How can I get current date / time in Lua embedded in Redis?
I need to have it in following format - YYYY-MM-DD, HH:MM:SS
Tried with os.date() but it does not recognize it.
Redis' Lua sandbox has only a handful of libraries, and os isn't one of these.
You can call the Redis TIME from Lua like so:
local t = redis.call('TIME')
However, you'll need to find a way to convert the epoch to the desired format and also note that it will stop you script from performing any writes (as it is a non-deterministic command).
Update: as of Redis v3.2, there is a new replication mode for scripts that is effect-based (rather than code-based). When using this mode you can actually call all the random, non-deterministic commands. More information is at EVAL's documentation page
This was already discussed in the comments, but the correct answer should have an answer:
The current time is non-deterministic i.e. it returns different values on repeated calls. This hurts replication. For this reason, the current time should be passed into your LUA script as a parameter.
Related
So I'm running a LUA script that executes every minute, this is controlled by software and I can't control the timing of the execution. I would like to check the time every day and trigger a function at a specific time. However, I would like to execute another script 5 minutes before that happens.
Right now I'm doing a string comparison using os.date() and parsing it to a string. It works, but the code isn't pretty and if the time changes, i have to manually change the time in two different variables, I'm pretty new to LUA so I've been having difficulty figuring out the best way to do this.
So the question is, how do I set a time variable, and compare that variable to os.date (or os.time) ?
There's no fancy way to do timers in pure Lua. What you can do to avoid os.date() strings, is to generate timestamps with preset dates, os.time accepts a key-value table to set a date:
timestamp = os.time({year=2019, month=1, day=n})
By iteratively increasing the n variable, you will receive a timestamp for every new n day after January 1st 2019. E.g.
os.date("%Y-%m-%d %H-%M-%S",os.time({year=2019,month=1,day=900})
--> 2021-06-18 12-00-00
If you can't save the day variable to keep track of (between application restarts), get the current "today" day and iterate from there:
os.date("%Y-%m-%d %H-%M-%S",
os.time({year=os.date("%Y"),month=os.date("%m"),day=os.date("%d")+n}
)
Using os.date with custom format and os.time makes your code independent of currently set date locale.
After you have determined the timestamp of the first task, offset the second actual task by five minutes secondTaskTimestamp = fistTaskTimestamp + 5*60 (or use os.time again). Your timer checker only should compare timestamps by now.
Now when you have to change the pre-configured time, you will only have to change the time-date of the first task, and the second task will be automatically offset.
Related: How do I execute a global function on a specific day at a specific time in Lua?
I'm wondering if it's possible to add timestamps to the journal file?
It appears that a date & time are recorded when SPSS is started, but if you have the program open for longer periods of time (i.e. days) it doesn't break it up if the program isn't closed.
Having timestamps would make it much easier to find what I'm looking for the times I look back to find things.
This is what I use to insert timestamps into my output:
HOST COMMAND=['echo %time%'].
However the journal file only shows the syntax.
The journal file is kept flushed and closed by Statistics, so you can probably write to it from another process. I don't think the suggestion above will work, because it will write the code but not the output to the journal. However, using Python you could do something like this.
begin program.
import time
open(r"full path to your journal file", "a").write("* " + time.asctime() + "\n")
end program
I can't see why it shouldn't work, unless you are not using a windows operating system.
On Unix-like system like Linux or Mac which run the bash (shell) you would rather use
HOST COMMAND =['date'].
If you have the Python extension installed you could also use Python code to to print the date and time (which would be a platform independent solution).
BEGIN PROGRAM.
import time
print time.ctime()
END PROGRAM.
I am setting up an app that has certain programs that a user can subscribe to. These programs start and stop consecutively (example program A runs form Jan-March, program B runs to April-June, program C runs July-September and program D runs from October-December).
I would like to only display the programs open for subscribers when the month is present for the specific program or programs.
What is the proper method (Built in or to create) to set up in my views? I have read through the Ruby Date Api Docs (http://api.rubyonrails.org/classes/Date.html) but didnt find anything to suffice
I assume your program model has a start date and an end date. You would compare the current time (DateTime.current or Time.current) against those to determine which program to show. You could define a helper method or a scope on Program to return the appropriate program based on the current time.
You can get your current quarter with this approach:
current_quarter = ((Time.now.month - 1) / 3) + 1
and then basically you leverage that to set the start and end date on a scope/helper method to filter your programs.
What I understand from your question is you want a program that run automatically on a special date, This gem may solve your problem.
https://github.com/javan/whenever
I have a cluster app that uses a distributed Redis back-end, with dynamically generated Lua scripts dispatched to the redis instances. The Lua component scripts can get fairly complex and have a significant runtime, and I'd like to be able to profile them to find the hot spots.
SLOWLOG is useful for telling me that my scripts are slow, and exactly how slow they are, but that's not my problem. I know how slow they are, I'd like to figure out which parts of them are slow.
The redis EVAL docs are clear that redis does not export any timekeeping functions to lua, which makes it seem like this might be a lost cause.
So, short a custom fork of Redis, is there any way to tell which parts of my Lua script are slower than others?
EDIT
I took Doug's suggestion and used debug.sethook - here's the hook routine I inserted at the top of my script:
redis.call('del', 'line_sample_count')
local function profile()
local line = debug.getinfo(2)['currentline']
redis.call('zincrby', 'line_sample_count', 1, line)
end
debug.sethook(profile, '', 100)
Then, to see the hottest 10 lines of my script:
ZREVRANGE line_sample_count 0 9 WITHSCORES
If your scripts are processing bound (not I/O bound), then you may be able to use the debug.sethook function with a count hook:
The count hook: is called after the interpreter executes every
count instructions. (This event only happens while Lua is executing a
Lua function.)
You'll have to build a profiler based on the counts you receive in your callback.
The PepperfishProfiler would be a good place to start. It uses os.clock which you don't have, but you could just use hook counts for a very crude approximation.
This is also covered in PiL 23.3 – Profiles
In standard Lua C, you can't. It's not a built-in function - it only returns seconds. So, there are two options available: You either write your own Lua extension DLL to return the time in msec, or:
You can do a basic benchmark using a millisecond-resolution time. You can access the current millisecond time with LuaSocket. Though this adds a dependency to your project, it's an effective way to do trivial benchmarking.
require "socket"
t = socket.gettime();
I have two files :
first file contains jobname and start time which looks like below:
ZPUDA13V STARTED - TIME=00.13.30
ZPUDM00V STARTED - TIME=03.26.54
ZPUDM01V STARTED - TIME=03.26.54
ZPUDM02V STARTED - TIME=03.26.54
ZPUDM03V STARTED - TIME=03.26.56
and the second file contains jobname and Endtime which looks like below:
ZPUDA13V ENDED - TIME=00.13.37
ZPUDM00V ENDED - TIME=03.27.38
ZPUDM01V ENDED - TIME=03.27.34
ZPUDM02V ENDED - TIME=03.27.29
ZPUDM03V ENDED - TIME=03.27.27
Now I am trying to combine these two files to get the report like JOBNAME START TIME ENDTIME.I have used ICETOOL to get the report If I get JOBNAME START TIME ,ENDTIME is SPACES .If I get Endtime ,JOBNAME START TIME gets spaces.
Please let me know how to code the outrec fields as I have coded with almost all possibilites to get the desired one.But still my output is not the same as I required
I have no idea what ICETOOL is (nor the inclination to even look it up in Google :-) but this is a classic COBOL data processing task.
Based on your simple data input, the algorithm would be:
for every record S in startfile:
for every record E in endfile:
if S.jobnname = E.jobname:
ouput S.jobname S.time E.time
next S
endif
endfor
endfor
However, you may need to take into account the fact that:
multiple jobs of the same name may run during the day (multiple entries in the file).
multiple jobs of the same name may run at the same time.
You could get around the first problem by ensuring the E record was the one immediately following the S record (based on time). The second problem is a doozy.
If you're running on z/OS (and you probably are, given the job names), have you considered using information from the SMF records to do this collection and analysis. I'm pretty certain SMF type 30 records hold everything you need.
And assuming this is a mainframe question, here's a shameless plug for a book one of my friends at work has written, check out What On Earth is a Mainframe? by David Stephens (ISBN-13 = 978-1409225355).
I know, i'm toooo late with my resolution, but may be helpful for new comers to stackoverflow
You can make use of JOINKEYS of DFSORT using JCL.
JOINKEYS F1 FIELDS=(01,08,CH,A)
JOINKEYS F2 FIELDS=(01,08,CH,A)
REFORMAT FIELDS=(F1:01,33,F2:25,08)
SORT FIELDS=COPY
OUTREC FIELDS=(01,08,25,08,34,08)
the outrec will hold the data as you need!