Why does map function return undefined but console.log logs out?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}

I want to return matching proprieties of two arrays of objects. But I got undefined from map function.

let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



fruits1.forEach((fruit1) => {

    fruits2.filter((fruit2) => {

        return fruit1.name === fruit2.name;

    }).map((newFruit) => {

        //console.log(newFruit.name);

        return newFruit.name;

    })

})

asked Jan 3 at 16:20

Ivan

65117

5

The .map() function returns a new array, but your code does not use the return value.

– Pointy
Jan 3 at 16:21

2

also Foreach returns undeifed developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Code Maniac
Jan 3 at 16:22

fruits2.filter(o=> fruits1.some(i=> i.name == o.name) ).map(o=> o.name);

– George Bailey
Jan 3 at 16:27

1

Is it just me, or is it a slight code smell that the elements have an "id", but that's not what is being used in the comparison? Maybe not.

– Taplar
Jan 3 at 16:27

add a comment |

I want to return matching proprieties of two arrays of objects. But I got undefined from map function.

let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



fruits1.forEach((fruit1) => {

    fruits2.filter((fruit2) => {

        return fruit1.name === fruit2.name;

    }).map((newFruit) => {

        //console.log(newFruit.name);

        return newFruit.name;

    })

})

asked Jan 3 at 16:20

Ivan

65117

5

The .map() function returns a new array, but your code does not use the return value.

– Pointy
Jan 3 at 16:21

2

also Foreach returns undeifed developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Code Maniac
Jan 3 at 16:22

fruits2.filter(o=> fruits1.some(i=> i.name == o.name) ).map(o=> o.name);

– George Bailey
Jan 3 at 16:27

1

Is it just me, or is it a slight code smell that the elements have an "id", but that's not what is being used in the comparison? Maybe not.

– Taplar
Jan 3 at 16:27

add a comment |

I want to return matching proprieties of two arrays of objects. But I got undefined from map function.

let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



fruits1.forEach((fruit1) => {

    fruits2.filter((fruit2) => {

        return fruit1.name === fruit2.name;

    }).map((newFruit) => {

        //console.log(newFruit.name);

        return newFruit.name;

    })

})

asked Jan 3 at 16:20

Ivan

65117

I want to return matching proprieties of two arrays of objects. But I got undefined from map function.

let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



fruits1.forEach((fruit1) => {

    fruits2.filter((fruit2) => {

        return fruit1.name === fruit2.name;

    }).map((newFruit) => {

        //console.log(newFruit.name);

        return newFruit.name;

    })

})

javascript

asked Jan 3 at 16:20

Ivan

65117

asked Jan 3 at 16:20

Ivan

65117

asked Jan 3 at 16:20

Ivan

65117

asked Jan 3 at 16:20

Ivan

65117

asked Jan 3 at 16:20

Ivan

65117

5

The .map() function returns a new array, but your code does not use the return value.

– Pointy
Jan 3 at 16:21

2

also Foreach returns undeifed developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Code Maniac
Jan 3 at 16:22

fruits2.filter(o=> fruits1.some(i=> i.name == o.name) ).map(o=> o.name);

– George Bailey
Jan 3 at 16:27

1

Is it just me, or is it a slight code smell that the elements have an "id", but that's not what is being used in the comparison? Maybe not.

– Taplar
Jan 3 at 16:27

add a comment |

5

The .map() function returns a new array, but your code does not use the return value.

– Pointy
Jan 3 at 16:21

2

also Foreach returns undeifed developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Code Maniac
Jan 3 at 16:22

fruits2.filter(o=> fruits1.some(i=> i.name == o.name) ).map(o=> o.name);

– George Bailey
Jan 3 at 16:27

1

Is it just me, or is it a slight code smell that the elements have an "id", but that's not what is being used in the comparison? Maybe not.

– Taplar
Jan 3 at 16:27

The .map() function returns a new array, but your code does not use the return value.

– Pointy
Jan 3 at 16:21

also Foreach returns undeifed developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Code Maniac
Jan 3 at 16:22

fruits2.filter(o=> fruits1.some(i=> i.name == o.name) ).map(o=> o.name);

– George Bailey
Jan 3 at 16:27

Is it just me, or is it a slight code smell that the elements have an "id", but that's not what is being used in the comparison? Maybe not.

– Taplar
Jan 3 at 16:27

add a comment |

3 Answers
3

active

oldest

votes

What are you looking for is an array intersection:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);

Usage:

console.log('inBoth:', inBoth(list1, list2));

Working Example:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);



let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



console.log('inBoth:', inBoth(fruits1, fruits2));

answered Jan 3 at 16:56

Mosè Raguzzini

4,7701226

add a comment |

You could use a Set and filter the names.

const names = ({ name }) => name;



var fruits1 = [{ id: 1, name: "apple" }, { id: 2, name: "dragon fruit" }, { id: 3, name: "banana" }, { id: 4, name: "kiwi" }, { id: 5, name: "pineapple" }, { id: 6, name: "watermelon" }, { id: 7, name: "pear" }],

    fruits2 = [{ id: 7, name: "pear" }, { id: 10, name: "avocado" }, { id: 5, name: "pineapple" }],

    common = fruits1

        .map(names)

        .filter(Set.prototype.has, new Set(fruits2.map(names)));



console.log(common);

answered Jan 3 at 17:00

Nina Scholz

198k15110181

add a comment |

What you want to do is this:

/* first we filter fruits1 (arbitrary) */

let matchingFruits = fruits1.filter(f1 => {

  /* then we filter the frut if it exists in frtuis2 */

  return fruits2.find(f2 => f2.name === f1.name)

}).map(fruit => fruit.name) // and now we map if we only want the name strings

If you're not using a polyfill Array.find will not work in IE. The alternative would be using Array.indexOf (thanks for pointing this out @JakobE).

Be aware that Array.forEach return value is undefined and that, in order to actually use the Array.map correctly, one has to consume the returned value somehow or assign it to a variable, as we just did with matchingFruits.

edited Jan 3 at 17:03

answered Jan 3 at 16:30

GMaiolo

1,9031024

Note! 'find' is not supported by IE

– Jakob E
Jan 3 at 16:34

let matchingFruits = fruits1.filter(item => fruits2.map(item => item.name).indexOf(item.name)!== -1).map(item => item.name);

– Jakob E
Jan 3 at 16:36

@JakobE just added that clarification, thanks.

– GMaiolo
Jan 3 at 17:03

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54026087%2fwhy-does-map-function-return-undefined-but-console-log-logs-out%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

What are you looking for is an array intersection:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);

Usage:

console.log('inBoth:', inBoth(list1, list2));

Working Example:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);



let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



console.log('inBoth:', inBoth(fruits1, fruits2));

answered Jan 3 at 16:56

Mosè Raguzzini

4,7701226

add a comment |

What are you looking for is an array intersection:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);

Usage:

console.log('inBoth:', inBoth(list1, list2));

Working Example:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);



let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



console.log('inBoth:', inBoth(fruits1, fruits2));

answered Jan 3 at 16:56

Mosè Raguzzini

4,7701226

add a comment |

What are you looking for is an array intersection:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);

Usage:

console.log('inBoth:', inBoth(list1, list2));

Working Example:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);



let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



console.log('inBoth:', inBoth(fruits1, fruits2));

answered Jan 3 at 16:56

Mosè Raguzzini

4,7701226

What are you looking for is an array intersection:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);

Usage:

console.log('inBoth:', inBoth(list1, list2));

Working Example:

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);



let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



console.log('inBoth:', inBoth(fruits1, fruits2));

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);



let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



console.log('inBoth:', inBoth(fruits1, fruits2));

// Generic helper function that can be used for the three operations:        

const operation = (list1, list2, isUnion = false) =>

    list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );



// Following functions are to be used:

const inBoth = (list1, list2) => operation(list1, list2, true),

      inFirstOnly = operation,

      inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);



let fruits1 = [

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

    {id: 2, name: "dragon fruit"},

    {id: 3, name: "banana"},

    {id: 4, name: "kiwi"},

    {id: 5, name: "pineapple"},

    {id: 6, name: "watermelon"},

    {id: 7, name: "pear"},

]

let fruits2 = [

    {id: 7, name: "pear"},

    {id: 10, name: "avocado"},

    {id: 5, name: "pineapple"},

]



console.log('inBoth:', inBoth(fruits1, fruits2));

answered Jan 3 at 16:56

Mosè Raguzzini

4,7701226

answered Jan 3 at 16:56

Mosè Raguzzini

4,7701226

answered Jan 3 at 16:56

Mosè Raguzzini

4,7701226

answered Jan 3 at 16:56

Mosè Raguzzini

4,7701226

add a comment |

You could use a Set and filter the names.

const names = ({ name }) => name;



var fruits1 = [{ id: 1, name: "apple" }, { id: 2, name: "dragon fruit" }, { id: 3, name: "banana" }, { id: 4, name: "kiwi" }, { id: 5, name: "pineapple" }, { id: 6, name: "watermelon" }, { id: 7, name: "pear" }],

    fruits2 = [{ id: 7, name: "pear" }, { id: 10, name: "avocado" }, { id: 5, name: "pineapple" }],

    common = fruits1

        .map(names)

        .filter(Set.prototype.has, new Set(fruits2.map(names)));



console.log(common);

answered Jan 3 at 17:00

Nina Scholz

198k15110181

add a comment |

You could use a Set and filter the names.

const names = ({ name }) => name;



var fruits1 = [{ id: 1, name: "apple" }, { id: 2, name: "dragon fruit" }, { id: 3, name: "banana" }, { id: 4, name: "kiwi" }, { id: 5, name: "pineapple" }, { id: 6, name: "watermelon" }, { id: 7, name: "pear" }],

    fruits2 = [{ id: 7, name: "pear" }, { id: 10, name: "avocado" }, { id: 5, name: "pineapple" }],

    common = fruits1

        .map(names)

        .filter(Set.prototype.has, new Set(fruits2.map(names)));



console.log(common);

answered Jan 3 at 17:00

Nina Scholz

198k15110181

add a comment |

You could use a Set and filter the names.

const names = ({ name }) => name;



var fruits1 = [{ id: 1, name: "apple" }, { id: 2, name: "dragon fruit" }, { id: 3, name: "banana" }, { id: 4, name: "kiwi" }, { id: 5, name: "pineapple" }, { id: 6, name: "watermelon" }, { id: 7, name: "pear" }],

    fruits2 = [{ id: 7, name: "pear" }, { id: 10, name: "avocado" }, { id: 5, name: "pineapple" }],

    common = fruits1

        .map(names)

        .filter(Set.prototype.has, new Set(fruits2.map(names)));



console.log(common);

answered Jan 3 at 17:00

Nina Scholz

198k15110181

You could use a Set and filter the names.

const names = ({ name }) => name;



var fruits1 = [{ id: 1, name: "apple" }, { id: 2, name: "dragon fruit" }, { id: 3, name: "banana" }, { id: 4, name: "kiwi" }, { id: 5, name: "pineapple" }, { id: 6, name: "watermelon" }, { id: 7, name: "pear" }],

    fruits2 = [{ id: 7, name: "pear" }, { id: 10, name: "avocado" }, { id: 5, name: "pineapple" }],

    common = fruits1

        .map(names)

        .filter(Set.prototype.has, new Set(fruits2.map(names)));



console.log(common);

const names = ({ name }) => name;



var fruits1 = [{ id: 1, name: "apple" }, { id: 2, name: "dragon fruit" }, { id: 3, name: "banana" }, { id: 4, name: "kiwi" }, { id: 5, name: "pineapple" }, { id: 6, name: "watermelon" }, { id: 7, name: "pear" }],

    fruits2 = [{ id: 7, name: "pear" }, { id: 10, name: "avocado" }, { id: 5, name: "pineapple" }],

    common = fruits1

        .map(names)

        .filter(Set.prototype.has, new Set(fruits2.map(names)));



console.log(common);

const names = ({ name }) => name;



var fruits1 = [{ id: 1, name: "apple" }, { id: 2, name: "dragon fruit" }, { id: 3, name: "banana" }, { id: 4, name: "kiwi" }, { id: 5, name: "pineapple" }, { id: 6, name: "watermelon" }, { id: 7, name: "pear" }],

    fruits2 = [{ id: 7, name: "pear" }, { id: 10, name: "avocado" }, { id: 5, name: "pineapple" }],

    common = fruits1

        .map(names)

        .filter(Set.prototype.has, new Set(fruits2.map(names)));



console.log(common);

answered Jan 3 at 17:00

Nina Scholz

198k15110181

answered Jan 3 at 17:00

Nina Scholz

198k15110181

answered Jan 3 at 17:00

Nina Scholz

198k15110181

answered Jan 3 at 17:00

Nina Scholz

198k15110181

add a comment |

What you want to do is this:

/* first we filter fruits1 (arbitrary) */

let matchingFruits = fruits1.filter(f1 => {

  /* then we filter the frut if it exists in frtuis2 */

  return fruits2.find(f2 => f2.name === f1.name)

}).map(fruit => fruit.name) // and now we map if we only want the name strings

If you're not using a polyfill Array.find will not work in IE. The alternative would be using Array.indexOf (thanks for pointing this out @JakobE).

edited Jan 3 at 17:03

answered Jan 3 at 16:30

GMaiolo

1,9031024

Note! 'find' is not supported by IE

– Jakob E
Jan 3 at 16:34

let matchingFruits = fruits1.filter(item => fruits2.map(item => item.name).indexOf(item.name)!== -1).map(item => item.name);

– Jakob E
Jan 3 at 16:36

@JakobE just added that clarification, thanks.

– GMaiolo
Jan 3 at 17:03

add a comment |

What you want to do is this:

/* first we filter fruits1 (arbitrary) */

let matchingFruits = fruits1.filter(f1 => {

  /* then we filter the frut if it exists in frtuis2 */

  return fruits2.find(f2 => f2.name === f1.name)

}).map(fruit => fruit.name) // and now we map if we only want the name strings

If you're not using a polyfill Array.find will not work in IE. The alternative would be using Array.indexOf (thanks for pointing this out @JakobE).

edited Jan 3 at 17:03

answered Jan 3 at 16:30

GMaiolo

1,9031024

Note! 'find' is not supported by IE

– Jakob E
Jan 3 at 16:34

let matchingFruits = fruits1.filter(item => fruits2.map(item => item.name).indexOf(item.name)!== -1).map(item => item.name);

– Jakob E
Jan 3 at 16:36

@JakobE just added that clarification, thanks.

– GMaiolo
Jan 3 at 17:03

add a comment |

What you want to do is this:

/* first we filter fruits1 (arbitrary) */

let matchingFruits = fruits1.filter(f1 => {

  /* then we filter the frut if it exists in frtuis2 */

  return fruits2.find(f2 => f2.name === f1.name)

}).map(fruit => fruit.name) // and now we map if we only want the name strings

If you're not using a polyfill Array.find will not work in IE. The alternative would be using Array.indexOf (thanks for pointing this out @JakobE).

edited Jan 3 at 17:03

answered Jan 3 at 16:30

GMaiolo

1,9031024

What you want to do is this:

/* first we filter fruits1 (arbitrary) */

let matchingFruits = fruits1.filter(f1 => {

  /* then we filter the frut if it exists in frtuis2 */

  return fruits2.find(f2 => f2.name === f1.name)

}).map(fruit => fruit.name) // and now we map if we only want the name strings

If you're not using a polyfill Array.find will not work in IE. The alternative would be using Array.indexOf (thanks for pointing this out @JakobE).

edited Jan 3 at 17:03

answered Jan 3 at 16:30

GMaiolo

1,9031024

edited Jan 3 at 17:03

answered Jan 3 at 16:30

GMaiolo

1,9031024

answered Jan 3 at 16:30

GMaiolo

1,9031024

answered Jan 3 at 16:30

GMaiolo

1,9031024

Note! 'find' is not supported by IE

– Jakob E
Jan 3 at 16:34

let matchingFruits = fruits1.filter(item => fruits2.map(item => item.name).indexOf(item.name)!== -1).map(item => item.name);

– Jakob E
Jan 3 at 16:36

@JakobE just added that clarification, thanks.

– GMaiolo
Jan 3 at 17:03

add a comment |

Note! 'find' is not supported by IE

– Jakob E
Jan 3 at 16:34

let matchingFruits = fruits1.filter(item => fruits2.map(item => item.name).indexOf(item.name)!== -1).map(item => item.name);

– Jakob E
Jan 3 at 16:36

@JakobE just added that clarification, thanks.

– GMaiolo
Jan 3 at 17:03

Note! 'find' is not supported by IE

– Jakob E
Jan 3 at 16:34

let matchingFruits = fruits1.filter(item => fruits2.map(item => item.name).indexOf(item.name)!== -1).map(item => item.name);

– Jakob E
Jan 3 at 16:36

@JakobE just added that clarification, thanks.

– GMaiolo
Jan 3 at 17:03

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

Search This Blog

Ufyukyu