Null return value from a async function

Question

1.00/5 (1 vote)

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How does this function returns null?

If I use function lsinfo() without async the code works fine but info['_myip'] return null.

when I use function lsinfo() with async then info['_myip'] return data but lsinfo() retuen 0 length

What I have tried:

JavaScript

function get_app_data() {

    var json_obj = lsinfo();
    var myJSON = JSON.stringify(json_obj);
}

async function lsinfo() {
    if (navigator.userAgent) {
        localStorage.lsuserAgent = navigator.userAgent;
    }

    if (navigator.geolocation) {
        navigator.geolocation.getCurrentPosition(displayLocationInfo);
    }

    var lsuser_id = localStorage.getItem("user_id")

    if (lsuser_id === null || lsuser_id.length === 0) {
        lsuser_id = "";
    }

    var info = new Object();
    info['_luserid'] = lsuser_id;
    info['_mypage'] = localStorage.getItem("lspage");
    info['_myip'] = localStorage.getItem("gioloc");
    info['_divid'] = localStorage.getItem("lsdvicid");
    info['_appver'] = localStorage.getItem("lsappver");
    info['_useragent'] = localStorage.getItem("lsuserAgent");
    info['_mymsg'] = "";
    return info;
}


function displayLocationInfo(position) {
    const lng = position.coords.longitude;
    const lat = position.coords.latitude;
    localStorage.gioloc = lat + ',' + lng;
}

Posted 27-Jun-22 0:48am

Shikha Gupta 2022

Updated 27-Jun-22 1:10am

Afzaal Ahmad Zeeshan

v2

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Afzaal Ahmad Zeeshan · Accepted Answer · 2022-06-27T01:10:00

Quote:
If I use function lsinfo() without async the code works fine but info['_myip'] return null.

The async keyword does nothing in your context because you are not using an await keyword in the function scope. You say that it returns null, the only plausible reason that I could think of — remember that I cannot access your machine, so debugging is a bit hard — is that your localStorage.getItem("gioloc"); returns a null object. Meaning, the item with key gioloc does not exist "yet" in your localStorage.

Quote:
when I use function lsinfo() with async then info['_myip'] return data but lsinfo() retuen 0 length

This is very unclear sentence. Async will not change the return type or the value of the return — other than making it a promise[^] that can be awaited, if anything. Secondly, your lsinfo is a method that you need to call. Applying the length to the return would provide the length of the return value (which in your method is a new Object() with some added attributes). Perhaps, you want to do lsinfo()['_myip'].length? Continue reading for a potential fix, I think you are not using the localStorage API correctly.

JavaScript

localStorage.gioloc = lat + ',' + lng;

Also, this is not how the localStorage API is supposed to be used. Here you are setting a field to the localStorage object "itself". To set a key in the localStorage you need to use the setItem function.

JavaScript

function displayLocationInfo(position) {
    const lng = position.coords.longitude;
    const lat = position.coords.latitude;
    localStorage.setItem("gioloc", lat + ',' + lng);
}

This should set the item and the next time you access it, you will be able to read the value.

Window.localStorage - Web APIs | MDN[^]