Matthew Moon - 1 year ago 26

Javascript Question

Here's what I've got. One array of objects:

`var teachers = [{`

Year: 2016,

FullName: "Matt",

Age: 39

},

{

Year: 2016,

FullName: "Sara",

Age: 25

},

...

];

And another array of objects. These would be nested like so:

`var students = [[`

{

Year: 2016,

FullName: "Zoe"

Age: 8

}

],

[

{

Year: 2016,

FullName: "Lulu"

Age: 9

},

{

Year: 2016,

FullName: "Leo",

Age: 13

}

],

[ // empty array here

],

[

{

Year: 2016,

FullName: "Lotta",

Age: 11

}

]

...

];

How they are organized is that students[0] is a student of teachers[0]. students[4] are students of teachers[4], and so forth.

What I was attempting to do what to take the FullName property, 'Students' in each student and put those values into an array of a new property of teachers called 'SundayStudents'. So what I'd end up with would be:

`teachers = [{`

Year: 2016,

FullName: "Matt",

Age: 39,

SundayStudents: ["Zoe"]

},

{

Year: 2016,

FullName: "Sara",

Age: 25,

SundayStudents: ["Lulu", "Leo"]

},

...

];

I tried a nested for-loop, but the students array has varying numbers of objects in each sub-array, and it doesn't create an array for the new property. I think I'm stuck.

`for (var j = 0, leng = teachers.length; j < leng; j++) {`

for (var k = 0, lent = students.length; k < lent; k++)

Teachers[i].SundayStudents = Students[j][k].FullName;

}

Any hints are welcome.

Answer

You could just iterate and check if the target element exist. Then you could make a new property with the mapped names.

```
var teachers = [{ Year: 2016, FullName: "Matt", Age: 39 }, { Year: 2016, FullName: "Sara", Age: 25 }],
students = [[{ Year: 2016, FullName: "Zoe", Age: 8 }], [{ Year: 2016, FullName: "Lulu", Age: 9 }, { Year: 2016, FullName: "Leo", Age: 13 }], [{ Year: 2016, FullName: "Lotta", Age: 11 }]];
students.forEach(function (a, i) {
if (Array.isArray(a) && teachers[i]) {
teachers[i].SundayStudents = a.map(function (b) {
return b.FullName;
});
}
});
console.log(teachers);
```

Source (Stackoverflow)