Python + Sqlite:无法确定提供错误的绑定数量的原因错误

时间:2012-12-09 22:02:46

标签: python sqlite

我会直接进入它 - 我首先为自己创建了一个本地数据库:

import sqlite3

conn = sqlite3.connect("tofire.db") #

cursor = conn.cursor()

# create a table

cursor.execute("""CREATE TABLE incidents
                  (Id INTEGER PRIMARY KEY, prime_street text, cross_street text, dispatch_time text, 
                   incident_number text, incident_type text, alarm_level text, area text, dispatched_units text, date_added text)
               """)

这顺便说一下 - 下一部分是我的功能,它使用漂亮的汤把桌子刮成一份清单。然后我尝试将每个子列表中的信息写入sqlite数据库。

# Toronto Fire Calls

import urllib2
import sqlite3
import time
import csv
import threading
from bs4 import BeautifulSoup

# Beautiful Soup imports the URL as html
def getincidents ():

    response = urllib2.urlopen('http://www.toronto.ca/fire/cadinfo/livecad.htm')

    html = response.read()

    # We give the html its own variable.

    soup = BeautifulSoup(html)

    # Find the table we want on the Toronto Fire Page

    table = soup.find("table", class_="info")

    # Find all the <td> tags in the table and assign them to variable.

    cols = table.find_all('td')

    # Find the length of rows, which is the number of <font> tags, and assign it to a variable num_cols.

    num_cols = len(cols)

    # Create an empty list to hold each of the <font> tags as an element

    colslist = []
    totalcols = 0
    # For each <font> in cols, append it to colslist as an element.

    for col in cols:
        colslist.append(col.string)
        totalcols = len(colslist)

    # Now colslist has every td as an element from [0] to totalcols = len(colslist)

    # The First 8 <font> entries are always the table headers i.e. Prime Street, Cross Street, etc.

    headers = colslist[0:8]

    # Prime Street
    # Cross Street
    # Dispatch Time
    # Incident Number
    # Incident Type
    # Alarm Level
    # Area
    # Dispatched Units

    # Get the indexes from 0 to the length of the original list, in steps of list_size, then create a sublist for each.
    # lists = [original_list[i:i+list_size] for i in xrange(0, len(original_list), list_size)]
    list_size = 8
    i = 0
    incidents = [colslist[i:i+list_size] for i in xrange(0, len(colslist), list_size)]

    # Works!

    num_inci = len(incidents) # Get the number of incidents
    added = time.strftime("%Y-%m-%d %H:%M")
    update = 'DB Updated @ ' + added

    # SQL TIME, Connect to our db.
    conn = sqlite3.connect("tofire.db")
    cursor = conn.cursor()
    lid = cursor.lastrowid

    # Now we put each incident into our database.

    for incident in incidents[1:num_inci]:
        incident.append(added)
        to_db = [(i[0:10]) for i in incident]
        import ipdb; ipdb.set_trace()
        cursor.executemany("INSERT INTO incidents (prime_street, cross_street, dispatch_time, incident_number, incident_type, alarm_level, area, dispatched_units, date_added) VALUES (?,?,?,?,?,?,?,?,?)", to_db)
    conn.commit()
    print update
    print "The last Id of the inserted row is %d" % lid
    threading.Timer(300, getincidents).start()

getincidents()

我总是最终收到错误消息“提供的绑定数量不正确” - 它声称我在我的声明中使用9时提供了10个。我试图缩小原因,但没有成功。

2 个答案:

答案 0 :(得分:1)

正如Ned Batchelder最近所说的那样,“调试的第一条规则:怀疑时,打印出更多。”将added追加到incident后,incident本身就有9个项目:

print(incident)
# [u'NORFINCH DR, NY', u'FINCH AVE W / HEPC', u'2012-12-09 17:32:57', u'F12118758', u'Medical - Other', u'0', u'142', u'\r\nP142, \r\n\r\n', '2012-12-09 17:46']

所以看起来你真正需要做的就是使用incident作为cursor.execute的第二个参数。或者,如果你想摆脱像u'\r\nP142, \r\n\r\n'这样的项目之外的一些空白, 你可以用

    to_db = [i.strip() for i in incident]

for incident in incidents[1:num_inci]:
    incident.append(added)
    to_db = [i.strip() for i in incident]
    import ipdb; ipdb.set_trace()
    cursor.execute(
        """INSERT INTO incidents
           (prime_street, cross_street, dispatch_time, incident_number,
            incident_type, alarm_level, area, dispatched_units, date_added)
           VALUES (?,?,?,?,?,?,?,?,?)""", to_db)
    lid = cursor.lastrowid

答案 1 :(得分:0)

您正在为已定义为仅包含9个元素的插入语句提供10个元素

或者

to_db = [(i[0:9]) for i in incident]  #or
to_db = [(i[1:10]) for i in incident]

会给你9个元素......它们的数量是多少? insert语句中的标记(以及要填充的字段数)......

我假设你想要什么