最新消息：20210816 当前crifan.com域名已被污染，为防止失联，请关注（页面右下角的）公众号
    你的位置：在路上 > 工作和技术 > Web > Crawl_EmulateLogin > pyspider > 【已解决】PySpider中如何强制让重复的url地址继续爬取

    【已解决】PySpider中如何强制让重复的url地址继续爬取

    pyspider crifan 3681浏览 0评论

    PySpider中，通过一个函数，实现了根据当前页面号码，递归获取下一个页面：

    相关部分代码是：

    <code>
    # @every(minutes=24 * 60)
    def on_start(self):
        self.crawl(
            'http://xxxa=audition&amp;act_id=3',
            callback=self.index_page)

    # @config(age=10 * 24 * 60 * 60)
    def index_page(self, response):
        # &lt;ul class="list-user list-user-1" id="list-user-1"&gt;
        for each in response.doc('ul[id^="list-user"] li  a[href^="http"]').items():
            self.crawl(each.attr.href, callback=self.showVideoCallback)

        curPageNum = 1
        self.getNextPageShowParentChild(response, curPageNum)
        self.getNextPageShowFriend(response, curPageNum)

    def getNextPageShowParentChild(self, response, curPageNum):
        """
            for 亲子组: recursively get next page shows until fail
        """
        # &lt;ul class="list-user list-user-1" id="list-user-1"&gt;
        self.getNextPageShow(response, curPageNum, 1)

    def getNextPageShowFriend(self, response, curPageNum):
        """
            for 好友组: recursively get next page shows until fail
        """
        # &lt;ul class="list-user list-user-2" id="list-user-2"&gt;
        self.getNextPageShow(response, curPageNum, 2)

    def getNextPageShow(self, response, curPageNum, order):
        """
            recursively get next page shows until fail
        """

        print("getNextPageShow: curPageNum=%s, order=%s" % (curPageNum, order))
        getShowsUrl = "http://xxxc=match_new&amp;a=get_shows"
        headerDict = {
            "Content-Type": "application/x-www-form-urlencoded"
        }

        dataDict = {
            "counter": curPageNum,
            "order": order,
            "match_type": 2,
            "match_name": "",
            "act_id": 3
        }

        curPageDict = {
            "curPageNum": curPageNum,
            "order": order
        }

        self.crawl(
            getShowsUrl,
            method="POST",
            headers=headerDict,
            data=dataDict,
            cookies=response.cookies,
            callback=self.parseGetShowsCallback,
            save=curPageDict
        )

    def parseGetShowsCallback(self, response):
        print("parseGetShowsCallback: self=%s, response=%s"%(self, response))
        respJson = response.json
        prevPageDict = response.save
        print("prevPageDict=%s, respJson=%s" % (prevPageDict, respJson))

        if respJson["status"] == 1:
            respData = respJson["data"]

            # recursively try get next page shows
            curPageNum = prevPageDict["curPageNum"] + 1
            self.getNextPageShow(response, curPageNum, prevPageDict["order"])

            for eachData in respData:
                # print("type(eachData)=" % type(eachData))
                showId = eachData["show_id"]
                href = eachData["href"]
                fullUrl = QupeiyinUrlRoot + href
                print("[%s] fullUrl=%s" % (showId, fullUrl))
                curShowInfoDict = eachData
                self.crawl(
                    fullUrl,
                    callback=self.showVideoCallback,
                    save=curShowInfoDict)
        else:
            print("!!! Fail to get shows json from %s" % response.url)
</code>

    但是后来发现：

    getNextPageShow

    parseGetShowsCallback

    在调试期间可以继续执行到：

    但是RUN运行时，就无法执行到了

    最终只保存出2页的60个result：

    而实际上希望是，几十几百个才对。

    然后也去注释掉了：

    <code># @every(minutes=24 * 60)
# @config(age=10 * 24 * 60 * 60)
</code>

    也还是无法执行到。

    然后才想到，估计是代码中此处是递归调用，函数和url没有变化：

    getNextPageShow

    中的：

    http:/xxxc=match_new&a=get_shows

    始终是一样的，虽然调用时POST参数不同

    估计就是这个原因导致无法继续运行的。

    所以要去想办法强制执行

    pyspider 函数 不继续执行

    pyspider 重复不继续执行

    Pyspider 函数不执行 – 足兆叉虫的回答 – SegmentFault 思否

    pyspider run状态下result没有数据，而且没有继续向下执行，为什么？ – 足兆叉虫的回答 – SegmentFault 思否

    那么抽空去加上随机的hash试试

    关于重复爬取出现问题 · Issue #598 · binux/pyspider

    找官网文档看看是否有关重复url的强制执行的设置

    self.crawl – pyspider中文文档 – pyspider中文网

    “age

    本参数用来指定任务的有效期，在有效期内不会重复抓取．默认值是-1（永远不过期，意思是只抓一次）”

    或许可以设置极低的age实现继续重复抓取？

    “itag

    任务标记值，此标记会在抓取时对比，如果这个值发生改变，不管有效期有没有到都会重新抓取新内容．多数用来动态判断内容是否修改或强制重爬．默认值是：None.”

    貌似用itag实现强制重复抓取。

    <code>
    def getNextPageShow(self, response, curPageNum, order):
        """
            recursively get next page shows until fail
        """

        print("getNextPageShow: curPageNum=%s, order=%s" % (curPageNum, order))
        getShowsUrl = "http://xxxmatch_new&amp;a=get_shows"
        headerDict = {
            "Content-Type": "application/x-www-form-urlencoded"
        }

        dataDict = {
            "counter": curPageNum,
            "order": order,
            "match_type": 2,
            "match_name": "",
            "act_id": 3
        }

        curPageDict = {
            "curPageNum": curPageNum,
            "order": order
        }

        fakeItagForceRecrawl = "%s_%s" % (curPageNum, order)

        self.crawl(
            getShowsUrl,
            itag=fakeItagForceRecrawl, # To force re-crawl for next page
            method="POST",
            headers=headerDict,
            data=dataDict,
            cookies=response.cookies,
            callback=self.parseGetShowsCallback,
            save=curPageDict
        )
</code>

    试试效果

    调试期间，可以看到itag值是每次都不同了：

    还是60个结果，没反应啊

    算了，清空之前数据，重新运行试试

    貌似可以了，至少超过60个了，就是等待继续爬取了：

    【总结】

    PySpider中，如果后续要爬取crawl的url，有重复，默认是不会重新爬取的。

    想要强制重新爬取的话，做法有：

    • 给URL后面加上无关紧要的hash值，比如：

      • http://xxx?m=home&c=match_new&a=get_shows#123

      • http://xxx?m=home&c=match_new&a=get_shows#456

      • 当然hash值的选取，也可以选取时间戳（带毫秒的，不容易重复）

    • 设置itag值

      • 默认为None，所以默认不强制抓取

      • 每次设置不同itag值，比如此处每次调用

      • getNextPageShow，要爬取的url是：

      • http://xxx?m=home&c=match_new&a=get_shows

      • 设置itag可以选择：random的值，或者此处和实际逻辑有关系的，比如此处的每次爬取的页面数curPageNum和order值，则使用：2_1，3_1，4_1等等

    参考代码：

    <code>
    def getNextPageShow(self, response, curPageNum, order):
        """
            recursively get next page shows until fail
        """

        print("getNextPageShow: curPageNum=%s, order=%s" % (curPageNum, order))
        getShowsUrl = "http://xxxmatch_new&amp;a=get_shows"
        ...
        fakeItagForceRecrawl = "%s_%s" % (curPageNum, order)

        self.crawl(
            getShowsUrl,
            itag=fakeItagForceRecrawl, # To force re-crawl for next page
            method="POST",
            headers=headerDict,
            data=dataDict,
            cookies=response.cookies,
            callback=self.parseGetShowsCallback,
            save=curPageDict
        )
</code>

    【后记】

    但是结果也不错，只有137个结果：

    应该是好几百，上千个才对

    pyspider 重复 不爬取

    算了，去加上#hash值试试

    <code>from datetime import datetime,timedelta
import time

        timestampStr = datetime.now().strftime("%Y%m%d_%H%M%S_%f")
        fakeItagForceRecrawl = "%s_%s_%s" % (timestampStr, curPageNum, order)
        getShowsUrlWithHash = getShowsUrl + "#" + timestampStr

        self.crawl(
            getShowsUrlWithHash,
            itag=fakeItagForceRecrawl, # To force re-crawl for next page
            method="POST",
            headers=headerDict,
            data=dataDict,
            cookies=response.cookies,
            callback=self.parseGetShowsCallback,
            save=curPageDict
        )
</code>

    既把itag更加复杂，不会重复了，又加了hash了。

    看看结果：

    调试时可以看到itag不同：

    <code>  "project": "xxx",
  "schedule": {
    "itag": "20180712_153856_329293_6_1"
  },
  "taskid": "36cb9d54f6a82215e66d268aaac65848",
  "url": "http://xxxa=get_shows#20180712_153856_329293"
}
</code>

    然后重新清空项目数据，重新运行，看看是否能强制执行所有的代码

     首次执行，就对了，生成2个url：

    <code>  "project": "xx",
  "schedule": {
    "itag": "20180712_154134_660231_1_1"
  },
  "taskid": "da109ba37f77ca5983d376c0f791cf72",
  "url": "http://xxxa=get_shows#20180712_154134_660231"
}
  "project": "xx",
  "schedule": {
    "itag": "20180712_154134_660436_1_2"
  },
  "taskid": "fc8d90bf8dff1ac7f9b7384cc779c4fd",
  "url": "http://xxxa=get_shows#20180712_154134_660436"
}
</code>

    然后后续再去调试看看

    调试看起来都是对的。

    再去DEBUG或RUN看看结果：

    目前有1000多个了，这才是对的：

    目前结果也有了24页了：

    是正常的。

    【总结2】

    此处，不仅要加上itag值，也要加上#hash值：

    <code>
from pyspider.libs.base_handler import *
import re
import os
import codecs
import json
from datetime import datetime,timedelta
import time

QupeiyinUrlRoot = "http://xxx"

OutputFullPath = "/Users/crifan/dev/dev_root/xxx/output"

class Handler(BaseHandler):
    crawl_config = {
    }

    # @every(minutes=24 * 60)
    def on_start(self):
        self.crawl(
            'http://xxxaudition&amp;act_id=3',
            callback=self.index_page)

    # @config(age=10 * 24 * 60 * 60)
    def index_page(self, response):
        # &lt;ul class="list-user list-user-1" id="list-user-1"&gt;
        for each in response.doc('ul[id^="list-user"] li  a[href^="http"]').items():
            self.crawl(each.attr.href, callback=self.showVideoCallback)

        curPageNum = 1
        self.getNextPageShowParentChild(response, curPageNum)
        self.getNextPageShowFriend(response, curPageNum)

    def getNextPageShowParentChild(self, response, curPageNum):
        """
            for 亲子组: recursively get next page shows until fail
        """
        # &lt;ul class="list-user list-user-1" id="list-user-1"&gt;
        self.getNextPageShow(response, curPageNum, 1)

    def getNextPageShowFriend(self, response, curPageNum):
        """
            for 好友组: recursively get next page shows until fail
        """
        # &lt;ul class="list-user list-user-2" id="list-user-2"&gt;
        self.getNextPageShow(response, curPageNum, 2)

    def getNextPageShow(self, response, curPageNum, order):
        """
            recursively get next page shows until fail
        """

        print("getNextPageShow: curPageNum=%s, order=%s" % (curPageNum, order))
        getShowsUrl = "http://xxxa=get_shows"
        headerDict = {
            "Content-Type": "application/x-www-form-urlencoded"
        }

        dataDict = {
            "counter": curPageNum,
            "order": order,
            "match_type": 2,
            "match_name": "",
            "act_id": 3
        }

        curPageDict = {
            "curPageNum": curPageNum,
            "order": order
        }

        timestampStr = datetime.now().strftime("%Y%m%d_%H%M%S_%f")
        fakeItagForceRecrawl = "%s_%s_%s" % (timestampStr, curPageNum, order)
        getShowsUrlWithHash = getShowsUrl + "#" + timestampStr

        self.crawl(
            getShowsUrlWithHash,
            itag=fakeItagForceRecrawl, # To force re-crawl for next page
            method="POST",
            headers=headerDict,
            data=dataDict,
            cookies=response.cookies,
            callback=self.parseGetShowsCallback,
            save=curPageDict
        )
</code>

    才可以，实现url不重复，使得实现强制重新抓取的效果。

    itag值和带hash的url是这种：

    <code>  "project": "xxx",
  "schedule": {
    "itag": "20180712_154134_660436_1_2"
  },
  "taskid": "fc8d90bf8dff1ac7f9b7384cc779c4fd",
  "url": "http://xxxget_shows#20180712_154134_660436"
}
</code>

    转载请注明：在路上 » 【已解决】PySpider中如何强制让重复的url地址继续爬取

    Post Views: 1,149

    与本文相关的文章

    发表我的评论
    取消评论

    表情

    Hi，您需要填写昵称和邮箱！

    • 昵称 (必填)
    • 邮箱 (必填)
    • 网址
    88 queries in 0.183 seconds, using 22.23MB memory