提取没有任何类或div的HTML源代码(python硒) [英] Extract HTML source code without any class or div (python selenium)
问题描述
我的列表不包含任何唯一的div或类.
I have list that does not contain any unique div or class.
我想复制下面每一行的HTML源代码.我找不到该行的班级.
I want to copy the HTML source code for each row below. I cannot find the class for the row.
当我打开编辑HTML"时,我看到以下代码:
When I open the 'Edit HTML' I see the following code:
<tr style="font-size: 11px">
<td class="center"><a href="/countries/1"><img alt="" src="/assets/flags/flag_1-1db156e1884c1b3d5614b55996cf96cd38843b290c7c43bdd5abbdb944b4075c.gif"></a></td>
<td><a href="/employees/9526577">Bernard Aarslev</a></td>
<td><a href="/clubs/1200094">Kirslev FC</a></td>
<td align="right" style="padding-right: 5px;">69</td>
<td>Talentspejder</td>
<td>Talentspejder</td>
<td align="right" style="padding-right: 5px;">24.000 C</td>
<td class="center" style="width: 120px;">
<div class="relative">
<div id="stats9526577" style="z-index: 99; position: absolute; top: -80px; right: 80px; display: none;" class="dark"></div>
<img src="/assets/detaljer-c83987d00da87f2fa8810793cc815a1659249440edee3c0d084333bc69323384.gif" alt="stats" onmouseout="hide_stats(9526577);" onmouseover="view_stats(9526577, 14, 13, 4, 7, 10, 8, 6, 3);">
</div>
</td>
</tr>
如何编写正确的find_element_by_xpath
函数以使其正常工作?
How do I write the correct find_element_by_xpath
function to make this work?
推荐答案
您可以使用:
node = driver.find_element_by_xpath("//table[@class='stretch']//tr[@style][1]")
首先,我们查找包含特定@class属性的表元素.然后,我们查找该表的第一个tr元素,其中包含@style属性.
First we look for the table element containing a specific @class attribute. Then we look for the first tr element of this table containing a @style attribute.
更多详细信息,因为它已被否决.将前面的表达式与element.get_attribute('outerHTML')组合(以保留标签).所以:
EDIT : More details, since it was downvoted. Combine the preceding expression with element.get_attribute('outerHTML') (to keep the tags). So :
data = node.get_attribute('outerHTML')
如果表的所有行都需要此,则:
If you need this for all the lines of the table, then :
node = driver.find_elements_by_xpath("//table[@class='stretch']//tr[@style]")
for elem in node :
data = elem.get_attribute('outerHTML')
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