所以在我的JS中我有一个像这样的对象;
[
{
firstname: "John",
lastname: "Smith"
},
{
firstname: "Peter",
lastname: "Gregory"
},
{
firstname: "John",
lastname: "Fisher"
},
{
firstname: "Sam",
lastname: "Fisher"
}
]
我想在逗号分隔的字符串中显示名字。现在这很简单,但我也想在必要时显示姓氏的第一个首字母,以区分具有相同名字的两个人。
所以,最后,我有这个:
John S., Peter, John F., Sam
到目前为止,我已经能够做一个记住过去的首字母的循环,但对我来说问题成了例子中的第四个条目;姓氏必须与其他人区分的人,但不与任何人分享姓名。
最聪明的方法是什么?
答案 0 :(得分:2)
Array.map + Array.join会这样做。如果你需要在底部,我可以为Array.map包含一个polyfill。
var people = [
{
firstname: "John",
lastname: "Smith"
},
{
firstname: "Peter",
lastname: "Gregory"
},
{
firstname: "John",
lastname: "Fisher"
},
{
firstname: "Sam",
lastname: "Fisher"
}
]
/* Count number of firstNames */
var firstnames = {};
for (var i = 0; i < people.length; i++) {
if (!firstnames[people[i].firstname]) {
firstnames[people[i].firstname] = 0;
}
firstnames[people[i].firstname] ++;
}
/* Create the string of names */
var peopleString = people.map(function (a) {
/* Check if we need a last name here */
var lastname = firstnames[a.firstname] > 1 ? (a.lastname ? ' ' + a.lastname.substr(0, 1) + '.' : '') : '';
return a.firstname + lastname;
}).join(', ');
document.write(peopleString);
/*Polyfill for Array.map taken from MDN: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map#Polyfill*/
// Production steps of ECMA-262, Edition 5, 15.4.4.19
// Reference: http://es5.github.io/#x15.4.4.19
if (!Array.prototype.map) {
Array.prototype.map = function(callback, thisArg) {
var T, A, k;
if (this == null) {
throw new TypeError(' this is null or not defined');
}
// 1. Let O be the result of calling ToObject passing the |this|
// value as the argument.
var O = Object(this);
// 2. Let lenValue be the result of calling the Get internal
// method of O with the argument "length".
// 3. Let len be ToUint32(lenValue).
var len = O.length >>> 0;
// 4. If IsCallable(callback) is false, throw a TypeError exception.
// See: http://es5.github.com/#x9.11
if (typeof callback !== 'function') {
throw new TypeError(callback + ' is not a function');
}
// 5. If thisArg was supplied, let T be thisArg; else let T be undefined.
if (arguments.length > 1) {
T = thisArg;
}
// 6. Let A be a new array created as if by the expression new Array(len)
// where Array is the standard built-in constructor with that name and
// len is the value of len.
A = new Array(len);
// 7. Let k be 0
k = 0;
// 8. Repeat, while k < len
while (k < len) {
var kValue, mappedValue;
// a. Let Pk be ToString(k).
// This is implicit for LHS operands of the in operator
// b. Let kPresent be the result of calling the HasProperty internal
// method of O with argument Pk.
// This step can be combined with c
// c. If kPresent is true, then
if (k in O) {
// i. Let kValue be the result of calling the Get internal
// method of O with argument Pk.
kValue = O[k];
// ii. Let mappedValue be the result of calling the Call internal
// method of callback with T as the this value and argument
// list containing kValue, k, and O.
mappedValue = callback.call(T, kValue, k, O);
// iii. Call the DefineOwnProperty internal method of A with arguments
// Pk, Property Descriptor
// { Value: mappedValue,
// Writable: true,
// Enumerable: true,
// Configurable: true },
// and false.
// In browsers that support Object.defineProperty, use the following:
// Object.defineProperty(A, k, {
// value: mappedValue,
// writable: true,
// enumerable: true,
// configurable: true
// });
// For best browser support, use the following:
A[k] = mappedValue;
}
// d. Increase k by 1.
k++;
}
// 9. return A
return A;
};
}
答案 1 :(得分:0)
我想你可以这样做:
var o = {};
var arr = [];
people.forEach(function(val) {
if (!o.hasOwnProperty(val.firstname)) {
// didn't see this name before,
// keeping it so we know we saw it
// and keeping its index, in case we see the same name again
var idx = arr.push(val.firstname);
o[val.firstname] = { idx: idx - 1, val: val };
}
else {
// we saw this name, adding the current with a last name
// also adding last name to the origninal we saved before
arr.push(val.firstname + " " + val.lastname.substr(0, 1));
arr[o[val.firstname].idx] = o[val.firstname].val.firstname + " " + o[val.firstname].val.lastname.substr(0, 1);
}
})
console.log(arr);
答案 2 :(得分:0)
如果您想要一个(几乎)内联解决方案......
var arr = [
{
firstname: "John",
lastname: "Smith"
},
{
firstname: "Peter",
lastname: "Gregory"
},
{
firstname: "John",
lastname: "Fisher"
},
{
firstname: "Sam",
lastname: "Fisher"
}
];
var out = arr.map (function(i){
return arr.filter(function(e){
return e.firstname == i.firstname && true;
}, []).length > 1 ? (i.firstname + " " + i.lastname.charAt(0)) : (i.firstname);
}, []).toString();
console.log(out);
小提琴http://jsfiddle.net/owqc35p3/2/
与其他许多人相比,这并不是那么高效,因为我们每次都在当前的arr地图中循环arr(同样,.toString()
可以被.join(" ,");
替换)