Adjust db struct to make faster and clearer #59
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web/api.py
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| commit_ids = [] | ||
| for commit in commits: | ||
| if commit['commitid'] not in commit_ids: | ||
| commit_ids.append(commit['commitid']) |
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注释里是
returns: list of CommitRecord
函数名应该叫 get_all，不需要单独取 commit_id
返回值应该是 [CommitRecord]
web/api.py
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| commit_ids.append(commit['commitid']) | ||
| return commit_ids | ||
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| def query_a_commit_info(self): |
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感觉差的比较多，之前好像一起定过接口，具体字段没有问题，但不需要暴露这么多接口。
下面的几个 method 就可以了。
class CommitRecord:
@staticmethod
def get_all():
''' Return list of CommitRecord '''
pass
def get_all_records(self):
''' Return list of TaskRecord '''
pass
class TaskRecord:
pass
class KpiRecord:
pass用户使用案例
- 获取所有的 commitid
commits = CommitRecord.get_all()
commitids = [c.commitid for c in commits]- 获取一个 commitid 下所有的 task 记录
# some commit
commit = commits[k]
task_records = commit.get_all_records()
- 遍历一个 commit 下所有的 kpi record
for task_record in commit.get_all_records():
for kpi_record in task_record.kpis:
print(kpi_record.values)
web/api.py
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| 'commitid': self.commit}) | ||
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| @staticmethod | ||
| def query_all_commit_infos(): |
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这个函数不需要，用户可以直观用
commit_records = CommitRecord.get_all()
# if want to get all commit id
commit_ids = [c.commit_id for c in commit_records]
web/api.py
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| 'commitid': self.commitid, 'task': self.name}) | ||
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| @staticmethod | ||
| def get_tasks_from_details(commit): |
web/api.py
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| taskobj.kpis = taskobj.get_kpi_details() | ||
| res[taskobj.name] = taskobj | ||
| return res | ||
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| def get_kpi_details(self): |
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这个函数是不是private 就可以了，用户需要显式调用吗
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脚本外目前没有调用的。就是拿1个task 的 kpi infos。
returns dict of KpiRecord
keys equal to kpi name,
values equal to KpiRecord'''
现在在commitRecord 类里面调用了这个函数， 改个名字 get_kpis？ 是不是可以暴露出去
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一个 TaskRecord 应该直接包含多个 KpiRecord，这个接口应该不需要额外调，所以不一定需要
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嗯啊 这是个现有鸡还是先有蛋的问题， 如果他仅仅初始化了一个TaskRecord 对象， 还没调用任何方法， 这个对象还是空的。
现在在CommitRecord 这个类里面用到了TaskRecord 里面的这个方法， 如果写成私有的CommitRecord类还能调用到它么
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感觉改个更general的名字， 比如get_kpis， 然后作为public 方法看着还行？
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