[英]Python SQLite3 - cursor.execute - no error
这是一段代码,需要执行以下功能:
该代码已经过测试,可以在单个数据库上工作。 但是,一旦将它放到另一段代码中,并用不同的数据库对其进行调用,它现在就不会进入第49行。但是也没有错误,因此我一直在努力寻找问题所在,因为我没有进行任何更改。
代码段第48行是最底行-
cursor.execute("SELECT name FROM sqlite_master WHERE type='table'")
print (cursor)
for tablerow in cursor.fetchall():
我在/ tmp /目录中运行此命令,原因是sqlite先前出现错误,无法在温度范围之外工作。
有任何问题请问他们。
谢谢!!
#!/usr/bin/python
# -*- coding: utf-8 -*-
import sqlite3
import os
import sys
filename = sys.argv[1]
def validateFile(filename):
filename, fileExt = os.path.splitext(filename)
print ("[Jconsole] Python: Filename being tested - " + filename)
if fileExt == '.db':
databases(filename)
elif fileExt == '.json':
jsons(fileExt)
elif fileExt == '':
blank()
else:
print ('Unsupported format')
print (fileExt)
def validate(number):
try:
number = float(number)
if -90 <= number <= 180:
return True
else:
return False
except ValueError:
pass
def databases(filename):
dbName = sys.argv[2]
print (dbName)
idCounter = 0
mainList = []
lat = 0
lon = 0
with sqlite3.connect(filename) as conn:
conn.row_factory = sqlite3.Row
cursor = conn.cursor()
cursor.execute("SELECT name FROM sqlite_master WHERE type='table'")
print (cursor)
for tablerow in cursor.fetchall():
print ("YAY1")
table = tablerow[0]
cursor.execute('SELECT * FROM {t}'.format(t=table))
for row in cursor:
print(row)
print ("YAY")
tempList = []
for field in row.keys():
tempList.append(str(field))
tempList.append(str(row[field]))
for i in tempList:
if i in ('latitude', 'Latitude'):
index = tempList.index(i)
if validate(tempList[index + 1]):
idCounter += 1
tempList.append(idCounter)
(current_item, next_item) = \
(tempList[index], tempList[index + 1])
lat = next_item
if i in ('longitude', 'Longitude'):
index = tempList.index(i)
if validate(tempList[index + 1]):
(current_item, next_item) = \
(tempList[index], tempList[index + 1])
lon = next_item
result = '{ "id": ' + str(idCounter) \
+ ', "content": "' + dbName + '", "title": "' \
+ str(lat) + '", "className": "' + str(lon) \
+ '", "type": "box"},'
mainList.append(result)
file = open('appData.json', 'a')
for item in mainList:
file.write('%s\n' % item)
file.close()
# {
# ...."id": 1,
# ...."content": "<a class='thumbnail' href='./img/thumbs/thumb_IMG_20161102_151122.jpg'>IMG_20161102_151122.jpg</><span><img src='./img/thumbs/thumb_IMG_20161102_151122.jpg' border='0' /></span></a>",
# ...."title": "50.7700721944444",
# ...."className": "-0.8727045",
# ...."start": "2016-11-02 15:11:22",
# ...."type": "box"
# },
def jsons(filename):
print ('JSON')
def blank():
print ('blank')
validateFile(filename)
固定。
问题在这里
filename, fileExt = os.path.splitext(filename)
没有文件扩展名的文件名变量被覆盖,因此当SQLite搜索时找不到文件。
奇怪的是没有错误出现,但是现在可以通过将文件名var更改为filename1来解决。
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