我是SQLAlchemy的新手（和SQL一样）。我无法弄清楚如何编写我脑子里的想法。
我正在创建一个性能测试结果数据库。
测试运行包含测试类型和数字（这是下面的TestRun类）
测试套件包含被测软件的版本字符串，以及一个或多个TestRun对象（下面是TestSuite类）。
测试版包含具有给定版本名称的所有测试套件。
这是我的代码，尽可能简单：
from sqlalchemy import *
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship, backref, sessionmaker
Base = declarative_base()
class TestVersion (Base):
__tablename__ = 'versions'
id = Column (Integer, primary_key=True)
version_name = Column (String)
def __init__ (self, version_name):
self.version_name = version_name
class TestRun (Base):
__tablename__ = 'runs'
id = Column (Integer, primary_key=True)
suite_directory = Column (String, ForeignKey ('suites.directory'))
suite = relationship ('TestSuite', backref=backref ('runs', order_by=id))
test_type = Column (String)
rate = Column (Integer)
def __init__ (self, test_type, rate):
self.test_type = test_type
self.rate = rate
class TestSuite (Base):
__tablename__ = 'suites'
directory = Column (String, primary_key=True)
version_id = Column (Integer, ForeignKey ('versions.id'))
version_ref = relationship ('TestVersion', backref=backref ('suites', order_by=directory))
version_name = Column (String)
def __init__ (self, directory, version_name):
self.directory = directory
self.version_name = version_name
# Create a v1.0 suite
suite1 = TestSuite ('dir1', 'v1.0')
suite1.runs.append (TestRun ('test1', 100))
suite1.runs.append (TestRun ('test2', 200))
# Create a another v1.0 suite
suite2 = TestSuite ('dir2', 'v1.0')
suite2.runs.append (TestRun ('test1', 101))
suite2.runs.append (TestRun ('test2', 201))
# Create another suite
suite3 = TestSuite ('dir3', 'v2.0')
suite3.runs.append (TestRun ('test1', 102))
suite3.runs.append (TestRun ('test2', 202))
# Create the in-memory database
engine = create_engine ('sqlite://')
Session = sessionmaker (bind=engine)
session = Session()
Base.metadata.create_all (engine)
# Add the suites in
version1 = TestVersion (suite1.version_name)
version1.suites.append (suite1)
session.add (suite1)
version2 = TestVersion (suite2.version_name)
version2.suites.append (suite2)
session.add (suite2)
version3 = TestVersion (suite3.version_name)
version3.suites.append (suite3)
session.add (suite3)
session.commit()
# Query the suites
for suite in session.query (TestSuite).order_by (TestSuite.directory):
print "\nSuite directory %s, version %s has %d test runs:" % (suite.directory, suite.version_name, len (suite.runs))
for run in suite.runs:
print " Test '%s', result %d" % (run.test_type, run.rate)
# Query the versions
for version in session.query (TestVersion).order_by (TestVersion.version_name):
print "\nVersion %s has %d test suites:" % (version.version_name, len (version.suites))
for suite in version.suites:
print " Suite directory %s, version %s has %d test runs:" % (suite.directory, suite.version_name, len (suite.runs))
for run in suite.runs:
print " Test '%s', result %d" % (run.test_type, run.rate)
该计划的输出：
Suite directory dir1, version v1.0 has 2 test runs:
Test 'test1', result 100
Test 'test2', result 200
Suite directory dir2, version v1.0 has 2 test runs:
Test 'test1', result 101
Test 'test2', result 201
Suite directory dir3, version v2.0 has 2 test runs:
Test 'test1', result 102
Test 'test2', result 202
Version v1.0 has 1 test suites:
Suite directory dir1, version v1.0 has 2 test runs:
Test 'test1', result 100
Test 'test2', result 200
Version v1.0 has 1 test suites:
Suite directory dir2, version v1.0 has 2 test runs:
Test 'test1', result 101
Test 'test2', result 201
Version v2.0 has 1 test suites:
Suite directory dir3, version v2.0 has 2 test runs:
Test 'test1', result 102
Test 'test2', result 202
这是不正确的，因为有两个名为“v1.0”的TestVersion对象。我通过添加一个TestVersion对象的私有列表以及一个找到匹配对象的函数来解决这个问题：
versions = []
def find_or_create_version (version_name):
# Find existing
for version in versions:
if version.version_name == version_name:
return (version)
# Create new
version = TestVersion (version_name)
versions.append (version)
return (version)
然后我修改了添加记录以使用它的代码：
# Add the suites in
version1 = find_or_create_version (suite1.version_name)
version1.suites.append (suite1)
session.add (suite1)
version2 = find_or_create_version (suite2.version_name)
version2.suites.append (suite2)
session.add (suite2)
version3 = find_or_create_version (suite3.version_name)
version3.suites.append (suite3)
session.add (suite3)
现在输出是我想要的：
Suite directory dir1, version v1.0 has 2 test runs:
Test 'test1', result 100
Test 'test2', result 200
Suite directory dir2, version v1.0 has 2 test runs:
Test 'test1', result 101
Test 'test2', result 201
Suite directory dir3, version v2.0 has 2 test runs:
Test 'test1', result 102
Test 'test2', result 202
Version v1.0 has 2 test suites:
Suite directory dir1, version v1.0 has 2 test runs:
Test 'test1', result 100
Test 'test2', result 200
Suite directory dir2, version v1.0 has 2 test runs:
Test 'test1', result 101
Test 'test2', result 201
Version v2.0 has 1 test suites:
Suite directory dir3, version v2.0 has 2 test runs:
Test 'test1', result 102
Test 'test2', result 202
这对我来说是不对的;我手动跟踪唯一版本名称并手动将套件添加到相应的TestVersion对象中感觉不对。
这段代码是否接近正确？
当我不是从头开始构建整个数据库时会发生什么，如本例所示。如果数据库已经存在，我是否必须查询数据库的TestVersion表以发现唯一的版本名称？
提前致谢。我知道这需要很多代码，我很感激帮助。
答案 0 :(得分：1)
我无法理解你的问题是什么，很大程度上是因为你没有改进它。您的问题可能是关于模式，也可能是其对应的对象关系模型。所以，这里的ORM剥离了它的核心：
class TestVersion(Base):
__tablename__ = 'versions'
id = Column(Integer, primary_key=True)
name = Column(String)
class TestSuite(Base):
__tablename__ = 'suites'
directory = Column(String, primary_key=True)
version = Column(Integer, ForeignKey ('versions.id'))
parent = relationship(TestVersion, backref=backref('suites',
order_by=directory))
class TestRun(Base):
__tablename__ = 'runs'
id = Column(Integer, primary_key=True)
directory = Column(String, ForeignKey ('suites.directory'))
parent = relationship(TestSuite, backref=backref('runs',
order_by=id))
我对你的声明采取了很多自由：抛出与你的问题无关的列，重新排序声明以使依赖链变得更加明显等等。也许对于这种简化模型你可以更好地描述你的问题。
另外，像PEP 8这样的编码标准存在是有原因的：如果你希望你的代码可以被其他人理解，可以使用4个空格缩进并避开名称和'（'，限制为79个字符的行之间的空格等等，是的，这看起来很迂腐，但是你遇到的情况是你的读者阅读你的代码比你想要的更困难。