python脚本单击按钮

Question

我让这个 cod 为一个游戏网站工作了很长时间，但现在他们改变了点击它的方式：-

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)

现在我必须单击一个按钮才能再次进入，该按钮的信息是：-

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

我如何进行点击：-

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

或在 :-

中使用的那个

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

点击那个按钮而不是它之前所做的？？

Answer 1

opener.open("http://"+server+".outwar.com/formraid.php?target=" + bossToRaidId + "&suid="+FORMER_ID, "formtime=3&message=&bomb=none&submit=Launch!" )

仅适用于 2.7 python