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app.run is not lazy

use click.echo for banner messages
pull/2707/head
David Lord 7 years ago
parent
commit
80a9e0edf6
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  1. 2
      flask/app.py
  2. 10
      flask/cli.py

2
flask/app.py

@ -933,7 +933,7 @@ class Flask(_PackageBoundObject):
options.setdefault('use_debugger', self.debug)
options.setdefault('threaded', True)
cli.show_server_banner(self.env, self.debug, self.name)
cli.show_server_banner(self.env, self.debug, self.name, False)
from werkzeug.serving import run_simple

10
flask/cli.py

@ -614,7 +614,7 @@ def load_dotenv(path=None):
return new_dir is not None # at least one file was located and loaded
def show_server_banner(env, debug, app_import_path, eager_loading=True):
def show_server_banner(env, debug, app_import_path, eager_loading):
"""Show extra startup messages the first time the server is run,
ignoring the reloader.
"""
@ -623,11 +623,13 @@ def show_server_banner(env, debug, app_import_path, eager_loading=True):
if app_import_path is not None:
message = ' * Serving Flask app "{0}"'.format(app_import_path)
if not eager_loading:
message += ' (lazy loading)'
print(message)
print(' * Environment: {0}'.format(env))
click.echo(message)
click.echo(' * Environment: {0}'.format(env))
if env == 'production':
click.secho(
@ -636,7 +638,7 @@ def show_server_banner(env, debug, app_import_path, eager_loading=True):
click.secho(' Use a production WSGI server instead.', dim=True)
if debug is not None:
print(' * Debug mode: {0}'.format('on' if debug else 'off'))
click.echo(' * Debug mode: {0}'.format('on' if debug else 'off'))
class CertParamType(click.ParamType):

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