问题
I am trying to resolve a method like below using ui-router
$stateProvider
.state('abc', {
url: 'xyz',
templateUrl: 'templateURL',
controller: ctrl,
resolve:{
data: function(someService){
data = someService.init();
return data;
}
}
})
And my service code looks like this
var someObject = {
data1: ...,
data2: ....,
...
}
return{
init: function(){
promise1 = ...
promise2 = ...
promise3 = $http.get('someurl').then(function(){
...
...
//do some manipulation with someObj
return someObject;
});
$q.all(promise1 , promise2 ).then(promise3);
}
}
When I debug the code, it is coming to line return someObject
but then it is not resolving.
What am I doing wrong?
回答1:
If you want promise3
to only execute after promise1
and promise2
then try
return $q.all([promise1, promise2])
.then(function(arrayOfData) {
return promise3;
});
Here is an example to illustrate the difference:
var promise1 = $timeout(function() {
return 'promise1Data';
}, 1000);
var promise2 = $timeout(function() {
return 'promise2Data';
}, 2000);
var promise3 = $timeout(function() {
return 'promise3Data';
}, 5000);
// This is what you're essentially doing when your route resolves
$q.all([promise1, promise2])
.then(promise3)
.then(function(data) {
console.log(data); // You get ["promise1Data", "promise2Data"]
});
// This is what I think you want
$q.all([promise1, promise2])
.then(function(arrayOfResolvedData) {
return promise3;
})
.then(function(data) {
console.log(data); // You get promise3Data
});
回答2:
You must pass callback functions to .then(…)
. Your promise3
is none.
I suspect you wanted to do
return $q.all([promise1, promise2, promise3])
来源:https://stackoverflow.com/questions/32000037/resolve-and-q-all-issue