python脚本点击按钮

Question

我有这个鳕鱼长时间工作的一个游戏网站，但现在他们改变了它应该被点击的方式： -

def formRaid(): 
    #print formerID 
    one = 1 
    while one == 1: 
     try: 
      global _startTime 
      _startTime = datetime.datetime.now() 

      formpage = alo("http://"+server+".outwar.com/formraid.php?target=" + bossToRaidId+"&suid="+FORMER_ID).read() 
      codeID = formpage.split('codeid" value="')[1].split('">')[0] 

      Form = alo2("http://"+server+".outwar.com/formraid.php?target=" + bossToRaidId, "target=" + bossToRaidId + "&codeid=" + codeID + "&formtime=3&submit=Launch!").read() 

      global RAID_ID, RAID_NAME 
      RAID_ID = getRaidID(Form) 
      RAID_NAME = getRaidName(Form) 

      print "" 
      print "***" + RAID_NAME + " Formed." 
      one = 0 
      return True 

     except IndexError: 
      print "Forming Error, retrying...." 
      time.sleep(1) 

现在我必须点击一个按钮，进入一个再次与此按钮的信息是： -

<input type="submit" value="Join this Raid!" name="submit"> 

我怎样才能让在使用的点击： -

formpage = alo("http://"+server+".outwar.com/formraid.php?target=" + bossToRaidId+"&suid="+FORMER_ID).read() 
codeID = formpage.split('codeid" value="')[1].split('">')[0] 

或用于： -

Form = alo2("http://"+server+".outwar.com/formraid.php?target=" + bossToRaidId, "target=" + bossToRaidId + "&codeid=" + codeID + "&formtime=3&submit=Launch!").read() 

单击该按钮，而不是它之前做过什么？

Answer 1

opener.open（ “HTTP：//” +服务器+ “？.outwar.com/formraid.php目标=” + bossToRaidId + “& SUID =” + FORMER_ID，“formtime = 3 &消息= &炸弹=无&提交=启动！“）

仅适用于2.7蟒蛇