我尝试在 Ubuntu 中创建服务,但出现错误。
我写了下面的代码-
在FileSystemWatcher.py
import requests
import json
import logging
import sys
import os
import datetime
from watchdog.events import PatternMatchingEventHandler
class DirectoryChangedHandler(PatternMatchingEventHandler):
patterns = ["*.json","*.csv"]
logFileName = datetime.datetime.now().strftime('%Y%m%d_log.log')
script_dir = os.path.dirname(__file__) # <-- absolute dir the script is in
rel_path = "log/"+logFileName
abs_logfile_path = os.path.join(script_dir, rel_path)
appConfigFilePath = os.path.abspath('config/app_config.json')
with open(appConfigFilePath) as data_file:
config = json.load(data_file)
logging.basicConfig(filename=abs_logfile_path, format="%(asctime)s - %(name)s - %(levelname)s - %(message)s", level=logging.NOTSET)
def process(self, event):
print event.src_path, event.event_type
try:
filelist = [('files', open(event.src_path, 'rb'))]
postUrl = self.config["apiBaseUrl"].encode('utf-8')+self.config["fileUplaodApiUrl"].encode('utf-8')
uploadResponse = requests.post(postUrl, files=filelist)
if uploadResponse.status_code == 200:
print "Upload Successful - ", event.src_path
else:
print "Upload Failed - ", event.src_path
except:
print "Unexpected error:", sys.exc_info()[0]
pass
def on_modified(self, event):
self.process(event)
def on_created(self, event):
self.process(event)
workflow.py-
#!/usr/bin/python2.7
import sys
import time
from watchdog.observers import Observer
import FileSystemWatcher as fSystemWatcher
if __name__ == '__main__':
args = sys.argv[1:]
observer = Observer()
observer.schedule(fSystemWatcher.DirectoryChangedHandler(), path=args[0] if args else '.', recursive=True)
observer.start()
try:
while True:
time.sleep(1)
except KeyboardInterrupt:
observer.stop()
observer.join()
和(/lib/systemd/system/FileWatcherSystemd.service)
[Unit]
Description=FileChangeService
[Service]
Type=forking
WorkingDirectory= /home/ashish/Documents/FileSystemWatcher
ExecStart= /home/ashish/Documents/FileSystemWatcher/workflow.py
Restart=on-failure
[Install]
WantedBy=multi-user.target
但我收到以下错误 -
FileWatcherSystemd.service - FileChangeService Loaded: loaded (/lib/systemd/system/FileWatcherSystemd.service; enabled; vendor preset: enabled) Active: inactive (dead) (Result: exit-code) since Wed 2017-07-19 10:44:39 IST; 8min ago Process: 6552 ExecStart=/home/ashish/Documents/FileSystemWatcher/FileSystemWatcher.py (code=exited, status=203/EXEC) Main PID: 6552 (code=exited, status=203/EXEC)
Jul 19 10:44:39 Ashish-PC systemd[1]: FileWatcherSystemd.service: Unit entered failed state. Jul 19 10:44:39 Ashish-PC systemd[1]: FileWatcherSystemd.service: Failed with result 'exit-code'. Jul 19 10:44:39 Ashish-PC systemd[1]: FileWatcherSystemd.service: Service hold-off time over, scheduling restart. Jul 19 10:44:39 Ashish-PC systemd[1]: Stopped FileChangeService. Jul 19 10:44:39 Ashish-PC systemd[1]: FileWatcherSystemd.service: Start request repeated too quickly. Jul 19 10:44:39 Ashish-PC systemd[1]: Failed to start FileChangeService.
我不知道我在 FileSytemWatcher.py
或 FileWatcherSystemd.service
中做错了什么
最佳答案
尝试将您的服务类型更改为简单。所以
type=simple
您的脚本似乎没有 fork 和退出。如果您的服务类型是 fork ,systemd 会假设已启动的进程 fork ,让某些内容在后台运行并退出。它会等到您的脚本运行完毕。
您的脚本不会退出,systemd 会检测到这一点并给出超时错误。您的脚本似乎处于无限循环等待键盘中断。这种脚本可以“简单”地运行。这意味着 systemd 只是在后台启动脚本并继续执行启动序列,而无需等待任何事情。
关于python - Linux (Ubuntu) 中的 systemd 服务出错,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45182651/