在帶有Q承諾的回調中返回值？

Question

在我的代碼中，我從數據庫中獲取hosts並在回調中對其進行處理。 如何退回已處理的hosts ？

var db = new sqlite3.Database(DB);
var all = Q.nbind(db.all, db);

function getHosts() {
    return all('SELECT host FROM hosts ORDER BY host DESC', function(err, rows){
        // rows:  [ { host: 'z' }, { host: 'a' } ]
        // transform into hosts: ['a','z']
        var hosts = [];
        var L = rows.length;

        for (var i=0; i<L; i++) {
            hosts.push(rows.pop().host);
        }
        // hosts = ['a','b', ... 'z']
        return hosts;  // <-- doesn't work!
    });
}

Answer 1

由於要將其轉換為基於promise的函數，因此需要按以下方式使用它：

function getHosts() {
    return all('SELECT host FROM hosts ORDER BY host DESC').then(function(rows) {
        // rows:  [ { host: 'z' }, { host: 'a' } ]
        // transform into hosts: ['a','z']
        var hosts = [];
        var L = rows.length;

        for (var i=0; i<L; i++) {
            hosts.push(rows.pop().host);
        }
        // hosts = ['a','b', ... 'z']
        return hosts;  // <-- doesn't work!
    });
}

我這里沒有catch()處理函數，因此您可以在某個地方處理它。