循環通過多個陣列遞增串的長度

Question

我有

$char=array("1","2","3","4","5","6","7","8","9","0","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","-"); 
$doma=array("aero","asia","biz","cat","com","coop","info","int","jobs","mobi","museum","name","net","org","pro","tel","travel","xxx","edu","gov","mil","co.uk","co.nr","co.au","au","ca","co.cc","cc","co","cn","co.jp","de","es","ie","in","it","jp","nl","nz","ru","co.tk","tk","tv","us") 

，我希望做什麼，是：

從1:1的長度可達32的長度佈置的字符成字符串，回顯字符串，然後再回到開頭。所以最終我的瀏覽器看起來是這樣的：

0.aero 
1.aero 
2.aero 
3.aero 
.... 
x.aero 
y.aero 
z.aero 
-.aero 
00.aero 
01.aero 
02.aero 
.... 
za.aero 
zb.aero 
zc.aero 
zd.aero 
.... 
50x90zx908.aero 
50x90zx909.aero 
50x90zx90a.aero 
50x90zx90b.aero 
.... 
50x90zx910.aero 
50x90zx911.aero 
ect; ect; 

我該如何創建for循環來做到這一點？將$ doma包含到每個循環的最後？

我知道這是巨大的，但是當我有一個想法，我得試試吧;）

Answer 1

如果你真的想與for循環做到這一點，然後讓他們的33，爲每個所需長度（1-32），一個用於$doma陣列。

但我不會這樣做。相反，請注意$char陣列中所需的字符組合實際上構成一棵樹。根節點表示空字符串。根的每個子節點代表1個字符的組合（「1」，「2」，「3」，...）。這些節點的每個子節點都表示一個2字符的組合，它具有其父節點的前綴（因此「1」節點的子節點都將以「1」開頭），依此類推。樹的葉子將是$char中字符的全部32個字符的組合。如果只有三個大字，A，B和C，它會是這個樣子：

然後，您可以實現這種樹的深度優先遍歷的遞歸函數，而不是在內存中生成樹並打印它，您可以讓該函數輸出每個節點到達它的內容。拋出一個參數，該參數允許您在節點內容之後放置後綴，並將該函數包裝在遍歷所有$doma中的元素的循環中，然後就完成了。

function f($array, $limit, $suffix, $prefix = "", $counter = 0) 
{ 
    if ($counter > 0) { 
     print "$prefix.$suffix\n"; 
    } 

    if ($counter < $limit) { 
     foreach ($array as $element) { 
      f($array, $limit, $suffix, $prefix . $element, $counter+1); 
     } 
    } 
} 

$char=array("1","2","3","4","5","6","7","8","9","0","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","-"); 
$doma=array("aero","asia","biz","cat","com","coop","info","int","jobs","mobi","museum","name","net","org","pro","tel","travel","xxx","edu","gov","mil","co.uk","co.nr","co.au","au","ca","co.cc","cc","co","cn","co.jp","de","es","ie","in","it","jp","nl","nz","ru","co.tk","tk","tv","us"); 

foreach ($doma as $d) { 
    f($char, 32, $d); 
} 

按照指定的順序將不完全，但這個順序與在陣列和深度優先遍歷順序元素的順序邏輯一致。

Answer 2

的代碼基本上像對待一個基35號的字符組合。它增加了一個循環。

<?php 

    $char=array("1","2","3","4","5","6","7","8","9","0","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","-"); 
    $doma=array("aero","asia","biz","cat","com","coop","info","int","jobs","mobi","museum","name","net","org","pro","tel","travel","xxx","edu","gov","mil","co.uk","co.nr","co.au","au","ca","co.cc","cc","co","cn","co.jp","de","es","ie","in","it","jp","nl","nz","ru","co.tk","tk","tv","us"); 

    $combo = array(0,0,0,0,0,0,0,0,0,0); //this stores the ten "digits" of the characters 
    $pval = 9; //this is the active "place value", the one that you increment 
    $list = ""; 

    for($i=0;$i<44;$i++) { //loops through doma 
     for($j=0;$j<2758547353515625;$j++) { //loops through character combos, that long number is the permutations 

      for($l=0;$l<10;$l++) { //loop displays combo 
       $list .= $char[$combo[$l]]; 
      } 
      $list .= "." . $doma[$i] . "<br/>"; //add combo to list 

      /*This next part check to see if the active digit is 35. It is is, it sets it 
      equal to zero and looks to the digit on the left. It repeats this until it 
      no longer finds a 35. This is like going from 09 to 10 in the decimal system. */ 
      if($combo[$pval] == 35) { 
       while($combo[$pval] == 35) { 
        $combo[$pval] = 0; 
        $pval--; 
       } 
      } 
      $combo[$pval]++; //whatever digit it left on gets incremented. 



      $pval = 9; //reset, go back to first place value 
     } 
     for($l=0;$l<10;$l++) { 
      $combo[$l] = 0; //reset combo for the next top level domain 
     } 
    } 
    echo $list; //print the compiled list. 
    ?>