pg-promise-将多个嵌套循环查询合并到父数组结果 [英] pg-promise - Combine multiple nested loop queries to parent array result

问题描述

我的问题基于合并嵌套循环查询到父数组结果-pg-promise .我也有类似的情况，但是有多个查询要组合才能得到最终结果.以下是我尝试用来实现我的要求的代码.但是我无法从与主查询结合的第二个查询中获得结果.我是一个初学者，想知道正确的实现方式.

My question is based on Combine nested loop queries to parent array result - pg-promise. I'm having a similar scenario but have multiple queries to be combined to get my final results. Following is my code with which I tried to implement my requirement. But I was not able to get the results from second query combined with the main query. I'm a beginner and would like to know the correct way of implementation.

  db.task(t => {
    const a = studies => t.any ('SELECT facility_contacts.name, facility_contacts.email, facility_contacts.phone FROM facility_contacts WHERE nct_id = $1', 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 = $1', 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);
});

谢谢.

推荐答案

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 = $1', 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 = $1', 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);
    });

可以有许多不同的方法，例如基于JSON的方法，它们的性能更好.

There can be many different approaches, some way better performing, like JSON-based.

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