如何使异步函数成为普通的同步函数

Question

i would use an async function as normal synchronous function, i tried use await but it still doesn't work. Where did i wrong?

const getScript = (url) => {
return new Promise((resolve, reject) => {
    const http = require('http'),
        https = require('https');

    let client = http;

    if (url.toString().indexOf("https") === 0) {
        client = https;
    }

    client.get(url, (resp) => {
        let data = '';

        // A chunk of data has been recieved.
        resp.on('data', (chunk) => {
            data += chunk;
        });

        // The whole response has been received. Print out the result.
        resp.on('end', () => {
            resolve(data);
        });

    }).on("error", (err) => {
        reject(err);
    });
});

};

async function getBody(url) {

(async (url) => {
    return ("qwe" + await getScript(url));
})(url);

}

console.log(getBody('http://myurl.com/myapi'));


console.log("end");

But it still doesn't load the data before the console.log "end". Where did i wrong? Thank you

Answer 1

Asynchronous function cannot be converted to synchronous. a in asynchronous means that it's not synchronous. async can converted to regular function, but it will still be asynchronous.

Asynchronicity is contagious. Once it's async (promise-based), it's preferable to stick to promise control flow.

There should be top-level async function at some point, and async IIFE inside getBody is redundant. It should be:

async function getBody(url) {
    return ("qwe" + await getScript(url));
}

(async () => {
    console.log(await getBody('http://myurl.com/myapi'));
    console.log("end");
})()

Answer 2

It will be await not async

async function getBody(url) {

(await (url) => {
    return ("qwe" + await getScript(url));
})(url);

Answer 3

It is because you are not awaiting for the result.Here when the IIFE is been executed (which is async) it returned a promise. You are not awaiting for the IIFE to be resolved. Thus you should simply add await in front of IIFE like -

async function getBody(url) {
    await (async (url) => { // add await before executing this function right away
        return ("qwe" + await getScript(url));
    })(url);

Answer 4

Try this (I write code from head)

async function getBody(url) {
    let r = await getScript(url));
    return ("qwe" + r);
}

(async () => {
    let b = await getBody('http://myurl.com/myapi')
    console.log(b);
    console.log("end");
}) ();