python- 如何根据公共值连接两个单独的查询
python- how to join two seperate queries based on common value
我有两个链接到不同数据库的查询
query = "select name ,ctry from xxxx where xxxx"
cursor.execute(query)
results1 = list(cursor.fetchall())
for row in results1:
query1 = "SELECT sessionname, country FROM xxx where and sessions.sessionname = '"+row[0] +"'"
cur.execute(query1)
results2.append(cur.fetchall())
如果它们具有公共值(会话名称和名称),如何连接它们并将其输出保存到文件。两个查询都位于不同的 dbo (oracle, postgresql)
我的代码在这里:
try:
query = """select smat.s_name "SQLITE name" ,smed.m_ctry as "Country", smed.m_name "HDD Label" from smart.smed2smat ss, smart.smed smed, smart.smat smat where ss.M2S_SMAT=smat.s_id and ss.m2s_smed=smed.m_id and smed.m_name like '{0}%' order by smat.s_name""" .format(line_name)
cursor.execute(query)
columns = [i[0] for i in cursor.description]
results1 = cursor.fetchall()
for row in results1:
query1 = "SELECT sessions.sessionname, projects.country , projects.projectname FROM momatracks.sessions, momatracks.projects, momatracks.sessionsgeo where sessions.projectid = projects.id and sessionsgeo.sessionname = sessions.sessionname and sessions.sessionname = '"+row[0] +"' order by sessions.sessionname"
cur.execute(query1)
results2 =cur.fetchall()
print "results1 -----> \n:", row
tmp=[]
output_items = []
for tmp in results2:
print "---> \n", tmp
try:
stations_dict = dict([(item[0], item[1:]) for item in tmp])
for item in row:
output_item = list(item) + stations_dict.get(item[0], [])
output_items.append(output_item)
except Exception, f:
print str (f)
cursor.close()
cur.close()
except Exception, g:
print str ( g )
except Exception, e:
print str ( e )
我从 row 和 tmp 得到的结果是:
row - WE246JP_2015_10_11__14_53_33', 'NLD', '031_025_SQLITE_NLD1510_03INDIA
和
tmp - WE246JP_2015_10_11__14_53_33', 'NLD', 'NLD15_N2C1-4_NL'
如何正确连接它们?我希望输出看起来像这样:
output_items - WE246JP_2015_10_11__14_53_33', 'NLD', '031_025_SQLITE_NLD1510_03INDIA', 'NLD15_N2C1-4_NL'
目前我收到这个错误:
can only concatenate list (not "str") to list
另外值 station_dict 看起来像这样 :( 这不是我想要做的
'W': 'E246JP_2015_10_11__15_23_33', 'N': 'LD15_N2C1-4_NL3'
我知道我的代码有问题,很容易加入。谁能给我解释一下?我使用了以下方法:
我想你可以在这里使用子查询。我没有办法测试它,但我认为它应该是这样的:
SELECT *
FROM (SELECT smat.s_name "SQLITE name" ,
smed.m_ctry as "Country",
smed.m_name "HDD Label"
FROM smart.smed2smat ss,
smart.smed smed,
smart.smat smat
WHERE ss.M2S_SMAT=smat.s_id
AND ss.m2s_smed=smed.m_id
AND smed.m_name like '{0}%'
ORDER BY smat.s_name) t1,
(SELECT sessions.sessionname,
projects.country ,
projects.projectname
FROM momatracks.sessions,
momatracks.projects,
momatracks.sessionsgeo
WHERE sessions.projectid = projects.id
AND sessionsgeo.sessionname = sessions.sessionname
AND sessions.sessionname = '"+row[0] +"'
ORDER BY sessions.sessionname) t2
WHERE t1."SQLITE name" = t2.sessionname ;
如果两个数据库中的会话完全相同,则只需压缩结果:
query = """
select
smat.s_name "SQLITE name",
smed.m_ctry as "Country",
smed.m_name "HDD Label"
from
smart.smed2smat ss
inner join
smart.smed smed on ss.M2S_SMAT = smat.s_id
inner join
smart.smat smat on ss.m2s_smed = smed.m_id
where smed.m_name like '{0}%'
order by smat.s_name
""".format(line_name)
cursor.execute(query)
results1 = cursor.fetchall()
query1 = """
select
sessions.sessionname,
projects.country,
projects.projectname
from
momatracks.sessions,
inner join
momatracks.projects on sessions.projectid = projects.id
inner join
momatracks.sessionsgeo on sessionsgeo.sessionname = sessions.sessionname
where sessions.sessionname in {}
order by sessions.sessionname
""".format(tuple([row[0] for row in results1]))
cur.execute(query1)
results2 = cur.fetchall()
zipped = zip(results1, results2)
output_list = [(m[0][0], m[0][1], m[0][2], m[1][2]) for m in zipped]
如果会话不同,则将每个结果作为要加入的字典。
我有两个链接到不同数据库的查询
query = "select name ,ctry from xxxx where xxxx"
cursor.execute(query)
results1 = list(cursor.fetchall())
for row in results1:
query1 = "SELECT sessionname, country FROM xxx where and sessions.sessionname = '"+row[0] +"'"
cur.execute(query1)
results2.append(cur.fetchall())
如果它们具有公共值(会话名称和名称),如何连接它们并将其输出保存到文件。两个查询都位于不同的 dbo (oracle, postgresql)
我的代码在这里:
try:
query = """select smat.s_name "SQLITE name" ,smed.m_ctry as "Country", smed.m_name "HDD Label" from smart.smed2smat ss, smart.smed smed, smart.smat smat where ss.M2S_SMAT=smat.s_id and ss.m2s_smed=smed.m_id and smed.m_name like '{0}%' order by smat.s_name""" .format(line_name)
cursor.execute(query)
columns = [i[0] for i in cursor.description]
results1 = cursor.fetchall()
for row in results1:
query1 = "SELECT sessions.sessionname, projects.country , projects.projectname FROM momatracks.sessions, momatracks.projects, momatracks.sessionsgeo where sessions.projectid = projects.id and sessionsgeo.sessionname = sessions.sessionname and sessions.sessionname = '"+row[0] +"' order by sessions.sessionname"
cur.execute(query1)
results2 =cur.fetchall()
print "results1 -----> \n:", row
tmp=[]
output_items = []
for tmp in results2:
print "---> \n", tmp
try:
stations_dict = dict([(item[0], item[1:]) for item in tmp])
for item in row:
output_item = list(item) + stations_dict.get(item[0], [])
output_items.append(output_item)
except Exception, f:
print str (f)
cursor.close()
cur.close()
except Exception, g:
print str ( g )
except Exception, e:
print str ( e )
我从 row 和 tmp 得到的结果是:
row - WE246JP_2015_10_11__14_53_33', 'NLD', '031_025_SQLITE_NLD1510_03INDIA
和
tmp - WE246JP_2015_10_11__14_53_33', 'NLD', 'NLD15_N2C1-4_NL'
如何正确连接它们?我希望输出看起来像这样:
output_items - WE246JP_2015_10_11__14_53_33', 'NLD', '031_025_SQLITE_NLD1510_03INDIA', 'NLD15_N2C1-4_NL'
目前我收到这个错误:
can only concatenate list (not "str") to list
另外值 station_dict 看起来像这样 :( 这不是我想要做的
'W': 'E246JP_2015_10_11__15_23_33', 'N': 'LD15_N2C1-4_NL3'
我知道我的代码有问题,很容易加入。谁能给我解释一下?我使用了以下方法:
我想你可以在这里使用子查询。我没有办法测试它,但我认为它应该是这样的:
SELECT *
FROM (SELECT smat.s_name "SQLITE name" ,
smed.m_ctry as "Country",
smed.m_name "HDD Label"
FROM smart.smed2smat ss,
smart.smed smed,
smart.smat smat
WHERE ss.M2S_SMAT=smat.s_id
AND ss.m2s_smed=smed.m_id
AND smed.m_name like '{0}%'
ORDER BY smat.s_name) t1,
(SELECT sessions.sessionname,
projects.country ,
projects.projectname
FROM momatracks.sessions,
momatracks.projects,
momatracks.sessionsgeo
WHERE sessions.projectid = projects.id
AND sessionsgeo.sessionname = sessions.sessionname
AND sessions.sessionname = '"+row[0] +"'
ORDER BY sessions.sessionname) t2
WHERE t1."SQLITE name" = t2.sessionname ;
如果两个数据库中的会话完全相同,则只需压缩结果:
query = """
select
smat.s_name "SQLITE name",
smed.m_ctry as "Country",
smed.m_name "HDD Label"
from
smart.smed2smat ss
inner join
smart.smed smed on ss.M2S_SMAT = smat.s_id
inner join
smart.smat smat on ss.m2s_smed = smed.m_id
where smed.m_name like '{0}%'
order by smat.s_name
""".format(line_name)
cursor.execute(query)
results1 = cursor.fetchall()
query1 = """
select
sessions.sessionname,
projects.country,
projects.projectname
from
momatracks.sessions,
inner join
momatracks.projects on sessions.projectid = projects.id
inner join
momatracks.sessionsgeo on sessionsgeo.sessionname = sessions.sessionname
where sessions.sessionname in {}
order by sessions.sessionname
""".format(tuple([row[0] for row in results1]))
cur.execute(query1)
results2 = cur.fetchall()
zipped = zip(results1, results2)
output_list = [(m[0][0], m[0][1], m[0][2], m[1][2]) for m in zipped]
如果会话不同,则将每个结果作为要加入的字典。