you can use Promise.all or for of plus async/await .

function pushData(arr, value) {
    return new Promise((reslove) => {
        setTimeout(() => {
            arr.push(value);
            console.log('push data over', value)
            reslove();
        }, 1000);
    })
}

async function longTimeDo() {
    let data = [];
    let sourse = [1, 2, 3, 4, 5];

    for (let value of sourse) {
        await pushData(data, value);
    }

    console.log(data)

}

longTimeDo();

this will not work:

function longTimeDo() {
    let data = [];
    let sourse = [1,2,3,4,5];
    sourse.forEach(value => {
        console.log('forEach-star')
        setTimeout(() => {
            data.push(value);
            console.log('forEach-async')
        }, 1000);
        console.log('forEach-end')
    });
    console.log('longTimeDo-end');
}
longTimeDo();

output is

from the results, we can see that the asynchrony in the loop only acts in the loop, but does not affect the outside of the loop. If you want this function to return the result after asynchronism, it is equivalent to asynchronously changing Synchronize, then you need to use Primise

async function longTimeDo () {
    let data = [];
    let sourse = [1,2,3,4,5];
    let allP = sourse.map(value => {
        return new Promise((res, rej) => {
            setTimeout(() => {
                data.push(value+1);
                res(data)
            }, 1000);
        });
    });
    await Promise.all(allP);

    return data;
}
longTimeDo().then(res => {
    log(res) // [2, 3, 4, 5, 6]
})

is to add a count, num=0;settimeout to add 1 at a time, and to judge return data when num==source.length