如何在javascript中按遞增順序在對象中添加屬性？

Question

我有對象數組。在對象中，有些具有shortkeys屬性，有些沒有shortkeys屬性。 我想為所有對象添加shortkeys屬性。 重要的是我需要為所有對象提供唯一的shortkeys 。

那些沒有shortkeys對象將遵循這些條件

• 如果對象沒有shortkeys ，則它會分配從1-9開始的shortkeys ，例如 'alt+1' 、'alt+2'...'alt+9'。
• 9之后，我需要將0屬性分配給 object 。
• 然后我需要像'alt+a','alt+b'一樣分配az
• 重要的是，我需要跳過已經定義的所有shortkeys ，例如“alt+m”。這樣每個對象都具有唯一的屬性。

這是我的代碼https://jsfiddle.net/krzz9zmf/

var arr=[
  {name:"abc",shortkeys:"alt+m"},
  {name:"c_1"},
  {name:"abc",shortkeys:"alt+t"},
  {name:"abc",shortkeys:"alt+c"},
  {name:"wes_2"},
  {name:"ncv_3"},
  {name:"sghb_4"},
  {name:"ijo_5"},
  {name:"nhio_6"},
  {name:"jion_7"},
  {name:"chudoi_8"},
  {name:"bdmki_9"},
  {name:"dssd_0"},
  {name:"sdfs_a"},
  {name:"abc",shortkeys:"alt+y"},
  {name:"abc",shortkeys:"alt+e"},
  {name:"sghb_b"},
  {name:"ijo_d"},
  {name:"gsha_e"},
  {name:"asdas_f"},
  {name:"bbb_g"},
  {name:"mko_h"},
  {name:"kioh_i"},
  {name:"qwee_j"},
  {name:"qwee_k"},
  {name:"qwee_l"},
  {name:"qwee_n"},

]

var j =1;
for(var i=0;i<arr.length;i++){
  var obj =arr[i];
  if (!'shortkeys' in myObj){
    //add shorkeys start from 1-9 then 0 and then a-z.In other words need to assign shortcut key like that 'alt+1','alt+2'....'alt+0'''alt+a','alt+b'...'alt+z';
    //some of the objects already define shortkeys example 'alt+m' .so I need to skip these shortcut key .so that each item have unique shortcut.
    myObj.shortkeys= 'alt+'+j;
    j++;
    if(j==10){
      j=0
    }
    if(j==1){
      myObj.shortkeys='alt+a';
    }
  }
}

預期產出

[
{name:"abc",shortkeys:"alt+m"},
{name:"c_1",,shortkeys:"alt+1"},
{name:"abc",shortkeys:"alt+t"},
{name:"abc",shortkeys:"alt+c"},
{name:"wes_2",shortkeys:"alt+2"},
{name:"ncv_3",,shortkeys:"alt+3"},
{name:"sghb_4",shortkeys:"alt+4"},
{name:"ijo_5",shortkeys:"alt+5"},
{name:"nhio_6",shortkeys:"alt+6"},
{name:"jion_7",shortkeys:"alt+7"},
{name:"chudoi_8",shortkeys:"alt+8"},
{name:"bdmki_9",shortkeys:"alt+9"},
{name:"dssd_0",shortkeys:"alt+0"},
{name:"sdfs_a",shortkeys:"alt+a"},
{name:"abc",shortkeys:"alt+y"},
{name:"abc",shortkeys:"alt+e"},
{name:"sghb_b",shortkeys:"alt+b"},
{name:"ijo_d",shortkeys:"alt+d"},
{name:"gsha_e",shortkeys:"alt+e"},
{name:"asdas_f",shortkeys:"alt+f"},
{name:"bbb_g",shortkeys:"alt+g"},
{name:"mko_h",shortkeys:"alt+h"},
{name:"kioh_i",shortkeys:"alt+i"},
{name:"qwee_j",shortkeys:"alt+j"},
{name:"qwee_k",shortkeys:"alt+k"},
{name:"qwee_l",shortkeys:"alt+l"},
{name:"qwee_n",shortkeys:"alt+n"},

]

Answer 1

您可以遍歷並刪除作為鍵存在的字符，然后返回並添加剩余的字符

https://jsfiddle.net/stevenkaspar/krzz9zmf/8/

var alpha_num_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'];

key_arr.map(function(k){
  if(!k.shortkeys) return;

  var key = k.shortkeys.split('+')[1];
  var key_index = alpha_num_array.indexOf(key);
  alpha_num_array.splice(key_index, 1);

})

key_arr = key_arr.map(function(k){
  if(k.shortkeys) return k;

  k.shortkeys = 'alt+'+alpha_num_array[0];
  alpha_num_array.shift();
  return k;
})

console.log(key_arr);

Answer 2

 var arr=[ {name:"abc",shortkeys:"alt+m"}, {name:"c_1"}, {name:"abc",shortkeys:"alt+t"}, {name:"abc",shortkeys:"alt+c"}, {name:"wes_2"}, {name:"ncv_3"}, {name:"sghb_4"}, {name:"ijo_5"}, {name:"nhio_6"}, {name:"jion_7"}, {name:"chudoi_8"}, {name:"bdmki_9"}, {name:"dssd_0"}, {name:"sdfs_a"}, {name:"abc",shortkeys:"alt+4"}, {name:"abc",shortkeys:"alt+e"}, {name:"sghb_b"}, {name:"ijo_d"}, {name:"gsha_e"}, {name:"asdas_f"}, {name:"bbb_g"}, {name:"mko_h"}, {name:"kioh_i"}, {name:"qwee_j"}, {name:"qwee_k"}, {name:"qwee_l"}, {name:"qwee_n"}, ]; //possible shortkeys reversed so we can pop them off var keys = [ '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'].reverse(); //elements without a shortkeys var tagsWithoutShortcuts = arr.filter(function(element){ return typeof element.shortkeys === 'undefined'; }); console.log(keys); console.log(tagsWithoutShortcuts); tagsWithoutShortcuts.forEach(function(tag){ var key = keys.pop(); //while key is already used, get another while (arr.filter(function(element){ return element.shortkeys === 'alt+'+ key; }).length) key = keys.pop(); //put the shortkeys on the tag tag.shortkeys = 'alt+'+ key; }); console.log(arr);

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

Answer 3

您現有嘗試的錯誤（除了沒有得到您想要的字母）是...

• 您在聲明var obj后使用了錯誤的變量名myObj
• 你有if (!'shortkeys' in myObj){而不是if (!('shortkeys' in myObj)){
• 在分配一個shortkeys之前，你沒有找到現有的shortkeys

在這段代碼中，我...

• 添加了一個跟蹤所有使用過的快捷鍵的對象，包括當前存在的快捷鍵
• 添加了一個循環，不斷尋找唯一的快捷鍵，直到找到為止
• 添加了一個用於獲取az字符的alpha變量
• 在主for循環中添加了一個OUTER標簽，以便我們可以在超出字母z中斷循環。 標簽是必需的，因為我們正在打破內部do-while循環。

 var arr=[ {name:"abc",shortkeys:"alt+m"}, {name:"c_1"}, {name:"abc",shortkeys:"alt+t"}, {name:"abc",shortkeys:"alt+c"}, {name:"wes_2"}, {name:"ncv_3"}, {name:"sghb_4"}, {name:"ijo_5"}, {name:"nhio_6"}, {name:"jion_7"}, {name:"chudoi_8"}, {name:"bdmki_9"}, {name:"dssd_0"}, {name:"sdfs_a"}, {name:"abc",shortkeys:"alt+y"}, {name:"abc",shortkeys:"alt+e"}, {name:"sghb_b"}, {name:"ijo_d"}, {name:"gsha_e"}, {name:"asdas_f"}, {name:"bbb_g"}, {name:"mko_h"}, {name:"kioh_i"}, {name:"qwee_j"}, {name:"qwee_k"}, {name:"qwee_l"}, {name:"qwee_n"} ] var j = 1; var alpha = 'a'.charCodeAt(0); // Number for the 'az' chars var z = 'z'.charCodeAt(0); // Number for the 'z' character var shorts = {}; // Holds all shortkeys to avoid duplicates var short = ""; // First gather all the existing shortkeys for (var i = 0; i < arr.length; i++) { if (arr[i].shortkeys) { shorts[arr[i].shortkeys] = arr[i].shortkeys } } OUTER: for (var i = 0; i < arr.length; i++) { var myObj = arr[i]; if (!('shortkeys' in myObj)) { // We want to make sure that our "short" is not yet used. do { if (j < 10) { short = 'alt+' + j j++; } else if (j == 10) { short = 'alt+0' j++; } else if (alpha <= z) { short = 'alt+' + String.fromCharCode(alpha); alpha++; } else { break OUTER; // We've moved beyond 'z', so just quit. } } while(short in shorts) // Track the new shortkey, and assign it shorts[short] = short myObj.shortkeys = short } } document.querySelector("pre").textContent = JSON.stringify(arr, null, 4)

 <pre></pre>

筆記...

• 您的預期輸出有誤； 它有兩次alt+e 。
• 由於字符映射到數字，因此無需手動創建值數組