I need to run a function \"searchBoats(boatType)\" in AngularJS 1.0.7 with a parameter. This parameter is the result of another function parseBoatType
You can return the resource $promise
from the BoatType
resource in the parseBoatTime
function and use the promise to resolve the parseUrl
deferred.
First return a promise from the parseBoatTime
function:
return BoatType.getBoatTypeByName({
name: boatTypeParsed
}, function success(result) {
return result;
}).$promise;
Then resolve the parseUrl
deferred
with the promise from the BoatType
resource:
parseBoatType().then(deferred.resolve);
Bellow is the full code taken from your question with the correction I mentioned.
var parseURL = function() {
var deferred = $q.defer();
var promise = deferred.promise;
promise.then(function success(result) {
console.log(result);
searchBoats(result);
});
parseBoatType().then(deferred.resolve);
};
parseURL();
var parseBoatType = function() {
// Do some stuff
// Calculate boatType calling a service that uses resource to call
// an API
// I can convert this callback into a promise but still facing same
// issue
// Code for ngResource@^1.2.0
/*return BoatType.getBoatTypeByName({
name: boatTypeParsed
}, function success(result) {
return result;
}).$promise;*/
// Code for ngResource lower than 1.2.0
var deferred = $q.defer(), promise = deferred.promise;
BoatType.getBoatTypeByName({
name: boatTypeParsed
}, deferred.resolve, deferred.reject);
return promise;
// The service method is called and the code is still running until
// the end of the function without waiting for the service result.
// Then the promise.then code in the parseURL is executed and
// searchBoats is run with boatType undefined.
};
// The service with the $resource call to the API
app.factory('BoatType',
function($resource, SERVER_URL) {
var boatTypes =
$resource('http://' + SERVER_URL + '/:action', {
action: 'boat_types'
}, {
query: {
method: 'GET',
isArray: true
},
getBoatTypeByName: {
method: 'GET',
params: {
action: 'getBoatTypeByName'
},
isArray: false
}
});
return boatTypes;
}
)