Javascript.find 使用一個元素返回另一個

Question

我是學生。 我正在嘗試基於另一個元素返回一個元素。 我需要使用.find 搜索結果“W”（運動隊獲勝的年份）並返回相應的年份。 為了滿足這一點，我預計會寫function (superbowlWin(record) 。

const record = [
{ year: "1971", result: "N/A"},
{ year: "1970", result: "N/A"},
{ year: "1969", result: "W"},
{ year: "1968", result: "N/A"},
{ year: "1967", result: "N/A"},
{ year: "1966", result: "L"},
{ year: "1965", result: "N/A"},
{ year: "1964", result: "N/A"},
{ year: "1963", result: "N/A"},
{ year: "1962", result: "N/A"},
{ year: "1961", result: "N/A"},
{ year: "1960", result: "N/A"}
]

Answer 1

數組方法 filter 和 find 可以很好地完成這項任務。 您將在下面找到兩者：

 const record = [ { year: "1971", result: "N/A"}, { year: "1970", result: "N/A"}, { year: "1969", result: "W"}, { year: "1968", result: "N/A"}, { year: "1967", result: "N/A"}, { year: "1966", result: "L"}, { year: "1965", result: "N/A"}, { year: "1964", result: "N/A"}, { year: "1963", result: "N/A"}, { year: "1962", result: "N/A"}, { year: "1961", result: "N/A"}, { year: "1960", result: "N/A"} ] let r = []; r1 = record.filter(x => x.result === 'W'); console.log('result filter: ',r1) r2 = record.find(x => x.result === 'W'); console.log('result find: ',r2)

Answer 2

為參考https://jsfiddle.net/ug70xL4y/ 添加 JSFiddle

const record = [{ year: "1971", result: "N/A"},
{ year: "1970", result: "N/A"},
{ year: "1969", result: "W"},
{ year: "1968", result: "N/A"},
{ year: "1967", result: "N/A"},
{ year: "1966", result: "L"},
{ year: "1965", result: "N/A"},
{ year: "1964", result: "N/A"},
{ year: "1963", result: "N/A"},
{ year: "1962", result: "N/A"},
{ year: "1961", result: "N/A"},
{ year: "1960", result: "N/A"}];

function superbowlWin(records) {
let el = records.find(e=>e.result==='W');
if(el){
return el.year;
}else{
return undefined;
}
}
superbowlWin(record);