我有很多文件(超过4000个),我想同时加载到PostgreSQL中.我已将它们分成4个不同的文件列表,我想要一个线程迭代加载数据的每个列表.
我遇到的问题是我使用os.system来调用加载程序但这会阻止其他线程同时运行.如果我使用subprocess.Popen然后它们同时运行但是线程认为它们已经完成了execeuting,所以移动到我的脚本的下一部分.
我这样做是对的吗?或者是否有更好的方法从线程内调用子进程.
def thread1Load(self, thread1fileList):
connectionstring = settings.connectionstring
postgreshost = settings.postgreshost
postgresdatabase = settings.postgresdatabase
postgresport = settings.postgresport
postgresusername = settings.postgresusername
postgrespassword = settings.postgrespassword
tablename = None
encoding = None
connection = psycopg2.connect(connectionstring)
for filename in thread1fileList:
load_cmd = #load command
run = subprocess.Popen(load_cmd, shell=True)
print "finished loading thread 1"
def thread2Load(self, thread2fileList):
connectionstring = settings.connectionstring
postgreshost = settings.postgreshost
postgresdatabase = settings.postgresdatabase
postgresport = settings.postgresport
postgresusername = settings.postgresusername
postgrespassword = settings.postgrespassword
tablename = None
connection = psycopg2.connect(connectionstring)
for filename in thread2fileList:
load_cmd = #load command
run = subprocess.Popen(load_cmd, shell=True)
print "finished loading thread 2"
def thread3Load(self, thread3fileList):
connectionstring = settings.connectionstring
postgreshost = settings.postgreshost
postgresdatabase = settings.postgresdatabase
postgresport = settings.postgresport
postgresusername = settings.postgresusername
postgrespassword = settings.postgrespassword
tablename = None
connection = psycopg2.connect(connectionstring)
for shapefilename in thread3fileList:
load_cmd = #load command
run = subprocess.Popen(load_cmd, shell=True)
print "finished loading thread 3"
def thread4Load(self, thread4fileList):
connectionstring = settings.connectionstring
postgreshost = settings.postgreshost
postgresdatabase = settings.postgresdatabase
postgresport = settings.postgresport
postgresusername = settings.postgresusername
postgrespassword = settings.postgrespassword
tablename = None
connection = psycopg2.connect(connectionstring)
for filename in thread4fileList:
load_cmd = #load command
run = subprocess.Popen(load_cmd, shell=True)
print "finished loading thread 4"
def finishUp(self):
print 'finishing up'
def main():
load = Loader()
thread1 = threading.Thread(target=(load.thread1Load), args=(thread1fileList, ))
thread2 = threading.Thread(target=(load.thread2Load), args=(thread2fileList, ))
thread3 = threading.Thread(target=(load.thread3Load), args=(thread3fileList, ))
thread4 = threading.Thread(target=(load.thread4Load), args=(thread4fileList, ))
threads = [thread1, thread2, thread3, thread4]
for thread in threads:
thread.start()
thread.join()
load.finishUp(connectionstring)
if __name__ == '__main__':
main()
最佳答案 >
Don’t repeat yourself.一个threadLoad方法就足够了.这样,如果您需要修改方法中的某些内容,则无需在4个不同的位置进行相同的修改.
>使用run.communicate()阻止,直到子进程完成.
>这将启动一个线程,然后阻塞直到该线程完成,然后
启动另一个线程等:
for thread in threads:
thread.start()
thread.join()
相反,首先启动所有线程,然后加入所有线程:
for thread in threads:
thread.start()
for thread in threads:
thread.join()
import subprocess
import threading
class Loader(object):
def threadLoad(self, threadfileList):
connectionstring = settings.connectionstring
...
connection = psycopg2.connect(connectionstring)
for filename in threadfileList:
load_cmd = # load command
run = subprocess.Popen(load_cmd, shell=True)
# block until subprocess is done
run.communicate()
name = threading.current_thread().name
print "finished loading {n}".format(n=name)
def finishUp(self):
print 'finishing up'
def main():
load = Loader()
threads = [threading.Thread(target=load.threadLoad, args=(fileList, ))
for fileList in (thread1fileList, thread2fileList,
thread3fileList, thread4fileList)]
for thread in threads:
thread.start()
for thread in threads:
thread.join()
load.finishUp(connectionstring)
if __name__ == '__main__':
main()