我可以使用Selenium
将文件上传为
from selenium import webdriver
driver = webdriver.Chrome()
driver.get(r'https://www.example.com')
a = driver.find_element_by_xpath("//input[@type='file']")
a.send_keys(r'C:\abc.jpg')
b = driver.find_element_by_xpath("//output[@id='result']").text
上述代码效果很好。
现在,我想将Requests
用于同一个工作。我试图在网上搜索并实现代码。但我无法得到结果。
import requests
from lxml import html
a = requests.get(r'https://www.example.com')
tree = html.fromstring(a.text)
b = tree.xpath("//input[@type='file']")
b.append(r'C:\abc.jpg')#Is it correct?
html代码如下:
<div id="test" class="abc def ghi">
<div class="xyz def ghi">
Drag or<br class="def ghi">Click to input file
</div>
<div class="pqr def ghi">
Upload file
</div>
<label id="select-file" for="input" class="def ghi"></label>
<input type="file" id="input" hidden="" class="def ghi">
</div>
代码中没有表格。如何解决问题?
答案 0 :(得分:1)
您需要使用files
参数。另外,表单的action
属性可以指定不同的上传网址,因此最好将其检出。
以下是完整示例:
import requests
from lxml import html
host = r'https://www.example.com'
url = '/'
filename = r'C:\abc.jpg'
req = requests.get(host + url)
tree = html.fromstring(req.text)
field = tree.xpath("//input[@type='file']")
form = next(f[0].iterancestors("form"), None)
action = form.attrib['action'] if form else url
requests.post(host + action,
files={filename: (open(filename, 'rb'), 'image/jpg')})