从引用到这里,How to get link from elements with Selenium and Python
我试过:
for a in driver.find_elements_by_xpath('.//span[contains(text(), "SPE8ED-21-Q-1288")]/a'):
print(a.get_attribute('href'))
结果表明:
https://www.dibbs.bsm.dla.mil/Default.aspx
https://www.dibbs.bsm.dla.mil/Solicitations/Default.aspx
https://www.dibbs.bsm.dla.mil/RFQ/Default.aspx
https://www.dibbs.bsm.dla.mil/RFQ/RFQNsn.aspx?value=8145014862449&category=post&Scope=
https://dibbs2.bsm.dla.mil/Downloads/RFQ/8/SPE8ED21Q1288.PDF
https://www.dibbs.bsm.dla.mil/rfq/rfqrec.aspx?sn=SPE8ED21Q1288
https://www.dibbs.bsm.dla.mil/RA/Quote/QuoteFrm.aspx?sn=SPE8ED21Q1288
我只对只包含Downloads
关键字的链接感兴趣
在这种情况下,
https://dibbs2.bsm.dla.mil/Downloads/RFQ/8/SPE8ED21Q1288.PDF
有没有办法做这个过滤
您可以像这样过滤下载关键字
代码说明:
我们得到一个
list of web elements
,然后(通过迭代)查找每个web元素,并查找每个web元素的属性href
,如果找到Downloads
,则打印href
试着这样做:
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