Retrieving a value from an object in an array of objects based on the condition of another object property

Question

Retrieving a value from an object in an array of objects based on the condition of another object property

Is there a way to save the value of naziv.value in var naziv within the find method without declaring another variable?

var naziv = obj.find(c => c.name === "naziv");
console.log(naziv)

The current output is test when using console.log(naziv.value), but I would prefer just using console.log(naziv).

var obj = [{
    name: "naziv",
    value: "test"
  },
  {
    name: "zzz",
    value: "xxx"
  }
]

var naziv = obj.find(c => c.name === "naziv");
console.log(naziv.value)

EDIT: And also, if the name is the same, how can we create an array of values? For example:

var obj = [{
    name: "naziv",
    value: "test"
  },
  {
    name: "zzz",
    value: "xxx"
  },
  {
    name: "Telefon[]",
    value: "tel1"
  }, {
    name: "Telefon[]",
    value: "tel2"
  }
]

var naziv = obj.find(c => c.name === "Telefon[]");
console.log(naziv.value)

The expected result is: [tel1,tel2]

javascript arrays object properties

Answer 1

Answer №1

To implement de-structuring, try using

const {result:brand} = items.filter(item => item.type === "brand");

Answer 2

To implement de-structuring, try using

const {result:brand} = items.filter(item => item.type === "brand");

Answer 3

Answer №2

If you're looking to gather multiple values in an array, this is one approach

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)

Similarly, if a single value is expected as the output, the following method can be used. Here, Optional Chaining is utilized

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)

In cases where older browser versions do not support Optional Chaining, another alternative is provided below

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);

Answer 4

If you're looking to gather multiple values in an array, this is one approach

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)

Similarly, if a single value is expected as the output, the following method can be used. Here, Optional Chaining is utilized

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)

In cases where older browser versions do not support Optional Chaining, another alternative is provided below

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);

Answer 5

Answer №3

let items = [{
    title: "apple",
    price: "$1"
  },
  {
    title: "banana",
    price: "$0.50"
  },
  {
    title: "apple",
    price: "$1.25"
  }
]

let apples = items.filter(item => item.title === 'apple').map(item => item.price)
console.log(apples)

Answer 6

let items = [{
    title: "apple",
    price: "$1"
  },
  {
    title: "banana",
    price: "$0.50"
  },
  {
    title: "apple",
    price: "$1.25"
  }
]

let apples = items.filter(item => item.title === 'apple').map(item => item.price)
console.log(apples)

Retrieving a value from an object in an array of objects based on the condition of another object property

Answer №1

Answer №2

Answer №3

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