從PHP數組

Question

刪除字符串我要搜索一個數組中包含「新」，「紐約」，「其他」的「東西」， 話並結合只有那些做了狀態的話。

在本例中，我想在陣列中要被組合用於狀態的匹配：

從這：

<?php 
$states = array("new york", "Nevada"); 
$names = array("something", "green", "something", "yellow", "new", "york", "new","jersey"); 

向該：

$names = ("something", "green", "something", "yellow", "new york", "new jersey"); 

因此改變的名稱數組來組合狀態。

Answer 1

之前觀看的解決方案，看看所作的兩個假設：1。 併發數組元素被檢查的$名稱。 2.檢查相對於$國家提出。

$states = array("new york", "Nevada"); 
$names = array("something", "green", "something", "yellow", "new", "york", "new","jersey"); 
foreach($names as $k => $v) { 
    $combined_names = $names[$k]." ".$names[$k+1]; 
    if (in_array($combined_names, $states)) { 
     $names[] = $combined_names; 
     $names[$k] = $names[$k+1] = ''; 
    } 
} 
$result = array_values(array_filter($names)); 

輸出：

Array 
(
    [0] => something 
    [1] => green 
    [2] => something 
    [3] => yellow 
    [4] => new 
    [5] => jersey 
    [6] => new york 
) 

Answer 2

假設你能解決註釋中的「新澤西」的問題，下面將單詞的一個陣列映射到目標陣列（在這種情況下狀態）。結果數組然後被去重複並重新索引。

如果名稱只需要匹配目標字符串中的整個單詞，則可以將stripos測試更改爲更高級的preg_match;

<?php 
$states = array("new york", "Nevada"); 
$names = array("something", "green", "something", "yellow", "new", "york", "new","jersey"); 

$result = array_map(function ($e) use ($states) { 
    foreach ($states as $state) { 
     // The moment a match is found, return it 
     if (stripos($state, $e) !== false) 
     { 
      return $state; 
     } 
    } 

    // If no match was found, fall back to the word in the original array 
    return $e; 
}, $names); 

print_r(array_values(array_unique($result))); 

=

Array 
(
    [0] => something 
    [1] => green 
    [2] => yellow 
    [3] => new york 
    [4] => jersey 
)