Question

我知道node.js是事件驱动的，我应该这样做异步但我无法在脑海中找到一种方法。

所以，我有这个

var querystr = "SELECT * FROM groups";
    var result = "";
    dbClient.query(querystr, function (err, res, fields) {
        if (err) {
            console.log(err);
            return;
        }
        for(var i in res) {
            result = result + "@" +res[i].name + " =";
            for (var j in res[i].members.split(",")) {
                var memberquery;
                if (j.substr(0,1) == "@") {
                    memberquery = "SELECT name FROM groups WHERE id = "+j.substr(1, j.length-1);
                } else {
                    memberquery = "SELECT username FROM users WHERE id= "+j;
                }
                dbClient.query(memberquery, function(err, memres, fields) {
                    var membername = "";
                    if (typeof memres[0].username == "undefined") {
                        membername = "@"+memres[0].name;
                    } else {
                        membername = memres[0].username;
                    }
                    result = result + " " + membername; 
                });

            }
            result = result + "\n"; 
        }
    });

使其同步的问题是内部问题。基本上我在结果变量中生成一个文档，我检查组并告诉成员所以预期的输出是

Group1 = member, member
Group2 = member, member

Answer 1

对于这类问题，我通常使用类似下面的模式。简而言之：获取一个事物列表然后调用一个函数来处理列表;该函数调用自身直到列表完成;你在累加器中收集结果;当列表为空时，返回通过回调累积的任何内容。这只是完成@Andrey Sidorov在回应中所表现出来的另一种方式。

//cb is (err, res) 
function getData(cb){
  var querystr = "SELECT * FROM groups";
  var result = "";

  dbClient.query(querystr, function (err, res, fields) {
      if (err)
          cb(err);
      else {
        var groups = [];
        for (var ndx in res)
          groups = groups.concat(res[ndx].members.split(","));

        getMembers(groups, [], cb);
      }
  });
}

function getMembers(members, sofar, cb){
  var member = members.shift();

  if (!member)
    cb(null, sofar);
  else {
    var memberquery;
    var params;

    if (member.substr(0,1) == "@") {
        memberquery = "SELECT name FROM groups WHERE id = ?";
        params = [member.substr(1, member.length-1)];
    } else {
        memberquery = "SELECT username FROM users WHERE id = ?";
        params = [member];
    }

    dbClient.query(memberquery, params, function(err, res) {
      if (err)
        cb(err);
      else {
        var membername = "";
        if (typeof res[0].username == "undefined") {
            membername = "@" + res[0].name;
        } else {
            membername = res[0].username;
        }

        sofar.push(membername);

        getMembers(members, sofar, cb);
      }
    });
  }
}

Answer 2

function do_queries( resultCallback )
{
    var querystr = "SELECT * FROM groups";
    var result = "";
    var num_queries_left = 0;
    dbClient.query(querystr, function (err, res, fields) {
        if (err) {
            console.log(err);
            return;
        }

        // calculate number of invocations of second sub-query
        for(var i in res)
            num_queries_left += res[i].members.split(",").length;

        for(var i in res) {
            result = result + "@" +res[i].name + " =";                 
            for (var j in res[i].members.split(",")) {
                var memberquery;
                if (j.substr(0,1) == "@") {
                    memberquery = "SELECT name FROM groups WHERE id = "+j.substr(1, j.length-1);
                } else {
                    memberquery = "SELECT username FROM users WHERE id= "+j;
                }
                dbClient.query(memberquery, function(err, memres, fields) {
                    var membername = "";
                    if (typeof memres[0].username == "undefined") {
                        membername = "@"+memres[0].name;
                    } else {
                        membername = memres[0].username;
                    }
                    result = result + " " + membername;
                    num_queries_left--;
                    if (num_queries_left == 0)
                    {
                        resultCallback(result);
                        return;
                    } 
                });

            }
            result = result + "\n"; 
        }
    });
}


do_queries( function(result) {
    console.log(result);
});

Node.js中的同步mysql

2 个答案: