read each line of a text file using cron schedule

问题

I want to use cron schedule to read each line of a text file.

cron:
- description: read a line of text file
  url: /read
  schedule: every 1 minutes

But GAE only allow to read a text file without modifying it.

How can I make the request read to know that the previous line was already read and it should read the next line. I dont want to store all the read line in datastore and then do query, check if the text line is in datastore or not, bla bla, because I have million of lines.

Please advice. I am thinking about Google Drive Api, any idea about that? Thanks

UPDATE: Can I do like this

with open('text.txt', 'r') as f:
        txt= f.readlines()
f.close()

t = txt[line_num]
DO SOMETHING
line_num = line_num+1
deferred.defer( read_deferred, line_num, _countdown = 60 ) # 60 sec = 1 min

and in somewhere else, I trigger deffered by calling

deferred.defer( read_deferred, 0, _countdown = 60 ) # 60 sec = 1 min

回答1:

Cron won't let you save the state, except in database. Since you don't want to use DB, you may use deferred task instead, start it once and it will keep going until the last line is read, something like:

def read_deferred(last_pos) :
    f = open( ... )

    if last_pos : f.seek(last_pos)

    # here you read the next line and do whatever you like,
    # just don't go further and return once you hit EOF

    last_pos = f.tell()

    deferred.defer( read_deferred, last_pos, _countdown = 60 ) # 60 sec = 1 min

and you call it first time like:

    deferred.defer( read_deferred, _countdown = 60 ) # 60 sec = 1 min

来源：https://stackoverflow.com/questions/15527617/read-each-line-of-a-text-file-using-cron-schedule