BeautifulSoup:“ResultSet”对象没有“find_all”属性

Question

from bs4 import BeautifulSoup
import urllib2

url = 'http://www.data.jma.go.jp/obd/stats/etrn/view/monthly_s3_en.php?block_no=47401&view=1'
html = urllib2.urlopen(url).read()        
soup = BeautifulSoup(html)
table = soup.find_all('table', class_='data2_s')
rows = table.find_all('tr')
print rows

 rows = table.find_all('tr')
AttributeError: 'ResultSet' object has no attribute 'find_all'

我想把表抓成CSV文件。怎么往前走？

这是表：

[<table class="data2_s"><caption class="m">WAKKANAI\xa0\xa0\xa0WMO Station ID:47401\xa0Lat\xa045<sup>o</sup>24.9'N\xa0\xa0Lon\xa0141<sup>o</sup>40.7'E</caption><tr><th scope="col">Year</th><th scope="col">Jan</th><th scope="col">Feb</th><th scope="col">Mar</th><th scope="col">Apr</th><th scope="col">May</th><th scope="col">Jun</th><th scope="col">Jul</th><th scope="col">Aug</th><th scope="col">Sep</th><th scope="col">Oct</th><th scope="col">Nov</th><th scope="col">Dec</th><th scope="col">Annual</th></tr><tr class="mtx" style="text-align:right;"><td style="text-align:center">1938</td><td class="data_0_0_0_0">-5.2</td><td class="data_0_0_0_0">-4.9</td><td class="data_0_0_0_0">-0.6</td><td class="data_0_0_0_0">4.7</td><td class="data_0_0_0_0">9.5</td><td class="data_0_0_0_0">11.6</td><td class="data_0_0_0_0">17.9</td><td class="data_0_0_0_0">22.2</td><td class="data_0_0_0_0">16.5</td><td class="data_0_0_0_0">10.7</td><td class="data_0_0_0_0">3.3</td><td class="data_0_0_0_0">-4.7</td><td class="data_0_0_0_0">6.8</td></tr>\n<tr class="mtx" style="text-align:right;"><td style="text-align:center">1939</td><td class="data_0_0_0_0">-7.5</td><td class="data_0_0_0_0">-6.6</td><td class="data_0_0_0_0">-1.4</td><td]

Answer 1

试试这个：

rows = table[0].find_all('tr')

由于find_all似乎返回了Pythonlist，因此您试图在list上调用find_all，而该列表没有这样的方法。

将结果列表（行）转换为CSV格式。。那要看情况了。如果您只想将其显示在屏幕上，可以执行以下操作：

','.join(rows)

如果你想把它放在一个文件中，你需要打开一个文件，然后把上面的一行写进去。但也有一些Python模块处理创建CSV内容。这是一个完全不同的问题。