javascript - Q.all 没有以正确的顺序解析

Question

我在 Node.js 中遇到 Q promise 的执行顺序问题。

代码应该执行以下操作:

a) 执行查询并使用生成的纬度/经度对

b) 计算最短路径(使用异步函数)和

c) 归还它们

  query() // Runs the query
    .then(function() {
      return computeRoutes(); // Calculates the routes
    })
    .then(function() {
      return returnRoutes(); // Returns the routes as JSON
    });

问题是，虽然 coordinates[] 数组在 query() 中填充并在 computeRoutes() 中填充/可用，但 routes[] 数组在 returnRoutes() 中保持为空。

奇怪的是，当我没有在 computeRoutes() 中遍历坐标 [] 而是只为坐标 [0] 计算一条路线时， promise 链成功(但当然我只看到一条路线作为 JSON 返回)

对这里可能出现的问题有什么想法吗？

谢谢!!

完整代码:

app.get('/',
function(req, res) {

  var id = parseInt(req.query.id);
  var radius = parseInt(req.query.radius) || 5000;
  var routes = [];
  var coordinates = [];

  function query() {
    var deferred = Q.defer();

    console.log('Starting query function...');
    var query = client.query('SELECT ST_X(ST_Transform(ST_SetSRID(a.geom, 3857),4326)) AS fromlat, ST_Y(ST_Transform(ST_SetSRID(a.geom, 3857),4326)) AS fromlon, ST_X(ST_Transform(ST_SetSRID(b.geom, 3857),4326)) AS tolat, ST_Y(ST_Transform(ST_SetSRID(b.geom, 3857),4326)) AS tolon FROM emme_veh AS c, emme_nodes3857 AS b, emme_nodes3857 AS a WHERE c.fid = a.id AND c.tid = b.id AND c.fid = $1 AND ST_Distance(a.geom, b.geom) < $2', [id, radius], function(err, result) {
      console.log('Inside query. result.rows.length:',result.rows.length);

      for(var i in result.rows) {
        coordinates.push({'from':[result.rows[i].fromlon,result.rows[i].fromlat], 'to':[result.rows[i].tolon,result.rows[i].tolat]});
      }

      deferred.resolve();
    });
    return deferred.promise;
  }

  function computeRoutes() {
    var the_promises = [];

    for(var i in coordinates) {
      var deferred = Q.defer();

      var query = {coordinates: [coordinates[i].from, coordinates[i].to], alternateRoute: false}
      osrm.route(query, function(err, result) {
        if(err) return res.json({"error": err.message});
        // console.log(result.route_geometry);
        routes.push(result.route_geometry);
        deferred.resolve();
      });
      the_promises.push(deferred.promise);
    }
    return Q.all(the_promises);
  }

  function returnRoutes() {
    return res.json(routes);
  }

  query()
    .then(function() {
      return computeRoutes();
    })
    .then(function() {
      return returnRoutes();
    });

});

Answer 1

经过反复试验找到了答案。

问题是我使用 for..in 循环在 computeRoutes() 函数中生成 promise 。

切换到带有函数的 forEach 循环就可以了。

工作代码片段(查看 computeRoutes() 部分):

app.get('/',
function(req, res) {

  var id = parseInt(req.query.id);
  var radius = parseInt(req.query.radius) || 5000;
  var routes = [];
  var coordinates = [];

  function query() {
    var deferred = Q.defer();

    console.log('Starting query function...');
    var query = client.query('SELECT ST_X(ST_Transform(ST_SetSRID(a.geom, 3857),4326)) AS fromlat, ST_Y(ST_Transform(ST_SetSRID(a.geom, 3857),4326)) AS fromlon, ST_X(ST_Transform(ST_SetSRID(b.geom, 3857),4326)) AS tolat, ST_Y(ST_Transform(ST_SetSRID(b.geom, 3857),4326)) AS tolon FROM emme_veh AS c, emme_nodes3857 AS b, emme_nodes3857 AS a WHERE c.fid = a.id AND c.tid = b.id AND c.fid = $1 AND ST_Distance(a.geom, b.geom) < $2', [id, radius], function(err, result) {
      console.log('Inside query. result.rows.length:',result.rows.length);

      for(var i in result.rows) {
        coordinates.push({'from':[result.rows[i].fromlon,result.rows[i].fromlat], 'to':[result.rows[i].tolon,result.rows[i].tolat]});
      }

      deferred.resolve();
    });
    return deferred.promise;
  }

  function computeRoutes() {
    var the_promises = [];

    console.log('Inside computeRoutes()');

    coordinates.forEach(function(coordinate) {
      var deferred = Q.defer();

      osrm.route({coordinates: [coordinate.from, coordinate.to], alternateRoute: false}, function(err, result) {      
        deferred.resolve(result);
        routes.push(result.route_geometry);
      });
      the_promises.push(deferred.promise);
    });

    return Q.all(the_promises);
  }

  function returnRoutes() {
    console.log('Inside returnRoutes()');
    return res.json(routes);
  }

  query()
    .then(function() {
      console.log('then() 1');
      return computeRoutes();
    })
    .then(function() {
      console.log('then() 2');
      return returnRoutes();
    });