javascriptでオブジェクトのプロパティを昇順に追加する方法は？

Question

オブジェクトの配列があります。オブジェクトにはshortkeysプロパティがあり、一部にはshortkeysプロパティがありません。すべてのオブジェクトにshortkeysプロパティを追加したいと思います。 重要なことは、すべてのオブジェクトにユニークなshortkeysを付ける必要があるということです。これらの条件に従いますshortkeysを持っていない

これらのオブジェクト

• それは「ALT + 1」、 'ALTよう1-9から始まるshortkeysを割り当てるよりも、オブジェクトがshortkeysを持っていない場合+2 '...' alt + 9 'を押します。
• 9の後にオブジェクトに0プロパティを割り当てる必要があります。
• その後、私は、私はすでに定義されているすべてのshortkeys例に「Alt + M」をスキップする必要がある重要
• に「Alt + b」が、それに「Alt + A」のようなa-zを割り当てる必要がありますが、すべてのオブジェクトは、ユニークな特性を持っていることの.so 。ここで

は私のコード 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'; 
    } 
    } 
} 

予想される出力である

[ 
{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

あなたが通過し、キーが戻って行くと残るものを追加として存在する文字を削除することができます

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

（余談したい文字を得ていないから）既存の試みとエラーが...

• に使用していたということでした間違った変数名myObj宣言後var obj
• あなたはこのコードでは、I 1

を割り当てる前に、既存のshortkeysを探すために失敗した代わりにif (!('shortkeys' in myObj)){

のif (!'shortkeys' in myObj){を持っていました...

• は現在、既存のもの
• を含め、すべての使用ShortKeysはを追跡するオブジェクトを追加したものが
• を取得するために使用されるalpha変数を追加発見されるまで、ユニークなshortkeyを探し続けてループを追加しましたa-z文字
• OUTERというラベルをメインのforループに追加して、ループを破ることができます。これは、文字がzを超えた場合に発生します。ラベルは内側のdo-whileループから壊れているので必要です。

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 'a-z' 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 { 
 
     \t 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>

注...

• あなたの予想出力は間違いがあります。 alt+eが2回あります。
• 文字はただ、何の後に `ALT + z`を不思議番号