javascript - Nodejs 结合了我的 sql 命令的两个结果

Question

我有两个 mySql 命令从数据库中获取结果，我试图用下面的代码获取它们，但它不能正常工作

例如:

从 userEwallets 表中获取结果:

function getMoneySenderUserInfo(data, callback) {
    var sourceEwalletNumber = data.transferToEwallet.substring(0, 5) + data.registeredMobileNumber.substring(1, data.registeredMobileNumber.length);
    var query               = "SELECT userEwallets.id as ewalletId, users.id as userId , money_repositories.money as money FROM userEwallets " +
                              "JOIN users ON users.id = userEwallets.userId " +
                              "JOIN money_repositories ON userEwallets.id = money_repositories.ewalletId " +
                              "WHERE ewalletNumber = '" + sourceEwalletNumber + "' ";
    connection.query(query, function (err, results) {
        if (err) return callback(false);

        if (results.length === 0) return callback(false);

        callback(true, results[0]);
    });
}

再次从userEwallets表中获取结果，得到其他结果:

function getDestinationTakeMoneyUserInfo(data, callback) {
    var query = "SELECT users.id as userId FROM userEwallets " +
                "JOIN users ON users.id = userEwallets.userId " +
                "WHERE ewalletNumber = '" + data.transferToEwallet + "' ";
    connection.query(query, function (err, results) {
        if (err) return callback(false);

        if (results.length === 0) return callback(false);

        callback(true, results[0]);
    });
}

现在我想将这个结果与这段代码结合起来:

usersInformation = {};
getMoneySenderUserInfo(data, function (success, result) {
    if (success) {
        usersInformation['senderId']  = result.userId;
        usersInformation['ewalletId'] = result.ewalletId;
        usersInformation['money']     = result.money;
        getDestinationTakeMoneyUserInfo(data, function (success, result) {
            usersInformation['destinationUserId'] = result.userId;
        });
    }
    log.info(usersInformation);
});

Answer 1

你应该使用 Promise

举个例子

function getMoneySenderUserInfo(data) {

 var promise = new Promise( function(resolve, reject) {
   var query = "..."
   connection.query(query, function (err, results) {
     if (err || results.length === 0) {
       reject();
     } else {
       resolve(results[0]);
     }
   });

 return promise;
}

您必须为 getDestinationTakeMoneyUserInfo 函数做同样的事情

然后你可以得到这样的最终结果:

var usersInformation = {};
Promise.all([getMoneySenderUserInfo(data), getDestinationTakeMoneyUserInfo(data)])
.then(function (results) {
    usersInformation['senderId']  = results[0].userId;
    usersInformation['ewalletId'] = results[0].ewalletId;
    usersInformation['money']     = results[0].money;
    usersInformation['destinationUserId'] = result[1].userId;
});

如您所见，Promise.all 将等待两个数据库请求，回调(在 then 函数中传递)将在两个 promise 执行时被调用已解决，因此您最终可以检索这两个结果。

你也可以像这样链接 promise :

getMoneySenderUserInfo(data)
.then(function (result) {
    usersInformation['senderId']  = result.userId;
    usersInformation['ewalletId'] = result.ewalletId;
    usersInformation['money']     = result.money;
    return getDestinationTakeMoneyUserInfo(data);
}).then(function (result) {
    usersInformation['destinationUserId'] = result.userId;
});