我正在使用Sequelize连接到Postgres数据库。我有这段代码:
return Promise.resolve()
.then(() => {
console.log('checkpoint #1');
const temp = connectors.IM.create(args);
return temp;
})
.then((x) => console.log(x))
.then((args) =>{
console.log(args);
args = Array.from(args);
console.log('checkpoint #2');
const temp = connectors.IM.findAll({ where: args }).then((res) => res.map((item) => item.dataValues))
return temp;
}
)
.then(comment => {
return comment;
})
.catch((err)=>{console.log(err);});
在检查点#1的第一个.then块中,新记录成功添加到Postgres数据库。在下一个块中的console.log(x)
中,这将记录到控制台:
{ dataValues:
{ id: 21,
fromID: '1',
toID: '2',
msgText: 'Test from GraphIQL',
updatedAt: Wed Oct 12 2016 09:52:05 GMT-0700 (PDT),
createdAt: Wed Oct 12 2016 09:52:05 GMT-0700 (PDT) },
_previousDataValues:
{ fromID: '1',
toID: '2',
msgText: 'Test from GraphIQL',
id: 21,
createdAt: Wed Oct 12 2016 09:52:05 GMT-0700 (PDT),
updatedAt: Wed Oct 12 2016 09:52:05 GMT-0700 (PDT) },
_changed:
{ fromID: false,
toID: false,
msgText: false,
id: false,
createdAt: false,
updatedAt: false },
'$modelOptions':
{ timestamps: true,
instanceMethods: {},
classMethods: {},
validate: {},
freezeTableName: false,
underscored: false,
underscoredAll: false,
paranoid: false,
rejectOnEmpty: false,
whereCollection: null,
schema: null,
schemaDelimiter: '',
defaultScope: {},
scopes: [],
hooks: {},
indexes: [],
name: { plural: 'IMs', singular: 'IM' },
omitNul: false,
sequelize:
{ options: [Object],
config: [Object],
dialect: [Object],
models: [Object],
modelManager: [Object],
connectionManager: [Object],
importCache: {},
test: [Object],
queryInterface: [Object] },
uniqueKeys: {},
hasPrimaryKeys: true },
'$options':
{ isNewRecord: true,
'$schema': null,
'$schemaDelimiter': '',
attributes: undefined,
include: undefined,
raw: undefined,
silent: undefined },
hasPrimaryKeys: true,
__eagerlyLoadedAssociations: [],
isNewRecord: false }
在检查点#2的.then((args) =>
代码块中,args
以未定义的形式出现。
如何让args包含来自检查点#1的结果数组?
答案 0 :(得分:5)
.then((x) => console.log(x))
.then((args) =>{
就像在做
.then((x) => {
console.log(x);
return undefined;
})
.then((args) =>{
因为console.log
返回undefined
。这意味着undefined
值将传递给下一个.then
。
最简单的方法是明确
.then((x) => {
console.log(x);
return x;
})
或使用逗号运算符的较短版本
.then((x) => (console.log(x), x))