Pull variable from Javascript function (Phonegap)

Question

I tried to get value of variable from function beyond it, but always get error: Uncaught ReferenceError: devuuid_raw is not defined

The task is: I need to get value of variable beyond function in another variable because that value need to be used further (send to the server).

Code:

document.addEventListener("deviceready", onDeviceReady, false);

function onDeviceReady() {
    console.log("Device is ready");
    var element = document.getElementById('devProps');
    devuuid_raw = device.uuid;
    return devuuid_raw;
    element.innerHTML = 'Идентификатор устройства: ' + device.uuid + '<br />';                                               
}

var devuuid = devuuid_raw;

How can I get device.uuid value outside the function?

Simon Hänisch · Accepted Answer · 2015-01-05 12:25:08Z

2

First create a variable

var devuuid;

then add the event listener

document.addEventListener("deviceready", onDeviceReady, false);

and simply change the variable inside the function:

function onDeviceReady() {
  console.log("Device is ready");
  var element = document.getElementById('devProps');
  devuuid_raw = device.uuid;
  devuuid = devuuid_raw;
  element.innerHTML = 'Идентификатор устройства: ' + device.uuid + '<br />';
}

If you don't use the 'var' keyword inside the function, it will reference to the variable defined before.

edited Jan 5, 2015 at 12:25

answered Apr 13, 2014 at 14:23

Simon Hänisch

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9 Comments

user3166813 Over a year ago

Thanks, your solution removed Uncaught ReferenceError, but I still cannot get value from variable, I tried this: var devuuid; document.addEventListener("deviceready", onDeviceReady, false); function onDeviceReady() { console.log("Device is ready"); devuuid = device.uuid; document.getElementById('devProps').value = devuuid; } But variable still not show value :(

Regent Over a year ago

@user3166813 how and when do you check devuuid value?

user3166813 Over a year ago

<p id="devProps">Load info...</p> So, js code above should get element by id and show value, but it dont do it, then I send this value with using jquery.ajax, but it also dont work because it cannot get value

Simon Hänisch Over a year ago

try console.log(devuuid) as last line inside the function. does it show the expected value? if you want to access the variable inside the javascript console, you have to create it globally without the 'var' keyword, e.g. devuuid = 0;

user3166813 Over a year ago

it don`t show the value, but I think it should not because device.uuid is specific phonegap feature and works only on phones

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