如何我使这个变量的结果?
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24-09-2019 - |
题
现在它设置为写入一个文件,但我希望它输出的值给一个变量。不知道如何。
from BeautifulSoup import BeautifulSoup
import sys, re, urllib2
import codecs
woof1 = urllib2.urlopen('someurl').read()
woof_1 = BeautifulSoup(woof1)
woof2 = urllib2.urlopen('someurl').read()
woof_2 = BeautifulSoup(woof2)
GE_DB = open('GE_DB.txt', 'a')
for row in woof_1.findAll("tr", { "class" : "row_b" }):
for col in row.findAll(re.compile('td')):
GE_DB.write(col.string if col.string else '')
GE_DB.write(" ")
GE_DB.write("\n")
GE_DB.close()
for row in woof_2.findAll("tr", { "class" : "row_b" }):
for col in row.findAll(re.compile('td')):
GE_DB.write(col.string if col.string else '')
GE_DB.write("\n")
GE_DB.close()
解决方案
values = []
for row in woof_1.findAll("tr", { "class" : "row_b" }):
for col in row.findAll(re.compile('td')):
if col.string:
values.append(col.string)
result = ''.join(values)
其他提示
摆脱所有提到GE_DB的。
做
outputtext = ""
向开头。
替换GE_DB.write(col.string if col.string else '')
与outputtext += col.string if col.string else ''
<击>也许是这样的。击>
<击>gedb = "";
for row in woof_1.findAll("tr", { "class" : "row_b" }):
for col in row.findAll(re.compile('td')):
if col.string:
gedb += col.string
击> <击> 撞击>
import cStringIO as StringIO # or import StringIO if on a fringe platform
buf = StringIO.StringIO()
for row in woof_1.findAll("tr", { "class" : "row_b" }):
for col in row.findAll(re.compile('td')):
buf.write(col.string if col.string else '')
result = buf.getvalue()
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