TypeScript - Take object out of array based on attribute value

Solution 1:

Use Array.find:

let array = [
    { id: 1, value: "itemname" },
    { id: 2, value: "itemname" }
];

let item1 = array.find(i => i.id === 1);

Array.find at MDN: https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/find

Solution 2:

I'd use filter or reduce:

let array = [
    { id: 1, value: "itemname" },
    { id: 2, value: "itemname" }
];

let item1 = array.filter(item => item.id === 1)[0];
let item2 = array.reduce((prev, current) => prev || current.id === 1 ? current : null);

console.log(item1); // Object {id: 1, value: "itemname"}
console.log(item2); // Object {id: 1, value: "itemname"}

(code in playground)

If you care about iterating over the entire array then use some:

let item;
array.some(i => {
    if (i.id === 1) {
        item = i;
        return true;
    }
    return false;
});

(code in playground)

Solution 3:

You can search a certain value in array of objects using TypeScript dynamically if you need to search the value from all fields of the object without specifying column

 var searchText = 'first';

let items = [
            { id: 1, name: "first", grade: "A" },
            { id: 2, name: "second", grade: "B" }
        ];

This below code will search for the value

var result = items.filter(item => 
             Object.keys(item).some(k => item[k] != null && 
             item[k].toString().toLowerCase()
             .includes(searchText.toLowerCase()))
             );

Same approach can be used to make a Search Filter Pipe in angularjs 4 using TypeScript

Solution 4:

You'll have to loop over the array, but if you make a hashmap to link each id to an index and save that, you only have to do it once, so you can reference any objeft after that directly:

var idReference = myArray.reduce(function( map, record, index ) {
    map[ record.id ] = index;
    return map;
}, {});

var objectWithId5 = myArray[ idReference["5"] ];

This does assume all ids are unique though.