Why in a simple loop I lose fetchAll's values of a sqlite3's table using python?-CodePudding

I need to print some rows from an sqlite3's table using python3, but into the loop I get none value. following my code:

import sqlite3
db=sqlite3.connect('db.db')
cdb=db.cursor()
cdb.execute('create table if not exists table1 (field1,field2,field3)')
cdb.execute('insert into table1 values (?,?,?)',(1,2,3))
db.commit()
cdb.execute('insert into table1 values (?,?,?)',(4,5,6))
db.commit()
cdb.execute('insert into table1 values (?,?,?)',(1,7,8))
db.commit()
cdb.execute('select * from table1')
for i in range(len(cdb.fetchall())):
    if cdb.fetchone()[0]==1:print(cdb.fetchone())
db.close()

my error message:
AttributeError: 'NoneType' object has no attribute 'replace'

Thanks

CodePudding user response：

Your call to cdb.fetchall() consumes the iterator and is pointless, so remove it:

for row in cdb.execute('select * from table1'):
    # do something with row

As described in the official documentation.

CodePudding user response：

After you fetchall the cursor point to nothing, so if you fetch anything it returns None.

to see what happend after commit the data just try:

cdb.execute('select * from table1')

print(cdb.fetchall()) # Return the data
print(cdb.fetchone()) # Return : None

db.close()

to fix this you could execute the sql again before the fetchone witch will be slow

I recommend doing this:

cdb.execute('select * from table1')

data = cdb.fetchall()
for i in data:
    if i[0] == 1: print(i)

db.close()