We are writing server code in nodejs and using Parse javascript sdk. Often we need to fetch or update various Parse objects. For example, fetch a user with username 'Example'.
function fetchUser(username)
{
var User = Parse.Object.extend("User");
var query = new Parse.Query("User");
query.equalTo("username",username);
query.first({
success: function(results) {
console.log("Successfully retrieved " + results.length + " scores.");
return results;
},
error: function(error) {
console.log("Error: " + error.code + " " + error.message);
}
});
}
This function might be called by some other function:
function test()
{
var user = fetchUser("example");
console.log(user); // would often print undefined
return user;
}
function top()
{
// some code
var user = test();
//some code
}
// and the function top() mmight be called by some other functions
The problem is that I would get undefined results because the find/first operations runs asynchronously. Is there any way to ensure that fetchUser() function returns only when the find/first operation is successful/complete?
JavaScript is asynchronous by its nature. You have to pass a callback to the fetchUser()
function and rewrite your client code to something like this:
function fetchUser(username, callback) {
// do request and call callback(null, user) on success
// or callback(some_error_object) on failure.
}
function test(callback) {
var onFetched = function(err, user) {
console.log(user);
callback(err, user);
}
var user = fetchUser("example", onFetched);
}
Also you can use promises or generators (since EcmaScript 6) approaches.
UPD: Although some APIs (DB drivers, fs and so on) allow one to use them in a synchronous manner, it's considered as bad way to write a code in JavaScript. Since JS runs your code in a single thread, blocking this thread by using synchronous calls will block all other tasks. So, if you develop a web server that handles a few clients simultaneously and decided to use synchronous calls to some API, only one client at a time will be served while other clients will be in a pending state.
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