这很奇怪(不过,这是我第一次尝试使用
python / sqlite),但是如果我执行fetchAll(),我似乎可以获得所有行,但除此之外 – 无论我尝试什么,总是会结束在db中只返回第一行 – 第二次迭代因为返回null而停止.想知道我是如何在python中编写代码的？ db好像..
con = sqlite3.connect('backup.db') con.row_factory = sqlite3.Row cur = con.cursor() cur.execute('select * from tb1;') for row in cur: try: # row = dataCur.fetchone() #if row == None: break print type(row) print ' Starting on: %i' % row[0] cleaner = Cleaner(scripts=True,remove_tags=['img'],embedded=True) try: cleaned = cleaner.clean_html(row[2]) #data stored in second col cur.execute('update tb1 set data = ? where id = ?;',(cleaned,row[0])) except AttributeError: print 'Attribute error' print ' Ended on: %i' % row[0] except IOError: print 'IOexception'
