Celery task queuing - flask

Question

I have created a flask application and it consist of 2 celery tasks.
Task 1: Generate a file through a process
Task 2: Email the generated file
Normally task one needs more time compared to task 2. I want to execute task 1 and then task 2. But the problem is both start to execute at the same time inside celery.
How can I resolve this issue.
#celery.task(name='celery_example.process')
def process(a,b,c,d,e,f):
command='rnx2rtkp -p '+a+' -f '+b+' -m '+c+' -n -o oout.pos '+d+' '+e+' '+f
os.system(command)
return 'Successfully created POS file'
#celery.task(name='celery_example.emailfile')
def emailfile(recipientemail):
email_user = ''
email_password = ''
subject = 'subject'
msg = MIMEMultipart()
msg['From'] = email_user
msg['To'] = recipientemail
msg['Subject'] = subject
body = 'This is your Post-Processed position file'
msg.attach(MIMEText(body,'plain'))
filename='oout.pos'
attachment =open(filename,'rb')
part = MIMEBase('application','octet-stream')
part.set_payload((attachment).read())
encoders.encode_base64(part)
part.add_header('Content-Disposition',"attachment; filename= "+filename)
msg.attach(part)
text = msg.as_string()
server = smtplib.SMTP('smtp.gmail.com',587)
server.starttls()
server.login(email_user,email_password)
server.sendmail(email_user,recipientemail,text)
server.quit()
return 'Email has been successfully sent'
This is the app.route
#app.route('/pp.php', methods=['GET', 'POST'])
def pp():
pp = My1Form()
target = os.path.join(APP_ROOT)
print(target)
for fileBase in request.files.getlist("fileBase"):
print(fileBase)
filename = fileBase.filename
destination = "/".join([target, filename])
print(destination)
fileBase.save(destination)
for fileObsRover in request.files.getlist("fileObsRover"):
print(fileObsRover)
filename = fileObsRover.filename
destination = "/".join([target, filename])
print(destination)
fileObsRover.save(destination)
for fileNavRover in request.files.getlist("fileNavRover"):
print(fileNavRover)
filename = fileNavRover.filename
destination = "/".join([target, filename])
print(destination)
fileNavRover.save(destination)
a=fileObsRover.filename
b=fileBase.filename
c=fileNavRover.filename
elevation=pp.ema.data
Freq=pp.frq.data
posMode=pp.pmode.data
emailAdd=pp.email.data
process.delay(posMode,Freq,elevation,a,b,c)
emailfile.delay(emailAdd)
return render_template('results.html', email=pp.email.data, Name=pp.Name.data, ema=elevation, frq=Freq, pmode=posMode, fileBase=a)
return render_template('pp.php', pp=pp)

Answer 1

As it currently stands your code does the following:
# schedule process to run asynchronously
process.delay(posMode,Freq,elevation,a,b,c)
# schedule emailfile to run asynchronously
emailfile.delay(emailAdd)
Both of these will immediately be picked up by workers and executed. You have provided nothing to inform celery that emailfile should wait until processfile is complete.
Instead you should:
alter the signature of emailfile to include another parameter that will be the output of a successful processfile call; then
call processfile using link.
For example:
deferred = processfile.apply_async(
(posMode,Freq,elevation,a,b,c),
link=emailfile.s())
deferred.get()
An alternative to using link, but semantically identical in this case, would be to use a chain.