pg-promise

Question

我的问题是基于。我遇到了类似的情况，但要组合多个查询以获得最终结果。以下是我试图实现我的要求的代码。但是我无法从第二个查询与主查询相结合中获得结果。我是初学者，想知道正确的实现方式。

  db.task(t => {
    const a = studies => t.any ('SELECT facility_contacts.name, facility_contacts.email, facility_contacts.phone FROM facility_contacts WHERE nct_id = ', studies.nct_id).then(facility_contacts => {
    studies.facility_contacts = facility_contacts;
    return studies
      });
    const b = studies => t.any ('SELECT eligibilities.criteria, eligibilities.gender FROM eligibilities WHERE nct_id = ', studies.nct_id).then(eligibilities => {
    studies.eligibilities = eligibilities;
    return studies;
      });
  return t.map('SELECT studies.nct_id, studies.official_title, studies.phase, facilities.country FROM studies, facilities WHERE facilities.country LIKE \'%Ireland%\' LIMIT 10',[], a, b).then(t.batch);
}).then(studies => {
  console.log(studies);
  res.send(studies);
}).catch(error => {
  console.log(error);
});

提前致谢。

Answer 1

你可以使用Promise.all得到所有结果

db.task(t => {
    const queries = [
        t.any('SELECT facility_contacts.name, facility_contacts.email, facility_contacts.phone FROM facility_contacts WHERE nct_id = ', studies.nct_id),
        t.any('SELECT eligibilities.criteria, eligibilities.gender FROM eligibilities WHERE nct_id = ', studies.nct_id),
        t.map('SELECT studies.nct_id, studies.official_title, studies.phase, facilities.country FROM studies, facilities WHERE facilities.country LIKE \'%Ireland%\' LIMIT 10', [], a, b)
    ];
    return Promise.all(queries);
}).then((ResultArrayOfQueries) => {
    console.log(ResultArrayOfQueries);
    // Combine as per your requirement
    res.send(ResultArrayOfQueries);
}).catch(error => {
    console.log(error);
});

Answer 2

这是一种可能性：

db.task(t => {
    const a = studies => t.any('SELECT facility_contacts.name, facility_contacts.email, facility_contacts.phone FROM facility_contacts WHERE nct_id = ', studies.nct_id)
        .then(facility_contacts => {
            studies.facility_contacts = facility_contacts;
            return studies
        });
    const b = studies => t.any('SELECT eligibilities.criteria, eligibilities.gender FROM eligibilities WHERE nct_id = ', studies.nct_id)
        .then(eligibilities => {
            studies.eligibilities = eligibilities;
            return studies;
        });
    const c = studies => t.batch([a(studies), b(studies)]);
    return t.map('SELECT studies.nct_id, studies.official_title, studies.phase, facilities.country FROM studies, facilities WHERE facilities.country LIKE \'%Ireland%\' LIMIT 10', [], c)
        .then(t.batch);
})
    .then(studies => {
        console.log(studies);
        res.send(studies);
    })
    .catch(error => {
        console.log(error);
    });

可以有很多不同的方法，有些方法性能更好，比如基于 JSON。

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node.js

express