來源格式與預期記錄
這是原創假設轉換例子。輸入限下列欄位的普通、無循環 JSON 形記錄,統一記錄格式及拒絕代碼由本例訂明。若換來源,先修改格式約定。
每個包裹須有不重複的正安全整數 number 及 contents 陣列。內容要有 sku 文字及非負安全整數數量。number 轉為 parcelNo,複製各自明細,再按包裹號排序。空包裹照留,空來源得空 consignment。
| 個案 | 輸入記錄(JSON) | 預期輸出(JSON) |
|---|---|---|
| 保留各箱內明細 | {"parcels":[{"number":2,"contents":[{"sku":"A","units":1}]},{"number":1,"contents":[]}]} | {"ok":true,"value":{"consignment":[{"parcelNo":1,"lines":[]},{"parcelNo":2,"lines":[{"sku":"A","units":1}]}]}} |
| 拒絕重複箱號 | {"parcels":[{"number":1,"contents":[]},{"number":1,"contents":[]}]} | {"ok":false,"error":"PARCEL_NUMBER"} |
| 保留無貨箱批次 | {"parcels":[]} | {"ok":true,"value":{"consignment":[]}} |
在本機執行轉換
把完整例子以 JavaScript 模組在 Node.js 執行,內含純轉換函數、三個輸入及寫明的預期輸出,對不上便由斷言停止。拒絕個案回傳 ok=false 及代碼,不回傳半份統一記錄。
import assert from 'node:assert/strict';
function parcelContents(i){
const fail=error=>({ok:false,error});if(!i||!Array.isArray(i.parcels))return fail('PARCEL_LIST');
const seen=new Set(),consignment=[];
for(const p of i.parcels){
if(!p||!Number.isSafeInteger(p.number)||p.number<1||seen.has(p.number)||!Array.isArray(p.contents))return fail('PARCEL_NUMBER');seen.add(p.number);
if(p.contents.some(c=>!c||typeof c.sku!=='string'||!c.sku||!Number.isSafeInteger(c.units)||c.units<0))return fail('PARCEL_CONTENT');
consignment.push({parcelNo:p.number,lines:p.contents.map(c=>({sku:c.sku,units:c.units}))});
}
consignment.sort((a,b)=>a.parcelNo-b.parcelNo);return {ok:true,value:{consignment}};
}
// Original synthetic cases; expected outputs are literal records.
const cases = [
{
"label": "Retain nested parcel contents",
"input": {
"parcels": [
{
"number": 2,
"contents": [
{
"sku": "A",
"units": 1
}
]
},
{
"number": 1,
"contents": []
}
]
},
"expected": {
"ok": true,
"value": {
"consignment": [
{
"parcelNo": 1,
"lines": []
},
{
"parcelNo": 2,
"lines": [
{
"sku": "A",
"units": 1
}
]
}
]
}
}
},
{
"label": "Reject repeated parcel number",
"input": {
"parcels": [
{
"number": 1,
"contents": []
},
{
"number": 1,
"contents": []
}
]
},
"expected": {
"ok": false,
"error": "PARCEL_NUMBER"
}
},
{
"label": "Preserve empty consignment",
"input": {
"parcels": []
},
"expected": {
"ok": true,
"value": {
"consignment": []
}
}
}
];
for (const sample of cases) {
const actual = parcelContents(sample.input);
assert.deepEqual(actual, sample.expected);
console.log(sample.label, JSON.stringify(actual));
}
判清邊界個案
包裹 1 排到包裹 2 前,A 是否也移到包裹 1?
查看參考答案
不會。包裹 1 仍 lines=[],包裹 2 仍有 A=1。排序只改記錄先後,沒有改所屬,也未證明已寄出。