将嵌套循环查询与父数组结果-电脑我帮您

将嵌套循环查询与父数组结果

我是node（express）和pg-promise的新手，并且无法弄清楚如何将每个嵌套查询（循环）的结果添加到主json数组结果查询中。

我有两张桌子：帖子和评论。

CREATE TABLE post(
id serial,
content text not null,
linkExterno text,
usuario VARCHAR(50) NOT NULL REFERENCES usuarios(alias) ON UPDATE cascade ON DELETE cascade,
multimedia text,
ubicacation VARCHAR(100),
likes integer default 0,
time VARCHAR default now(),
reported boolean default false,
PRIMARY KEY (id)  );

CREATE TABLE comment(
id serial,
idPost integer NOT NULL REFERENCES post(id) ON UPDATE cascade ON DELETE cascade,
acount VARCHAR(50) NOT NULL REFERENCES users(alias) ON UPDATE cascade ON DELETE cascade,
content text NOT NULL,
date date default now(),
PRIMARY KEY (id));

所以我想将每条评论的结果添加到每个帖子并返回帖子。我有这个，但不起作用：

con.task(t => {
    return t.any('select *, avatar from post, users where user= $1 and user = alias ORDER BY time DESC LIMIT 10 OFFSET $2', [username, pos])
    .then(posts => {
        if(posts.length > 0){
            for (var post of posts){
                postents = t.any('select * from comment where idPost = $1 ', post.id);
            }
        }
    });
}).then(posts => {
    res.send(posts);
}).catch(error => {
    console.log(error);
});

有什么建议？ PD：我认为我的问题与此类似：get JOIN table as array of results with PostgreSQL/NodeJS

解答：

选项1（最佳选择）：

通过JSON对psql进行单个查询（JSON查询）

请参阅@ vitaly-t的回答

要么

使用ajax异步获取嵌套数据。

选项2：

function buildTree(t) {
        return t.map("select *, avatar from publicacion, usuarios where usuario = $1 and usuario = alias ORDER BY hora DESC LIMIT 10 OFFSET $2", [username, cantidad], posts => {
                return t.any('select * from comentario where idPublicacion = $1', posts.id)
                    .then(coments => {
                        postsents = coments;
                        console.log(postsents);
                        return posts;
                    });
        }).then(t.batch); // settles the array of generated promises
    }

    router.get('/publicaciones', function (req, res) {
        cantidad = req.query.cantidad || 0; //num de publicaciones que hay
        username = req.session.user.alias;

        con.task(buildTree)
        .then(data => {
            res.send(data);
        })
        .catch(error => {
            console.log(error);
        });
    });

选项3（异步）：

try{
    var posts = await con.any('select *, avatar from post, users where user = $1 and user = alias ORDER BY time DESC LIMIT 10 OFFSET $2', [username, q])
    for (var post of posts){
        postents = await con.any('select * from comment where idPublictcion = $1', post.id);
    }
}catch(e){
    console.log(e);
}

回答如下：

我是pg-promise的作者;）

con.task(t => {
    const a = post => t.any('SELECT * FROM comment WHERE idPost = $1', post.id)
        .then(comments => {
            postments = comments;
            return post;
        });
    return t.map('SELECT *, avatar FROM post, users WHERE user = $1 AND user = alias ORDER BY time DESC LIMIT 10 OFFSET $2', [username, pos], a)
        .then(t.batch);
})
    .then(posts => {
        res.send(posts);
    })
    .catch(error => {
        console.log(error);
    });

另见这个问题：get JOIN table as array of results with PostgreSQL/NodeJS。

UPDATE

如果您不想一直使用JSON查询方法，那么以下内容将比原始解决方案更好地扩展，因为我们连接所有子查询，然后将它们作为一个查询执行：

con.task(async t => {
    const posts = await t.any('SELECT *, avatar FROM post, users WHERE user = $1 AND user = alias ORDER BY time DESC LIMIT 10 OFFSET $2', [username, pos]);
    const a = post => ({query: 'SELECT * FROM comment WHERE idPost = ${id}', values: post});
    const queries = pgp.helpers.concat(posts.map(a));
    await t.multi(queries)
        .then(comments => {
            posts.forEach((p, index) => {
                pments = comments[index];
            });
        });
    return posts;
})
    .then(posts => {
        res.send(posts);
    })
    .catch(error => {
        console.log(error);
    });

见API：

helpers.concat
Database.multi

将嵌套循环查询与父数组结果

我是node（express）和pg-promise的新手，并且无法弄清楚如何将每个嵌套查询（循环）的结果添加到主json数组结果查询中。

我有两张桌子：帖子和评论。

CREATE TABLE post(
id serial,
content text not null,
linkExterno text,
usuario VARCHAR(50) NOT NULL REFERENCES usuarios(alias) ON UPDATE cascade ON DELETE cascade,
multimedia text,
ubicacation VARCHAR(100),
likes integer default 0,
time VARCHAR default now(),
reported boolean default false,
PRIMARY KEY (id)  );

CREATE TABLE comment(
id serial,
idPost integer NOT NULL REFERENCES post(id) ON UPDATE cascade ON DELETE cascade,
acount VARCHAR(50) NOT NULL REFERENCES users(alias) ON UPDATE cascade ON DELETE cascade,
content text NOT NULL,
date date default now(),
PRIMARY KEY (id));

所以我想将每条评论的结果添加到每个帖子并返回帖子。我有这个，但不起作用：

con.task(t => {
    return t.any('select *, avatar from post, users where user= $1 and user = alias ORDER BY time DESC LIMIT 10 OFFSET $2', [username, pos])
    .then(posts => {
        if(posts.length > 0){
            for (var post of posts){
                postents = t.any('select * from comment where idPost = $1 ', post.id);
            }
        }
    });
}).then(posts => {
    res.send(posts);
}).catch(error => {
    console.log(error);
});

有什么建议？ PD：我认为我的问题与此类似：get JOIN table as array of results with PostgreSQL/NodeJS

解答：

选项1（最佳选择）：

通过JSON对psql进行单个查询（JSON查询）

请参阅@ vitaly-t的回答

要么

使用ajax异步获取嵌套数据。

选项2：

function buildTree(t) {
        return t.map("select *, avatar from publicacion, usuarios where usuario = $1 and usuario = alias ORDER BY hora DESC LIMIT 10 OFFSET $2", [username, cantidad], posts => {
                return t.any('select * from comentario where idPublicacion = $1', posts.id)
                    .then(coments => {
                        postsents = coments;
                        console.log(postsents);
                        return posts;
                    });
        }).then(t.batch); // settles the array of generated promises
    }

    router.get('/publicaciones', function (req, res) {
        cantidad = req.query.cantidad || 0; //num de publicaciones que hay
        username = req.session.user.alias;

        con.task(buildTree)
        .then(data => {
            res.send(data);
        })
        .catch(error => {
            console.log(error);
        });
    });

选项3（异步）：

try{
    var posts = await con.any('select *, avatar from post, users where user = $1 and user = alias ORDER BY time DESC LIMIT 10 OFFSET $2', [username, q])
    for (var post of posts){
        postents = await con.any('select * from comment where idPublictcion = $1', post.id);
    }
}catch(e){
    console.log(e);
}

回答如下：

我是pg-promise的作者;）

con.task(t => {
    const a = post => t.any('SELECT * FROM comment WHERE idPost = $1', post.id)
        .then(comments => {
            postments = comments;
            return post;
        });
    return t.map('SELECT *, avatar FROM post, users WHERE user = $1 AND user = alias ORDER BY time DESC LIMIT 10 OFFSET $2', [username, pos], a)
        .then(t.batch);
})
    .then(posts => {
        res.send(posts);
    })
    .catch(error => {
        console.log(error);
    });

另见这个问题：get JOIN table as array of results with PostgreSQL/NodeJS。

UPDATE

如果您不想一直使用JSON查询方法，那么以下内容将比原始解决方案更好地扩展，因为我们连接所有子查询，然后将它们作为一个查询执行：

con.task(async t => {
    const posts = await t.any('SELECT *, avatar FROM post, users WHERE user = $1 AND user = alias ORDER BY time DESC LIMIT 10 OFFSET $2', [username, pos]);
    const a = post => ({query: 'SELECT * FROM comment WHERE idPost = ${id}', values: post});
    const queries = pgp.helpers.concat(posts.map(a));
    await t.multi(queries)
        .then(comments => {
            posts.forEach((p, index) => {
                pments = comments[index];
            });
        });
    return posts;
})
    .then(posts => {
        res.send(posts);
    })
    .catch(error => {
        console.log(error);
    });

见API：

helpers.concat
Database.multi

电脑我帮您（www.dn580.com）

将嵌套循环查询与父数组结果

将嵌套循环查询与父数组结果

将嵌套循环查询与父数组结果

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