我遇到以下问题:
我的 process_query 函数如下:
def process_query(query, parameters):
print query
print parameters
if(parameters is None):
cursor1.execute(query)
return cursor1.fetchall()
else:
cursor1.execute(query, parameters)
return cursor1.fetchall()
我在以下函数中调用它:
def recieving_sumrpt():
theList=[]
theList=cursor1.execute('SELECT DISTINCT SSN FROM Employee').fetchall()
print theList
templist=[['SSN','Lname','FName','MInit','Known Hours','Unknown hours','Overtime']]
for i in theList:
basics= process_query('SELECT SSN, LName, FName, MInit FROM Employee WHERE SSN=?', (i))
templist.append(basics)
known_hours = process_query('SELECT sum(Coalesce(Hours,0)) FROM Works_On WHERE ESSN=?', (i))
print templist, known_hours
templist.append(known_hours[0])
count=cursor1.execute('SELECT Distinct ESSN FROM Works_On WHERE ISNULL(Hours) AND ESSN=?', (i)).rowcount
if count >=1:
templist.append('Yes')
else:
templist.append('No')
问题是当我打印临时列表时,我有以下外观:
[['SSN', 'Lname', 'FName', 'MInit', 'Known Hours', 'Unknown hours', 'Overtime'],
[('123456789', 'Smith', 'John', 'B')], [(Decimal('40.0'), )], 'No', 'Yes',
[('333445555', 'Wong', 'Franklin', 'T')], [(Decimal('40.0'), )], 'No', 'Yes',
[('453453453', 'English', 'Joyce', 'A')], [(Decimal('40.0'), )], 'No', 'Yes',
[('666884444', 'Narayan', 'Ramesh', 'K')], [(Decimal('40.0'), )], 'No', 'Yes',
[('888665555', 'Borg', 'James', 'E')], [(Decimal('0.0'), )], 'Yes', 'Yes',
[('987654321', 'Wallace', 'Jennifer', 'S')], [(Decimal('35.0'), )], 'No', 'Yes',
[('987987987', 'Jabbar', 'Ahmad', 'V')], [(Decimal('40.0'), )], 'No', 'Yes',
[('999887777', 'Zelaya', 'Alicia', 'J')], [(Decimal('40.0'), )], 'No', 'Yes']
我相信稍后当我想打印结果时会出现问题:
def printing(list):
for i in list:
print '%-10s %-10s %-10s %-10s %-15s %-10s %-15s' % (i[0],i[1],i[2],i[3],i[4], i[5], i[6])
因为这会给出错误:“列表索引超出范围”
我认为这是因为我没有以下外观:
['999887777', 'Zelaya', 'Alicia', 'J', Decimal('40.0'), 'No', 'Yes']
对于每一行。 (不过可能是错的,这是我的第一个 Python 实验室)。
打印函数的输出仅仅是:
SSN Lname FName MInit Known Hours Unknown hours Overtime
仅此而已。
我怎样才能将所有这些变成可以打印的东西?列表中出现的小数('40.0')是否可以被删除?我尝试过使用append(float(known_hours[0]),但这给了我一个错误,说known_hours必须是数字或字符串。另外,我还没有真正弄清楚如何打印;有什么问题吗?那个函数,还是 process_query 函数,因为我已经从原始的给定函数中配置了很多东西?
最佳答案
list = [['SSN', 'Lname', 'FName', 'MInit', 'Known Hours', 'Unknown hours', 'Overtime'],
[('123456789', 'Smith', 'John', 'B')],
[(Decimal('40.0'), )],
'No',
'Yes',
[('333445555', 'Wong', 'Franklin', 'T')],
[(Decimal('40.0'), )]
]
def printing(list):
for i in list:
print '%-10s %-10s %-10s %-10s %-15s %-10s %-15s' % (i[0],i[1],i[2],i[3],i[4], i[5], i[6])
最明显的是列表项没有 7 个条目... 事实上,在正确格式化数据时,您可能会看到列表的内容甚至不一致。有列表、元组列表、字符串等...
<小时/>我会尽力让您走上正确的道路来清理困惑:
def recieving_sumrpt():
theList = cursor1.execute('SELECT DISTINCT SSN FROM Employee').fetchall()
# You get a list of tupple containing each one SSN
templist=[['SSN','Lname','FName','MInit','Known Hours','Unknown hours','Overtime']]
# ok: a list of list of 7 strings.
for i in theList: # for each SSN
# here i contains a tuple of *1* element
basics= process_query('SELECT SSN, LName, FName, MInit FROM Employee WHERE SSN=?', i)
# basics contains now a list on *one* tupple of *4* items
templist.append(basics)
...
正如您将在上面的评论中看到的那样,在这一点上,我认为您已经走在错误的轨道上了。我的猜测是您想要一致的列表。包含 4 或 7 个项目,因为您打算将结果显示为某种表格。
最后一句话:也许这里的主要困难是你不明白 fetch all 返回一个“元组列表”。列表中每行的每个项目。元组的每一列包含一个项目。例如,在查询 SSN 时,您应该获得类似的信息:
[(1,), (2,), (3,)]
^ ^
\ \ each "row" is a tuple of 1 item because I selected only one *column*
\
\ a list of 3 items because there was 3 rows in my test table
关于python - 将我的查询结果放入字符串中以便可以打印,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18230899/