Test if html5 geolocation permission already has been granted

Question

Does anyone know if there is a way to test if prior html5 geolocation permission has been granted?

I try to make a script that does not request the geolocation unles

Answer 1

The first part actually answers your question, at least on Chrome 43+ and Firefox 46+ (see https://developer.mozilla.org/en-US/docs/Web/API/Permissions_API):

navigator.permissions &&
navigator.permissions.query({name: 'geolocation'}).then(function(PermissionStatus) {
    if('granted' === PermissionStatus.state) {
        navigator.geolocation.getCurrentPosition(function(geoposition) {
            console.log(geoposition) /* You can use this position without prompting the user if the permission had already been granted */
        })
    }
})

Answer 2

According to the Geolocation API Specification there isn't any function for this. Even the option to save it in the browser preferences isn't pushed to your code.

If it's realy important and you realy have to beeing aware of this, you have to write a browser addon for every common browser and ask the user to install it. But i wouldn't doubt of it.

Answer 3

Best you can do is to keep track of this yourself...

// In chrome you can now do this    
navigator.permissions.query({name: 'geolocation'}).then(function(PermissionStatus){
    console.log(PermissionStatus.state) // prompt, granted, denied
    // even listen for changes
    PermissionStatus.onchange = function(){
        console.log(this.state)
    }
})

fallback method:

// initialization
if( sessionStorage.getItem("geo_access") === null ){
    // just assume it is prompt
    sessionStorage.setItem("geo_access", "prompt");
}

function ask(){
    navigator.geolocation.getCurrentPosition(function(){
        sessionStorage.setItem("geo_access", "granted");

    }, function(err){
        if(err.code == 1){ // PERMISSION_DENIED
            sessionStorage.setItem("geo_access", "denied");
        }
        sessionStorage.setItem("geo_access", "prompt");
    });
};

// Then somewhere
ask();