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Don’t catch exceptions if API fails to create job

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There is no good reason why the API should fail to create a job at this point.
If it’s returning a 404, this is an error, and we should be monitoring for it.
So we should let it raise, and throw a 500.
This commit is contained in:
Chris Hill-Scott
2016-03-08 17:50:18 +00:00
parent 93b4419bb2
commit c1bf48f91e
1 changed files with 8 additions and 14 deletions
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22
app/main/views/send.py
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@@ -247,25 +247,19 @@ def start_job(service_id, upload_id):
upload_data = session['upload_data'] upload_data = session['upload_data']
services_dao.get_service_by_id_or_404(service_id) services_dao.get_service_by_id_or_404(service_id)
if request.files or not session['upload_data'].get('valid'): if request.files or not upload_data.get('valid'):
# The csv was invalid, validate the csv again # The csv was invalid, validate the csv again
return send_messages(service_id, upload_data.get('template_id')) return send_messages(service_id, upload_data.get('template_id'))
session.pop('upload_data') session.pop('upload_data')
try: job_api_client.create_job(
job_api_client.create_job( upload_id,
upload_id, service_id,
service_id, upload_data.get('template_id'),
upload_data.get('template_id'), upload_data.get('original_file_name'),
upload_data.get('original_file_name'), upload_data.get('notification_count')
upload_data.get('notification_count') )
)
except HTTPError as e:
if e.status_code == 404:
abort(404)
else:
raise e
return redirect( return redirect(
url_for('main.view_job', service_id=service_id, job_id=upload_id) url_for('main.view_job', service_id=service_id, job_id=upload_id)