我有一个HTML文件(来自Newegg),他们的HTML组织如下.其规格表中的所有数据均为" desc ",而每个部分的标题均为" 名称".'以下是来自Newegg页面的两个数据示例.
<tr>
<td class="name">Brand</td>
<td class="desc">Intel</td>
</tr>
<tr>
<td class="name">Series</td>
<td class="desc">Core i5</td>
</tr>
<tr>
<td class="name">Cores</td>
<td class="desc">4</td>
</tr>
<tr>
<td class="name">Socket</td>
<td class="desc">LGA 1156</td>
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<tr>
<td class="name">Brand</td>
<td class="desc">AMD</td>
</tr>
<tr>
<td class="name">Series</td>
<td class="desc">Phenom II X4</td>
</tr>
<tr>
<td class="name">Cores</td>
<td class="desc">4</td>
</tr>
<tr>
<td class="name">Socket</td>
<td class="desc">Socket AM3</td>
</tr>
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最后,我希望有一个CPU(已经设置好)的类,它包含Brand,Series,Cores和Socket类型,用于存储每个数据.这是我能想到的唯一方法:
if(parsedDocument.xpath(tr/td[@class="name"])=='Brand'):
CPU.brand = parsedDocument.xpath(tr/td[@class="name"]/nextsibling?).text
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并为其余的值执行此操作.我如何完成nextsibling并且有更简单的方法吗?
我的朋友和我都很难过.在这两个代码块中,为什么第一个内循环比第二个内循环快?这是某种JVM优化吗?
public class Test {
public static void main(String[] args) {
int[] arr = new int[100000000];
arr[99999999] = 1;
long t1, t2, t3;
for (int ndx = 0; ndx < 10; ndx++) {
t1 = System.currentTimeMillis();
for (int i = 0; i < arr.length; i++)
if (0 < arr[i])
System.out.print("");
t2 = System.currentTimeMillis();
for (int i = 0; i < arr.length; i++)
if (arr[i] > 0)
System.out.print("");
t3 = System.currentTimeMillis();
System.out.println(t2 - t1 +" "+(t3 - t2));
}
}
} …Run Code Online (Sandbox Code Playgroud) 是否有可能获得Math.random()的Big O估计值?
我正在编写一个简单的Python CGI脚本来抓取网页并在Web浏览器中显示HTML文件(就像代理一样).这是脚本:
#!/usr/bin/env python3.0
import urllib.request
site = "http://reddit.com/"
site = urllib.request.urlopen(site)
site = site.read()
site = site.decode('utf8')
print("Content-type: text/html\n\n")
print(site)
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从命令行运行时,此脚本可以正常工作,但是当它通过Web浏览器查看时,它会显示一个空白页面.这是我在Apache的error_log中得到的错误:
Traceback (most recent call last):
File "/home/public/projects/proxy/script.cgi", line 11, in <module>
print(site)
File "/usr/local/lib/python3.0/io.py", line 1491, in write
b = encoder.encode(s)
File "/usr/local/lib/python3.0/encodings/ascii.py", line 22, in encode
return codecs.ascii_encode(input, self.errors)[0]
UnicodeEncodeError: 'ascii' codec can't encode character '\u2019' in position 33777: ordinal not in range(128)
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