Turning array with objects into array with specific key from that objects? [duplicate]

-1

This question already has an answer here:

From an array of objects, extract value of a property as array

12 answers

how to get a list of key values from array of objects - JavaScript [duplicate]

2 answers

What I have is this:

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ]

And I want this:

const names = [

    "John",

    "Marry",

    "Jack"

  ]

How? Thanks!

asked Nov 19 at 11:14

FatBoyGebajzla

327

marked as duplicate by trincot javascript
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Nov 19 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you made any attempt at all yourself yet?
– CertainPerformance
Nov 19 at 11:15

Clearly he asked question without even trying himself
– siddhant sankhe
Nov 19 at 11:18

There's no need for that, I am fetching some objects from server, doing this for hours and lost my self... Actually the first answer is correct and I already tried that but did not realize that it work... No need for minuses, I am just tired... Thanks guys
– FatBoyGebajzla
Nov 19 at 11:22

add a comment |

-1

This question already has an answer here:

From an array of objects, extract value of a property as array

12 answers

how to get a list of key values from array of objects - JavaScript [duplicate]

2 answers

What I have is this:

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ]

And I want this:

const names = [

    "John",

    "Marry",

    "Jack"

  ]

How? Thanks!

asked Nov 19 at 11:14

FatBoyGebajzla

327

marked as duplicate by trincot javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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Nov 19 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you made any attempt at all yourself yet?
– CertainPerformance
Nov 19 at 11:15

Clearly he asked question without even trying himself
– siddhant sankhe
Nov 19 at 11:18

There's no need for that, I am fetching some objects from server, doing this for hours and lost my self... Actually the first answer is correct and I already tried that but did not realize that it work... No need for minuses, I am just tired... Thanks guys
– FatBoyGebajzla
Nov 19 at 11:22

add a comment |

-1

This question already has an answer here:

From an array of objects, extract value of a property as array

12 answers

how to get a list of key values from array of objects - JavaScript [duplicate]

2 answers

What I have is this:

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ]

And I want this:

const names = [

    "John",

    "Marry",

    "Jack"

  ]

How? Thanks!

asked Nov 19 at 11:14

FatBoyGebajzla

327

This question already has an answer here:

From an array of objects, extract value of a property as array

12 answers

how to get a list of key values from array of objects - JavaScript [duplicate]

2 answers

What I have is this:

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ]

And I want this:

const names = [

    "John",

    "Marry",

    "Jack"

  ]

How? Thanks!

This question already has an answer here:

From an array of objects, extract value of a property as array

12 answers

how to get a list of key values from array of objects - JavaScript [duplicate]

2 answers

javascript

asked Nov 19 at 11:14

FatBoyGebajzla

327

asked Nov 19 at 11:14

FatBoyGebajzla

327

asked Nov 19 at 11:14

FatBoyGebajzla

327

asked Nov 19 at 11:14

FatBoyGebajzla

327

asked Nov 19 at 11:14

FatBoyGebajzla

327

marked as duplicate by trincot javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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Nov 19 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by trincot javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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Nov 19 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you made any attempt at all yourself yet?
– CertainPerformance
Nov 19 at 11:15

Clearly he asked question without even trying himself
– siddhant sankhe
Nov 19 at 11:18

There's no need for that, I am fetching some objects from server, doing this for hours and lost my self... Actually the first answer is correct and I already tried that but did not realize that it work... No need for minuses, I am just tired... Thanks guys
– FatBoyGebajzla
Nov 19 at 11:22

add a comment |

Have you made any attempt at all yourself yet?
– CertainPerformance
Nov 19 at 11:15

Clearly he asked question without even trying himself
– siddhant sankhe
Nov 19 at 11:18

There's no need for that, I am fetching some objects from server, doing this for hours and lost my self... Actually the first answer is correct and I already tried that but did not realize that it work... No need for minuses, I am just tired... Thanks guys
– FatBoyGebajzla
Nov 19 at 11:22

Have you made any attempt at all yourself yet?
– CertainPerformance
Nov 19 at 11:15

Clearly he asked question without even trying himself
– siddhant sankhe
Nov 19 at 11:18

There's no need for that, I am fetching some objects from server, doing this for hours and lost my self... Actually the first answer is correct and I already tried that but did not realize that it work... No need for minuses, I am just tired... Thanks guys
– FatBoyGebajzla
Nov 19 at 11:22

add a comment |

3 Answers
3

active

oldest

votes

Use Array.prototype.map() to return only the name property.

The map() method creates a new array with the results of calling a provided function on every element in the calling array.

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ] 



const names = objArr.map(p => p.name);



console.log(names);

answered Nov 19 at 11:17

Mamun

24.7k71428

1

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

here you go ;)

const names = objArr.map(person => person.name);

answered Nov 19 at 11:16

Nikita Malyschkin

1136

add a comment |

Just use map method in combination with destructuring by passing a callback provided function as argument.

const objArr = [{name: "John", id: 1}, {name: "Marry", id: 2}, {name: "Jack", id: 3}] 

console.log(objArr.map(({name}) => name));

edited Nov 19 at 11:23

answered Nov 19 at 11:16

Mihai Alexandru-Ionut

29.4k63769

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Use Array.prototype.map() to return only the name property.

The map() method creates a new array with the results of calling a provided function on every element in the calling array.

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ] 



const names = objArr.map(p => p.name);



console.log(names);

answered Nov 19 at 11:17

Mamun

24.7k71428

1

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

Use Array.prototype.map() to return only the name property.

The map() method creates a new array with the results of calling a provided function on every element in the calling array.

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ] 



const names = objArr.map(p => p.name);



console.log(names);

answered Nov 19 at 11:17

Mamun

24.7k71428

1

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

Use Array.prototype.map() to return only the name property.

The map() method creates a new array with the results of calling a provided function on every element in the calling array.

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ] 



const names = objArr.map(p => p.name);



console.log(names);

answered Nov 19 at 11:17

Mamun

24.7k71428

Use Array.prototype.map() to return only the name property.

The map() method creates a new array with the results of calling a provided function on every element in the calling array.

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ] 



const names = objArr.map(p => p.name);



console.log(names);

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ] 



const names = objArr.map(p => p.name);



console.log(names);

const objArr = [

    {name: "John", id: 1}, 

    {name: "Marry", id: 2}, 

    {name: "Jack", id: 3}

  ] 



const names = objArr.map(p => p.name);



console.log(names);

answered Nov 19 at 11:17

Mamun

24.7k71428

answered Nov 19 at 11:17

Mamun

24.7k71428

answered Nov 19 at 11:17

Mamun

24.7k71428

answered Nov 19 at 11:17

Mamun

24.7k71428

1

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

1

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

here you go ;)

const names = objArr.map(person => person.name);

answered Nov 19 at 11:16

Nikita Malyschkin

1136

add a comment |

here you go ;)

const names = objArr.map(person => person.name);

answered Nov 19 at 11:16

Nikita Malyschkin

1136

add a comment |

here you go ;)

const names = objArr.map(person => person.name);

answered Nov 19 at 11:16

Nikita Malyschkin

1136

here you go ;)

const names = objArr.map(person => person.name);

answered Nov 19 at 11:16

Nikita Malyschkin

1136

answered Nov 19 at 11:16

Nikita Malyschkin

1136

answered Nov 19 at 11:16

Nikita Malyschkin

1136

answered Nov 19 at 11:16

Nikita Malyschkin

1136

add a comment |

Just use map method in combination with destructuring by passing a callback provided function as argument.

const objArr = [{name: "John", id: 1}, {name: "Marry", id: 2}, {name: "Jack", id: 3}] 

console.log(objArr.map(({name}) => name));

edited Nov 19 at 11:23

answered Nov 19 at 11:16

Mihai Alexandru-Ionut

29.4k63769

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

Just use map method in combination with destructuring by passing a callback provided function as argument.

const objArr = [{name: "John", id: 1}, {name: "Marry", id: 2}, {name: "Jack", id: 3}] 

console.log(objArr.map(({name}) => name));

edited Nov 19 at 11:23

answered Nov 19 at 11:16

Mihai Alexandru-Ionut

29.4k63769

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

Just use map method in combination with destructuring by passing a callback provided function as argument.

const objArr = [{name: "John", id: 1}, {name: "Marry", id: 2}, {name: "Jack", id: 3}] 

console.log(objArr.map(({name}) => name));

edited Nov 19 at 11:23

answered Nov 19 at 11:16

Mihai Alexandru-Ionut

29.4k63769

Just use map method in combination with destructuring by passing a callback provided function as argument.

const objArr = [{name: "John", id: 1}, {name: "Marry", id: 2}, {name: "Jack", id: 3}] 

console.log(objArr.map(({name}) => name));

const objArr = [{name: "John", id: 1}, {name: "Marry", id: 2}, {name: "Jack", id: 3}] 

console.log(objArr.map(({name}) => name));

const objArr = [{name: "John", id: 1}, {name: "Marry", id: 2}, {name: "Jack", id: 3}] 

console.log(objArr.map(({name}) => name));

edited Nov 19 at 11:23

answered Nov 19 at 11:16

Mihai Alexandru-Ionut

29.4k63769

edited Nov 19 at 11:23

answered Nov 19 at 11:16

Mihai Alexandru-Ionut

29.4k63769

answered Nov 19 at 11:16

Mihai Alexandru-Ionut

29.4k63769

answered Nov 19 at 11:16

Mihai Alexandru-Ionut

29.4k63769

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

You should really have voted to close as duplicate
– trincot
Nov 19 at 11:30

add a comment |

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