javascript return property value from nested array of objects based on condition

Question

i have an array of objects, in which each object could have an array of objects inside.

var mylist = [
    {
        "email" : null, 
        "school" : "schoolA",
        "courses": [
            {
                "name" : 'ABC', 
                "type" : "chemistry"
            }, 
            {
                "name" : 'XYZ',
                "type": "math"
            }
        ]
    }, 
    {
        "email" : null,
        "school": "schoolB"
    }
];

i want to return course name if one of the course type is chemistry. The course types are unique and even if they are some duplicates, we return the first one.

var result = mylist.some(function (el) {
            el.courses && el.courses.some(function(u) {
              if (u.type === 'chemistry') {
                 return u.name;
              };    
            })
        });

console.log('outcome:', result);

my code is not working at this stage.

You must use the find function to find the right course object and then return the name of the found course. The function some only returns a boolean value with the value of the given predicate. — Gilad Bar
– Gilad Bar, Commented May 7, 2018 at 9:35

T.J. Crowder · Accepted Answer · 2018-05-07 09:43:24Z

The some callback should return a truthy or falsy value, which tells some whether to keep going (true = stop), and some returns a boolean, not a callback return value.

Probably simplest in this case just to assign directly to result:

var result;
mylist.some(function(el) {
    return (el.courses || []).some(function(course) {
        if (course.type === "chemistry") {
            result = course.name;
            return true;
        }
        return false;
    });
});

Live Example:

var mylist = [
    {
        "email" : null, 
        "school" : "schoolA",
        "courses": [
            {
                "name" : 'ABC', 
                "type" : "chemistry"
            }, 
            {
                "name" : 'XYZ',
                "type": "math"
            }
        ]
    }, 
    {
        "email" : null,
        "school": "schoolB"
    }
];

var result;
mylist.some(function(el) {
    return (el.courses || []).some(function(course) {
        if (course.type === "chemistry") {
            result = course.name;
            return true;
        }
        return false;
    });
});
console.log(result);

I stuck to ES5 syntax since you didn't use any ES2015+ in your question, but in ES2015+, simplest probably to use nested for-of loops:

let result;
outer: for (const el of mylist) {
    for (const course of el.courses || []) {
        if (course.type === "chemistry") {
            result = course.name;
            break outer;
        }
    }
}

Live Example:

const mylist = [
    {
        "email" : null, 
        "school" : "schoolA",
        "courses": [
            {
                "name" : 'ABC', 
                "type" : "chemistry"
            }, 
            {
                "name" : 'XYZ',
                "type": "math"
            }
        ]
    }, 
    {
        "email" : null,
        "school": "schoolB"
    }
];

let result;
outer: for (const el of mylist) {
    for (const course of el.courses || []) {
        if (course.type === "chemistry") {
            result = course.name;
            break outer;
        }
    }
}
console.log(result);

Nenad Vracar · Accepted Answer · 2018-05-07 09:40:59Z

2

You could use reduce() method to iterate through each object in array and then find() method to find if some course matches type.

var mylist = [{"email":null,"school":"schoolA","courses":[{"name":"ABC","type":"chemistry"},{"name":"XYZ","type":"math"}]},{"email":null,"school":"schoolB"}]

const course = mylist.reduce((r, {courses}) => {
  if (courses && !r) {
    const course = courses.find(({type}) => type == 'chemistry');
    if (course) r = course.name;
  }
  return r;
}, null)

console.log(course)

answered May 7, 2018 at 9:40

Nenad Vracar

122k16 gold badges160 silver badges184 bronze badges

2 Comments

Nenad Vracar Over a year ago

Yes, reduce keeps going but at least find doesn't starts.

T.J. Crowder Over a year ago

Yeah, it's a really quick callback for the remaining elements. :-)

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