Javascript/NodeJS : html rendering before . 在 forEach 内查找完成

Question

此循环应该检查数据库中与从 salesforce api 结果中提取的 opp 相匹配的 opp，然后创建一个新的 opp 或查找现有的 opp 并将其推送到数组。看起来 res.render 在找到 opp 之前就已经运行了。它创建一个新的 opp，但在页面呈现时数组返回空。

 Account.find({owner:req.user._id, prospect:'false'}).sort({ammount:'desc'}).populate({path: "notes", options:{ sort:{ 'date': -1 } } }).exec(function(err, allAccounts) {
        let callGoal = req.user.callGoal;
        if(err){
            res.send(err);
        }else{
            // if auth has not been set, redirect to index
            if (!req.session.accessToken || !req.session.instanceUrl) { res.redirect('/'); }
        //SOQL query
            let q = "SELECT Id,Amount,CloseDate,LastActivityDate,Name,StageName,account.Name FROM Opportunity WHERE CloseDate < 2018-10-01 AND OwnerId = '0050a00000J12PdAAJ' AND IsClosed = false AND StageName != 'Stage 6: Won'";
        //instantiate connection
            let conn = new jsforce.Connection({
                oauth2 : {oauth2},
                accessToken: req.session.accessToken,
                instanceUrl: req.session.instanceUrl
           });
        //set records array
            let softopps = [];
            let sfOpps = [];
            let query = conn.query(q)
               .on("record", function(record) {
                 sfOpps.push(record);
               })
               .on("end", function() {
                 console.log("total in database : " + query.totalSize);
                 console.log("total fetched : " + query.totalFetched);
                 let user = req.user;
                 sfOpps.forEach(function(sfopp){
                     if(err){
                         res.send(err);
                     }else{
                         Opp.findOne({sfid:sfopp.Id}).exec(function(err, opp){
                         if(!opp.length){
                            Opp.create(req.body, function(err, newOpp) {
                                if(err){
                                    res.send(err)
                                }else{
                                    newOpp.sfid = sfopp.Id;
                                    newOpp.name = sfopp.Name;
                                    newOpp.owner = user.sfid;
                                    newOpp.save();
                                    return softopps.push(newOpp)
                                }
                            })
                         }else{
                             return softopps.push(opp);
                         }
                     })
                     }
                 })
                 res.render("myaccounts", {accounts:allAccounts, callGoal:callGoal, user:user, sfOpps:sfOpps, opps:softopps});
               })
               .on("error", function(err) {
                 console.error(err);
               })
               .run({ autoFetch : true, maxFetch : 4000 });
        }
    });

Answer 1

您的 Opp.findOne() 和 Opp.create() 调用是异步的，因此它们在 res.render() 之后触发。

<小时/>

另一方面，你几乎会因为所有不必要的请求而杀死你的 mongodb。

尝试这个逻辑(从 .forEach 开始)

1. 按 sfOpps id 查找所有 Opp
2. 找出未找到的 Opp 并创建它们
3. 连接所有找到和未找到的Opps
4. 然后，只有那时回应

我没有测试该代码，但它可以让您大致了解我的意思

Opp.find({ sfid : { $in: sfOpps.map(opp => opp.id) } })
  .then(found => {

    const foundIds = found.map(opp => opp.sfid)
    const notFound = sfOpps.filter(opp => !foundIds.includes(opp.sfid)).map(sfopp => {
      return {
        sfid: sfopp.sfid,
        name: sfopp.name,
        owner: user.sfid
      }
    })

    Opp.insertMany(notFound)
      .then((insertResult) => {

        res.render("myaccounts", {
          accounts: allAccounts, 
          callGoal: callGoal, 
          user: user, 
          sfOpps: found.concat(notFound), 
          opps: softopps
        });

      }).catch(handleError)

  }).catch(handleError)