3 回答

TA貢獻2037條經(jīng)驗 獲得超6個贊
如果您希望多個值作為一個數(shù)組,這可以是其中一種方法
var obj = [{name:"naziv",value:"test"},{name:"zzz",value:"xxx"},{name:"Telefon[]",value:"tel1"},{name:"Telefon[]",value:"tel2"}]
var naziv = obj.filter(c => c.name === "Telefon[]").map(res => res.value);
console.log(naziv)
同樣,如果該值本身就是預期輸出,并且如果只期望單個值,則以下是其中一種方法。在這里,我用過Optional Chaining
var obj = [{name:"naziv",value:"test"},{name:"zzz",value:"xxx"},{name:"Telefon[]",value:"tel1"}]
var naziv = obj.find(c => c.name === "Telefon[]")?.value;
console.log(naziv)
如果您需要此代碼來運行某些舊版本的瀏覽器并且不支持,則可能有機會Optional Chaining下面是另一種方式
var obj = [{name: "naziv",value: "test"},{name: "zzz",value: "xxx"},{name: "Telefon[]",value: "tel1"}]
var naziv = (obj.find(c => c.name === "Telefon[]") || {}).value;
console.log(naziv)
var notFound = (obj.find(c => c.name === "not_found") || {}).value;
console.log(notFound);

TA貢獻1883條經(jīng)驗 獲得超3個贊
var obj = [{
name: "naziv",
value: "test"
},
{
name: "zzz",
value: "xxx"
},
{
name: "naziv",
value: "xxx"
}
]
var naziv = obj.filter(item => item.name === 'naziv').map(item => item.value)
console.log(naziv)
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