SQLAlchemy-utils: create view “if not exists”
When I use the create_view ()
method of the sqlalchemy-utils
module, everything works just fine the first time I run my script. However, every time after that first call, I encounter this error :
sqlalchemy.exc.OperationalError: (sqlite3.OperationalError) table XXX already exists
.
I am currently capturing the exception and pass
to avoid my script to stop but this seems messy to me.
Is there a way to avoid that behavior ?
python sqlalchemy sqlalchemy-utils
add a comment |
When I use the create_view ()
method of the sqlalchemy-utils
module, everything works just fine the first time I run my script. However, every time after that first call, I encounter this error :
sqlalchemy.exc.OperationalError: (sqlite3.OperationalError) table XXX already exists
.
I am currently capturing the exception and pass
to avoid my script to stop but this seems messy to me.
Is there a way to avoid that behavior ?
python sqlalchemy sqlalchemy-utils
There is SQLAlchemy.inspect().get_table_names(), which gives you a list of - surprise - table names. You could then check if the table you want to create is already in there. However, the try:except-method you mentioned is perfectly valid, cleaner and less expensive.
– joppich
Nov 21 '18 at 10:40
add a comment |
When I use the create_view ()
method of the sqlalchemy-utils
module, everything works just fine the first time I run my script. However, every time after that first call, I encounter this error :
sqlalchemy.exc.OperationalError: (sqlite3.OperationalError) table XXX already exists
.
I am currently capturing the exception and pass
to avoid my script to stop but this seems messy to me.
Is there a way to avoid that behavior ?
python sqlalchemy sqlalchemy-utils
When I use the create_view ()
method of the sqlalchemy-utils
module, everything works just fine the first time I run my script. However, every time after that first call, I encounter this error :
sqlalchemy.exc.OperationalError: (sqlite3.OperationalError) table XXX already exists
.
I am currently capturing the exception and pass
to avoid my script to stop but this seems messy to me.
Is there a way to avoid that behavior ?
python sqlalchemy sqlalchemy-utils
python sqlalchemy sqlalchemy-utils
asked Nov 21 '18 at 10:27
SebDeclercqSebDeclercq
161
161
There is SQLAlchemy.inspect().get_table_names(), which gives you a list of - surprise - table names. You could then check if the table you want to create is already in there. However, the try:except-method you mentioned is perfectly valid, cleaner and less expensive.
– joppich
Nov 21 '18 at 10:40
add a comment |
There is SQLAlchemy.inspect().get_table_names(), which gives you a list of - surprise - table names. You could then check if the table you want to create is already in there. However, the try:except-method you mentioned is perfectly valid, cleaner and less expensive.
– joppich
Nov 21 '18 at 10:40
There is SQLAlchemy.inspect().get_table_names(), which gives you a list of - surprise - table names. You could then check if the table you want to create is already in there. However, the try:except-method you mentioned is perfectly valid, cleaner and less expensive.
– joppich
Nov 21 '18 at 10:40
There is SQLAlchemy.inspect().get_table_names(), which gives you a list of - surprise - table names. You could then check if the table you want to create is already in there. However, the try:except-method you mentioned is perfectly valid, cleaner and less expensive.
– joppich
Nov 21 '18 at 10:40
add a comment |
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There is SQLAlchemy.inspect().get_table_names(), which gives you a list of - surprise - table names. You could then check if the table you want to create is already in there. However, the try:except-method you mentioned is perfectly valid, cleaner and less expensive.
– joppich
Nov 21 '18 at 10:40