Python代码获取源页面中表格的html数据

Question

我是Python新手，我正在尝试抓取一个网站。我可以登录网站并获取 html 页面，但我不需要整个页面，我只需要指定表格中的超链接。

我编写了以下代码，但这获取了所有超链接。

soup = BeautifulSoup(the_page)
for table in soup.findAll('table',{'id':'ctl00_Main_lvMyAccount_Table1'} ):
        for link in soup.findAll('a'):
                print link.get('href')

谁能帮我看看我哪里出错了？

下面是表格的 html 文本

<table id="ctl00_Main_lvMyAccount_Table1" width="680px">
 <tr id="ctl00_Main_lvMyAccount_Tr1">
    <td id="ctl00_Main_lvMyAccount_Td1">
                        <table id="ctl00_Main_lvMyAccount_itemPlaceholderContainer" border="1" cellspacing="0" cellpadding="3">
        <tr id="ctl00_Main_lvMyAccount_Tr2" style="background-color:#0090dd;">
            <th id="ctl00_Main_lvMyAccount_Th1"></th>
            <th id="ctl00_Main_lvMyAccount_Th2">

                                    <a id="ctl00_Main_lvMyAccount_SortByAcctNum" href="javascript:__doPostBack('ctl00$Main$lvMyAccount$SortByAcctNum','')">
                                        <font color=white>
                                            <span id="ctl00_Main_lvMyAccount_AcctNum">Account number</span>
                                        </font>

                                        </a>
                                </th>
            <th id="ctl00_Main_lvMyAccount_Th4">
                                    <a id="ctl00_Main_lvMyAccount_SortByServAdd" href="javascript:__doPostBack('ctl00$Main$lvMyAccount$SortByServAdd','')">
                                    <font color=white>
                                        <span id="ctl00_Main_lvMyAccount_ServiceAddress">Service address</span>
                                    </font>
                                    </a>
                                </th>
            <th id="ctl00_Main_lvMyAccount_Th5">
                                    <a id="ctl00_Main_lvMyAccount_SortByAcctName" href="javascript:__doPostBack('ctl00$Main$lvMyAccount$SortByAcctName','')">
                                    <font color=white>
                                        <span id="ctl00_Main_lvMyAccount_AcctName">Name</span>
                                    </font>
                                    </a>
                                </th>
            <th id="ctl00_Main_lvMyAccount_Th6">
                                    <a id="ctl00_Main_lvMyAccount_SortByStatus" href="javascript:__doPostBack('ctl00$Main$lvMyAccount$SortByStatus','')">
                                    <font color=white>
                                        <span id="ctl00_Main_lvMyAccount_AcctStatus">Account status</span>
                                    </font>
                                    </a>
                                </th>
            <th id="ctl00_Main_lvMyAccount_Th3"></th>
        </tr>


            <tr>
                <td>

提前致谢。

Answer 1

嗯，这是正确的方法。

soup = BeautifulSoup(the_page)
for table in soup.findAll('table',{'id':'ctl00_Main_lvMyAccount_Table1'} ): 
        for link in table.findAll('a'): #search for links only in the table
                print link['href'] #get the href attribute

此外，您可以跳过父循环，因为指定的 id 只有一个匹配项:

soup = BeautifulSoup(the_page)
table = soup.find('table',{'id':'ctl00_Main_lvMyAccount_Table1'})
for link in table.findAll('a'): #search for links only in the table
                print link['href'] #get the href attribute

更新:注意到@DSM 所说的。修复了表格分配中缺少的引号。