1.善于利用soup节点的parent属性
比如对于已经得到了如下html代码:
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< td style = "padding-left:0" width = "60%" >< label >November</ label >
< input type = "Hidden" id = "cboMonth1" name = "cboMonth1" value = "11" >
</ td >< td style = "padding-right:0;" width = "40%" >
< label >2012</ label >
< input type = "Hidden" id = "cboYear1" name = "cboYear1" value = "2012" >
</ td >
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的soup变量eachMonthHeader了。
想要提取其中的
Month的label的值:November
和Year的label的值:2012
最简单,也是最省事的办法是,直接搜两个label,然后肯定会找到这两个label,然后分别对应着Month和Year的label,然后获得对应的string即可:
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foundTwoLabel = eachMonthHeader.findAll( "label" );
print "foundTwoLabel=" ,foundTwoLabel;
monthLabel = foundTwoLabel[ 0 ];
yearLabel = foundTwoLabel[ 1 ];
monthStr = monthLabel.string;
yearStr = yearLabel.string;
print "monthStr=" ,monthStr; # monthStr= November
print "yearStr=" ,yearStr; # yearStr= 2012
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但是很明显,这样的逻辑性很不好,而且万一处理多个这样的soup变量,而且两者的顺便颠倒了,那么结果也就错误了。
此时,可以考虑利用soup变量的parent属性,从一个soup变量本身,获得其上一级的soup变量。
示例代码如下:
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# <td style="padding-left:0" width="60%"><label>November</label>
# <input type="Hidden" id="cboMonth1" name="cboMonth1" value="11">
# </td><td style="padding-right:0;" width="40%">
# <label>2012</label>
# <input type="Hidden" id="cboYear1" name="cboYear1" value="2012">
# </td>
foundCboMonth = eachMonthHeader.find( "input" , { "id" :re. compile ( "cboMonth\d+" )});
#print "foundCboMonth=",foundCboMonth;
tdMonth = foundCboMonth.parent;
#print "tdMonth=",tdMonth;
tdMonthLabel = tdMonth.label;
#print "tdMonthLabel=",tdMonthLabel;
monthStr = tdMonthLabel.string;
print "monthStr=" ,monthStr;
foundCboYear = eachMonthHeader.find( "input" , { "id" :re. compile ( "cboYear\d+" )});
#print "foundCboYear=",foundCboYear;
tdYear = foundCboYear.parent;
#print "tdYear=",tdYear;
tdYearLabel = tdYear.label;
#print "tdYearLabel=",tdYearLabel;
yearStr = tdYearLabel.string;
print "yearStr=" ,yearStr;
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我们再来看一个例子:
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from BeautifulSoup import BeautifulSoup
doc = [ '<html><head><title>Page title</title></head>' ,
'<body><p id="firstpara" align="center">This is paragraph <b>one</b>.' ,
'<p id="secondpara" align="blah">This is paragraph <b>two</b>.' ,
'</html>' ]
soup = BeautifulSoup(''.join(doc))
print soup.prettify()
# <html>
# <head>
# <title>
# Page title
# </title>
# </head>
# <body>
# <p id="firstpara" align="center">
# This is paragraph
# <b>
# one
# </b>
# .
# </p>
# <p id="secondpara" align="blah">
# This is paragraph
# <b>
# two
# </b>
# .
# </p>
# </body>
# </html>
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这个例子中,<HEAD> Tag的parent是<HTML> Tag. <HTML> Tag 的parent是BeautifulSoup 剖析对象自己。 剖析对象的parent是None. 利用parent,你可以向前遍历剖析树。
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soup.head.parent.name
# u'html'
soup.head.parent.parent.__class__.__name__
# 'BeautifulSoup'
soup.parent = = None
# True
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2.当解析非UTF-8或ASCII编码类型的HTML时,需要指定对应的字符编码
当html为ASCII或UTF-8编码时,可以不指定html字符编码,便可正确解析html为对应的soup:
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#这里respHtml是ASCII或UTF-8编码,此时可以不指定编码类型,即可正确解析出对应的soup
soup = BeautifulSoup(respHtml);
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当html为其他类型编码,比如GB2312的话,则需要指定相应的字符编码,BeautifulSoup才能正确解析出对应的soup:
比如:
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#此处respHtml是GB2312编码的,所以要指定该编码类型,BeautifulSoup才能解析出对应的soup
htmlCharset = "GB2312" ;
soup = BeautifulSoup(respHtml, fromEncoding = htmlCharset);
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