Question

我有这个鳕鱼长时间工作的游戏网站，但现在他们改变了应该点击的方式： -

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)

现在我必须再次单击一个按钮，此按钮的信息为： -

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

如何使用中的点击： -

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： -

的那个

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

单击该按钮而不是之前的操作？

Answer 1

opener.open（“http：//”+ server +“。outwar.com / formraid.php？target =”+ bossToRaidId +“＆amp; suid =”+ FORMER_ID，“formtime = 3＆amp; message =＆amp; bomb =无＆安培;！提交=启动“）

仅适用于2.7 python

单击按钮的python脚本

1 个答案: