itself elseif($value == $needle) { // This time append $value, not $key $prefix = $currentStack[$i]['prefix']; $prefix[] = $value; return $prefix; } } } } // At this point we searched the entire array and didn't find anything, so we return an empty array return []; }循环：

$path = findMyThing('Alabama', $array);

然后就像

一样使用它

remoteView.setViewVisibility(...)

Answer 4

@Siguza

避免那些邪恶的递归函数

递归不是邪恶的（或eval），并且适用于堆栈

function df($v,array &$in,array &$stack,$search) {
    $stack[] = $v;
    if ( $v == $search ) {
        return [true,$stack];
    }
    if ( is_array($in) ) {
        foreach ($in as $vv => $k) {
            if ( is_array($k) ) {
                $r = df($vv, $k, $stack, $search);
                if ($r[0]) {
                    return $r;
                }
            }
            else if ($k == $search) {
                $stack[] = $k;
                return [true,$stack];
            }
        }
    }
    array_pop($stack);
    return [false,null];
}

用法：

$s = [];
$r = df('',$in,$s,'Bonn');
print_r($r);
$s = [];
$r = df('',$in,$s,'West Germany');
print_r($r);
$s = [];
$r = df('',$in,$s,'NtFound');
print_r($r);

输出：

Array
(
    [0] => 1
    [1] => Array
(
    [0] =>
        [1] => Germany
            [2] => West Germany
            [3] => Bonn
        )

)
Array
(
    [0] => 1
    [1] => Array
(
    [0] =>
        [1] => Germany
            [2] => West Germany
        )

)
Array
(
    [0] =>
        [1] =>
)

Answer 5

根据你的数据结构。

$data['USA'] = ['Alabama' => ['Montgomery','Birmingham'],'Arizona' => ['Phoenix','Mesa','Gilbert']];
$data['Germany'] = ['West Germany' => ['Bonn','Cologne']];

function getHierarchy($location, $data){
    $totalCountries = count($data);

    //Get Array Keys of rows eg countries.
    $keys = array_keys($data);
    $hierarchy= [];

    //Loop Through Countries
    for($i = 0; $i < $totalCountries; $i++){
        //If we have found the country then return it.
        if($location == $keys[$i]) return [$keys[$i]];
        $hierarchy[] = $keys[$i];
        foreach($data[$keys[$i]] as $city => $places){

            // if we have found the city then return it with country.
            if($city == $location){
                $hierarchy[] = $city;
                return $hierarchy;
            }

            // if we have found the place in our places array then return it with country -> city -> place.
            if(in_array($location, $places)){
                $hierarchy[] = $city;
                $hierarchy[] = $location;
                return $hierarchy;
            }
        }
        // Reset Hirarcy if we do not found our location in previous country.
        $hierarchy = [];
    }
}

$found = getHierarchy('Birmingham', $data);
if($found){
    echo implode(' -> ', $found);
    // Output will be USA -> Alabama -> Birmingham
}

它只能找到一个国家的城市和地点，如果找到任何位置，它将打破整个功能并返回第一个位置与城市和地点。

这是更多改进版本，也可以找到多个位置。 https://gist.github.com/touqeershafi/bf89351f3b226aae1a29

希望它对你有所帮助。

在PHP中搜索分层数据

5 个答案: