Node.js 返回结果后执行Graphql postgresql承诺
我正在尝试graphql。我正在从postgresql中提取数据,除了一个问题,在查询完成之前,解析就完成了,所有的一切似乎都正常工作。我认为我的查询承诺工作正常。这里是我为查询数据库而创建的类:Node.js 返回结果后执行Graphql postgresql承诺,node.js,es6-promise,graphql,Node.js,Es6 Promise,Graphql,我正在尝试graphql。我正在从postgresql中提取数据,除了一个问题,在查询完成之前,解析就完成了,所有的一切似乎都正常工作。我认为我的查询承诺工作正常。这里是我为查询数据库而创建的类: class PgPoolClient { constructor(options) { this.connection = options.connection; } query(sql, data = []) { return new P
class PgPoolClient {
constructor(options) {
this.connection = options.connection;
}
query(sql, data = []) {
return new Promise((resolve, reject) => {
var pool = new pg.Pool(this.connection);
pool.connect((err, client, done) => {
if (err) {
reject({
code: 505,
message: err.message,
entrys: []
});
} else {
client.query(sql, data, (err, results) => {
done();
if (err) {
reject({
code: 404,
message: err.message,
entrys: []
});
} else {
resolve({
code: 200,
message: "ok",
entrys: results
});
}
});
}
});
pool.on("error", (err, client) => {
reject({
code: 505,
message: err.message,
entrys: {}
});
});
});
}
}
以下是我的graphql查询:
const Query = new GraphQLObjectType({
name: 'Query',
description: 'Root query object',
fields: () => ({
accounts: {
type: new GraphQLList(Account),
resolve(root, args) {
const parameters = [];
pg.query(`SELECT
id, token, user_identifier as userIdentifier, hash_password as hashPassword,
is_enabled as isEnabled, security_type_id as securityTypeID,
is_admin as isAdmin, ad_path as adPath, date_created as dateCreated,
date_modified as dateModified, modified_by as modifiedBy
FROM accounts.account`, parameters)
.then((result) =>
{
console.log(result.entrys.rows);
//return result.entrys.rows;
return [
{
id: '33333',
token: '111',
userIdentifier: 'test'
}
]
})
.catch((err) =>
{
console.log(err);
return err.message;
});
console.log('finished');
}
})
});
我对我的结果进行了注释,只是想了解一下如何返回静态内容,似乎当它在then()内时,结果不会返回到graphisql(意味着解析在then()之前完成)。输出:
{
"data": {
"accounts": null
}
}
我可以通过将静态返回数组放在console.log('finished')下面来验证这一点,它正确地返回了数据
console.log('finished');
return [
{
id: '33333',
token: '111',
userIdentifier: 'test'
}
]
它显示在graphiql中:
{
"data": {
"accounts": [
{
"id": "33333",
"token": "111",
"userIdentifier": "test"
}
]
}
}
所以我的承诺似乎没有得到遵守。我的查询函数中是否缺少导致解析不等待结果的内容?任何帮助都将不胜感激
更新1
似乎如果我将查询包装成一个承诺,它就会起作用:
return new Promise((resolve, reject) => {
pg.query(`SELECT
id, token, user_identifier as userIdentifier, hash_password as hashPassword,
is_enabled as isEnabled, security_type_id as securityTypeID,
is_admin as isAdmin, ad_path as adPath, date_created as dateCreated,
date_modified as dateModified, modified_by as modifiedBy
FROM accounts.account`, parameters)
.then((result) =>
{
console.log(result.entrys.rows);
//return result.entrys.rows;
resolve( [
{
id: '33333',
token: '111',
userIdentifier: 'test'
}
]);
})
.catch((err) =>
{
console.log(err);
reject(err.message);
});
我想我的问题是,如果我的查询函数已经包含在一个承诺中,为什么我还要再做一次呢?您的
resolve
函数需要返回承诺。当你明确地把它包装在一个承诺中时,它就起作用了,因为你还了它。如果您返回调用pg.query
的结果,则它应该可以不使用该选项
const Query = new GraphQLObjectType({
name: 'Query',
description: 'Root query object',
fields: () => ({
accounts: {
type: new GraphQLList(Account),
resolve(root, args) {
const parameters = [];
return pg.query(`query`, parameters)
.then((result) =>
{
console.log(result.entrys.rows);
//return result.entrys.rows;
return [
{
id: '33333',
token: '111',
userIdentifier: 'test'
}
]
})
.catch((err) =>
{
console.log(err);
return err.message;
});
console.log('finished');
}
})
});
谢谢你的建议