不能在單個陣列上使用levenshtein（）來比較和查找最短距離的元素

Question

無法用levenshtein（）將我的頭圍繞在這個元素上。

可以說我有一個數組，看起來像這樣：

Array 
(
    [0] => Array 
     (
      [Gammal URL] => /bil-och-garage 
      [Ny URL] => /catalog/verktyg-och-maskiner 
     ) 

    [1] => Array 
     (
      [Gammal URL] => /bil-och-garage/12-v-utrustning 
      [Ny URL] => /catalog/verktyg-och-maskiner/handverktyg 
     ) 

    [2] => Array 
     (
      [Gammal URL] => /bil-och-garage/12-v-utrustning/antenn 
      [Ny URL] => /catalog/verktyg-och-maskiner/handverktyg/slag-brytverktyg 
     ) 
) 

我想要做的就是打印陣列，但有增加。我想要做的是通過數組和每個'Gammal URL'做levenshtein（）並找到'Ny URL'，並且距離當前的'Gammal URL'最近。

如果沒有完全匹配（0），則打印最短的一個。我一直在嘗試使用foreach的不同用法，嵌套，但無法將我的頭圍繞在如何可以一次檢查1個URL的其餘部分。

總之，我想打印整個數組作爲上面，但第三列與最短距離的URL。如果上述不是最佳解決方案，那麼使用兩個陣列的任何建議也是受歡迎的

編輯

有了這個我仍然得到錯誤的URL爲 「匹配URL」 - 任何想法？

foreach ($import as $key => $arr) { 
     $shortest = ''; 
     foreach ($import as $key2 => $arr2) { 
     if ($shortest != '') { 
      // If the distance between the current Ny URL is shorted than the previously shortest one : 
      // -> it's the new shortest one, otherwise, I keep the previous one 
      $shortest = (levenshtein($arr['Gammal URL'], $arr2['Ny URL']) < levenshtein($arr['Gammal URL'], $shortest)) ? $arr2['Ny URL'] : $shortest; 
     } else { // First attempt is set as the shortest aby default 
      $shortest = $arr2['Ny URL']; 
     } 
     } 
     // I found the shortest one for that Gammal URL 
     $import[$key]['shortest'] = $shortest; 

      echo'<tr>'; 
      echo'<td>'. $arr['Gammal URL']."</td>"; 
      echo'<td>'. $arr['Ny URL'].'</td>'; 
      echo'<td>'. $shortest .'</td>'; 

    } 

的完整代碼

<?php 

//debug 

ini_set('display_errors', 'On'); 
error_reporting(E_ALL); 
ini_set('auto_detect_line_endings', TRUE); 
ini_set('max_execution_time', 300); 
?> 


<?php 

//import 

function csv_import($filename='', $delimiter=';') 
{ 
    if(!file_exists($filename) || !is_readable($filename)) 
    return FALSE; 

    $header = NULL; 
    $data = array(); 
    if (($handle = fopen($filename, 'r')) !== FALSE) 
    { 
    while (($row = fgetcsv($handle, 1000, $delimiter)) !== FALSE) 
    { 
     if(!$header) 
     $header = $row; 
     else 
     $data[] = array_combine($header, $row); 
    } 
    fclose($handle); 
    } 
    return $data; 
} 

$import = csv_import('urler.csv'); 

//output 

//print_r($import); 

echo '<table>'; 
echo '<thead>'; 
echo '<tr>'; 
echo "<th>Gammal URL</th>"; 
echo "<th>Ny URL</th>"; 
echo "<th>Match URL</th>"; 
echo "</tr>"; 
echo "</thead>"; 
echo "</tbody>"; 

foreach ($import as $key => $arr) { 
    $shortest = ''; 

    foreach ($import as $key2 => $arr2) { 
    if ($shortest != '') { 
     // If the distance between the current Ny URL is shorted than the previously shortest one : 
     // -> it's the new shortest one, otherwise, I keep the previous one 
     $shortest = (levenshtein($arr['Gammal URL'], $arr2['Ny URL']) < levenshtein($arr['Gammal URL'], $shortest)) ? $arr2['Ny URL'] : $shortest; 
    } else { // First attempt is set as the shortest aby default 
     $shortest = $arr2['Ny URL']; 
    } 
    } 
    // I found the shortest one for that Gammal URL 
    $import[$key]['shortest'] = $shortest; 


     echo'<tr>'; 
     echo'<td>'. $arr['Gammal URL']."</td>"; 
     echo'<td>'. $arr['Ny URL'].'</td>'; 
     echo'<td>'. $shortest .'</td>'; 

    } 

echo "</tbody>"; 
echo "</table>"; 

?> 

Answer 1

這應該做的工作：

foreach ($array as $key => $arr) { 
    $shortest = ''; 
    foreach ($array as $key2 => $arr2) { 
    if ($shortest != '') { 
     // If the distance between the current Ny URL is shorted than the previously shortest one : 
     // -> it's the new shortest one, otherwise, I keep the previous one 
     $shortest = (levenshtein($arr['Gammal URL'], $arr2['Ny URL']) < levenshtein($arr['Gammal URL'], $shortest)) ? $arr2['Ny URL'] : $shortest; 
    } else { // First attempt is set as the shortest by default 
     $shortest = $arr2['Ny URL']; 
    } 
    } 
    // I found the shortest one for that Gammal URL 
    $array[$key]['shortest'] = $shortest; 
} 

我猜你是不是太遠，因爲你不得不嵌套的foreach，而只是問如果您需要更多解釋，請點評