javascript - How to filter array of objects and then return a specific attribute? - Stack Overflow

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consider the data :

let orders = {
        "data": [
            {
                "email": "[email protected]", "orders": [
                    { "orderName": "something", "price": "43$" },
                    { "orderName": "anotherthing", "price": "4$" }
                ]
            },{
                "email": "[email protected]", "orders": [
                    { "orderName": "fish", "price": "43$" },
                    { "orderName": "parrot", "price": "4$" }
                ]
            }
        ]
    };

I'm trying to filter the orders of the object with some email like:

email = '[email protected]'
x=orders.data.filter(o =>{if (o.email === email) return o.orders});

but the whole return value is the whole matching object, with email and orders, and I don't want the whole object , I want only the orders.

consider the data :

let orders = {
        "data": [
            {
                "email": "[email protected]", "orders": [
                    { "orderName": "something", "price": "43$" },
                    { "orderName": "anotherthing", "price": "4$" }
                ]
            },{
                "email": "[email protected]", "orders": [
                    { "orderName": "fish", "price": "43$" },
                    { "orderName": "parrot", "price": "4$" }
                ]
            }
        ]
    };

I'm trying to filter the orders of the object with some email like:

email = '[email protected]'
x=orders.data.filter(o =>{if (o.email === email) return o.orders});

but the whole return value is the whole matching object, with email and orders, and I don't want the whole object , I want only the orders.

• javascript

Share Improve this question Follow edited Jun 11, 2019 at 13:50 Harel.Lebel asked Jun 11, 2019 at 13:34 Harel.LebelHarel.Lebel 28966 silver badges1717 bronze badges 3

• 3 why not just map after filter? – Austaras Commented Jun 11, 2019 at 13:36
• Possible duplicate of From an array of objects, extract value of a property as array – Heretic Monkey Commented Jun 11, 2019 at 13:42
• filter just checks whether the value returned by the function is truthy or not, it doesn't use it as the result. – Barmar Commented Jun 11, 2019 at 13:51

Add a ment  | 

4 Answers 4

Sorted by: Reset to default 6

You can't do with filter alone, you also need to map:

orders.data.filter(o => o.email === '[email protected]').map(o => o.orders)

You cannot do this with .filter as the method will only return a subsection if the array, while you want to also transform it.

You can chain Array#filter with Array#map to produce the result:

let orders = { "data": [{ "email": "[email protected]", "orders": [{ "orderName": "something", "price": "43$" }, { "orderName": "anotherthing", "price": "4$" } ] }, { "email": "[email protected]", "orders": [{ "orderName": "fish", "price": "43$" }, { "orderName": "parrot", "price": "4$" } ] }]};

email = '[email protected]';
x = orders.data
  .filter(o => o.email === email)
  .map(o => o.orders);
  
console.log(x);

If you expect a single element here, you can use Array#find instead:

let orders = { "data": [{ "email": "[email protected]", "orders": [{ "orderName": "something", "price": "43$" }, { "orderName": "anotherthing", "price": "4$" } ] }, { "email": "[email protected]", "orders": [{ "orderName": "fish", "price": "43$" }, { "orderName": "parrot", "price": "4$" } ] }]};

email = '[email protected]';
x = orders.data
  .find(o => o.email === email)
  .orders;
  
console.log(x);

An alternative is to use find and then just reference orders when needed. Adding || {'orders': 'Email not found'}; after will catch if the email isn't found

let orders = {
  "data": [{
    "email": "[email protected]",
    "orders": [{
        "orderName": "something",
        "price": "43$"
      },
      {
        "orderName": "anotherthing",
        "price": "4$"
      }
    ]
  }, {
    "email": "[email protected]",
    "orders": [{
        "orderName": "fish",
        "price": "43$"
      },
      {
        "orderName": "parrot",
        "price": "4$"
      }
    ]
  }]
};

email = '[email protected]'
x = orders.data.find(o => {
  return o.email === email
}) || {
  'orders': 'Email not found'
};
console.log(x.orders);
email = '[email protected]'
x = orders.data.find(o => {
  return o.email === email
}) || {
  'orders': 'Email not found'
};
console.log(x.orders);

let orders = {
        "data": [
            {
                "email": "[email protected]", "orders": [
                    { "orderName": "something", "price": "43$" },
                    { "orderName": "anotherthing", "price": "4$" }
                ]
            },{
                "email": "[email protected]", "orders": [
                    { "orderName": "fish", "price": "43$" },
                    { "orderName": "parrot", "price": "4$" }
                ]
            }
        ]
    };
    
const filteredOrders = orders.data.map((o) => o.email === '[email protected]' ? o.orders : null).filter(o => o);

console.log(filteredOrders)

You can map first too and filter after that the valid results.

consider the data :

let orders = {
        "data": [
            {
                "email": "[email protected]", "orders": [
                    { "orderName": "something", "price": "43$" },
                    { "orderName": "anotherthing", "price": "4$" }
                ]
            },{
                "email": "[email protected]", "orders": [
                    { "orderName": "fish", "price": "43$" },
                    { "orderName": "parrot", "price": "4$" }
                ]
            }
        ]
    };

I'm trying to filter the orders of the object with some email like:

email = '[email protected]'
x=orders.data.filter(o =>{if (o.email === email) return o.orders});

but the whole return value is the whole matching object, with email and orders, and I don't want the whole object , I want only the orders.

consider the data :

let orders = {
        "data": [
            {
                "email": "[email protected]", "orders": [
                    { "orderName": "something", "price": "43$" },
                    { "orderName": "anotherthing", "price": "4$" }
                ]
            },{
                "email": "[email protected]", "orders": [
                    { "orderName": "fish", "price": "43$" },
                    { "orderName": "parrot", "price": "4$" }
                ]
            }
        ]
    };

I'm trying to filter the orders of the object with some email like:

email = '[email protected]'
x=orders.data.filter(o =>{if (o.email === email) return o.orders});

but the whole return value is the whole matching object, with email and orders, and I don't want the whole object , I want only the orders.

• javascript

Share Improve this question Follow edited Jun 11, 2019 at 13:50 Harel.Lebel asked Jun 11, 2019 at 13:34 Harel.LebelHarel.Lebel 28966 silver badges1717 bronze badges 3

• 3 why not just map after filter? – Austaras Commented Jun 11, 2019 at 13:36
• Possible duplicate of From an array of objects, extract value of a property as array – Heretic Monkey Commented Jun 11, 2019 at 13:42
• filter just checks whether the value returned by the function is truthy or not, it doesn't use it as the result. – Barmar Commented Jun 11, 2019 at 13:51

Add a ment  | 

4 Answers 4

Sorted by: Reset to default 6

You can't do with filter alone, you also need to map:

orders.data.filter(o => o.email === '[email protected]').map(o => o.orders)

You cannot do this with .filter as the method will only return a subsection if the array, while you want to also transform it.

You can chain Array#filter with Array#map to produce the result:

let orders = { "data": [{ "email": "[email protected]", "orders": [{ "orderName": "something", "price": "43$" }, { "orderName": "anotherthing", "price": "4$" } ] }, { "email": "[email protected]", "orders": [{ "orderName": "fish", "price": "43$" }, { "orderName": "parrot", "price": "4$" } ] }]};

email = '[email protected]';
x = orders.data
  .filter(o => o.email === email)
  .map(o => o.orders);
  
console.log(x);

If you expect a single element here, you can use Array#find instead:

let orders = { "data": [{ "email": "[email protected]", "orders": [{ "orderName": "something", "price": "43$" }, { "orderName": "anotherthing", "price": "4$" } ] }, { "email": "[email protected]", "orders": [{ "orderName": "fish", "price": "43$" }, { "orderName": "parrot", "price": "4$" } ] }]};

email = '[email protected]';
x = orders.data
  .find(o => o.email === email)
  .orders;
  
console.log(x);

An alternative is to use find and then just reference orders when needed. Adding || {'orders': 'Email not found'}; after will catch if the email isn't found

let orders = {
  "data": [{
    "email": "[email protected]",
    "orders": [{
        "orderName": "something",
        "price": "43$"
      },
      {
        "orderName": "anotherthing",
        "price": "4$"
      }
    ]
  }, {
    "email": "[email protected]",
    "orders": [{
        "orderName": "fish",
        "price": "43$"
      },
      {
        "orderName": "parrot",
        "price": "4$"
      }
    ]
  }]
};

email = '[email protected]'
x = orders.data.find(o => {
  return o.email === email
}) || {
  'orders': 'Email not found'
};
console.log(x.orders);
email = '[email protected]'
x = orders.data.find(o => {
  return o.email === email
}) || {
  'orders': 'Email not found'
};
console.log(x.orders);

let orders = {
        "data": [
            {
                "email": "[email protected]", "orders": [
                    { "orderName": "something", "price": "43$" },
                    { "orderName": "anotherthing", "price": "4$" }
                ]
            },{
                "email": "[email protected]", "orders": [
                    { "orderName": "fish", "price": "43$" },
                    { "orderName": "parrot", "price": "4$" }
                ]
            }
        ]
    };
    
const filteredOrders = orders.data.map((o) => o.email === '[email protected]' ? o.orders : null).filter(o => o);

console.log(filteredOrders)

You can map first too and filter after that the valid results.

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