我有对象数组。在对象中，有些属性具有shortkeys属性，有些属性没有shortkeys属性。我想将shortkeys属性添加到所有对象。
重要的是我需要为所有对象提供唯一的shortkeys。
那些没有shortkeys的对象将遵循这些条件
shortkeys，则从 1-9 开始分配shortkeys，例如'alt + 1'，'alt + 2'......的 'ALT + 9'。9之后我需要将0属性分配给对象。a-z，例如'alt + a'，'alt + b'shortkeys示例'alt + m'。所以每个对象都有唯一的属性。这是我的代码 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';
}
}
}
预期输出
[
{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"},
]
答案 0 :(得分：1)
您可以浏览并删除作为键存在的字符，然后返回并添加剩余的
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);
答案 1 :(得分：1)
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>
答案 2 :(得分：1)
您现有尝试的错误（除了没有得到您想要的字母）就是......
myObj var obj
if (!'shortkeys' in myObj){而不是if (!('shortkeys' in myObj)){ shortkeys
在这段代码中，我......
alpha变量以用于获取a-z字符OUTER循环中添加了for标签，以便我们可以在超过字母z时中断循环。标签是必需的，因为我们正在从内部do-while循环中断开。
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 {
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... 的
alt+e两次。