<html>
<header><title>This is title</title></header>
<body>
Hello world
</body>
</html>
I have this above content in a file("/home/usr/hello.html" ) in my local disc. I want to display this file path link in my console, so that user can click it to open in browser. Can anybody suggest a way to do this in python.
import webbrowser
webbrowser.open_new(url)
Related
I am currently working on a code that downloads an excel file from a website. The respective file is actually hidden behind an Export button (see website: https://www.amundietf.co.uk/retail/product/view/LU1437018838). However, I have already identified the link behind which is the following: https://www.amundietf.co.uk/retail/ezaap/service/ExportDataProductSheet/Fwd/en-GB/718/LU1437018838/object/export/repartition?idFonctionnel=export-repartition-pie-etf&exportIndex=3&hasDisclaimer=1. Since the link does not directly guide to the file but rather executes some Java widget, I am not able to download the file via python. I have tried the folling code:
import re
import requests
link = 'https://www.amundietf.co.uk/retail/ezaap/service/ExportDataProductSheet/Fwd/en-GB/718/LU1437018838/object/export/repartition?idFonctionnel=export-repartition-pie-etf&exportIndex=3&hasDisclaimer=1'
r = requests.get(link, verify= False)
However, I am not able to connect to the file. Does somebody has an idea for doing this?
I would recommend using HTML:
<html lang=en>
<body>
Click here to download
</body>
</html>
In the href attribute to tag, you can put the path to your own excel file. I used an external link to an example file I found on google. To open in new tab, use target="_blank" as attribute to .
Hope it works!
I am trying to get the source code of a particular site with the help of Selenium with:
Python code:
driver.page_source
But it returns it after it has been encoded.
The raw file:
<html>
<head>
<title>AAAAAAAA</title>
</head>
<body>
</body>
When press 'View page source' inside Chrome, I saw the correct source raw without encoding.
How can this be achieved?
You can try using Javascript instead of Python builtin code to get the page source.
javascriptPageSource = driver.execute_script("return document.body.outerHTML;")
I am getting the error:
Refused to frame 'http://localhost:8888/' because an ancestor violates the following Content Security Policy directive: "frame-ancestors 'self'".
When trying to create an iframe from a local HTML file that contains an iframe of another local HTML file.
For example test-outer.html:
<!DOCTYPE html>
<html>
<body>
<iframe sandbox="allow-scripts" id="0" src="./test-inner.html" width="100" height="100" frameborder="0"></iframe>
</body>
</html>
With the Jupyter Notebook Python code:
from IPython.display import IFrame
IFrame(src='test-outer.html', width=200, height=200)
test-inner.html could contain anything.
test-outer.html loads perfectly on a separate browser tab. The issue is loading it inside Jupyter Notebook.
Although this is an old question, but to help new seekers.
You need to add headers in your jupyter_notebook_config.py
c.NotebookApp.tornado_settings={'headers': {'Content-Security-Policy': "frame-ancestors self https://my-website-url.com/jupyter "}}
https://my-website-url.com/jupyter is the url where you're loading jupyter notebooks in iFrame.
This works in Firefox and Chrome. Not tested in Safari.
I see that my selenium cannot execute codes except to launch Chrome.
I don't know why my selenium is not working. It just open the browser (Chrome) with the URL and then doing nothing even to maximize the window, not even inserting the form.
Is there anything wrong with my code:
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
import re, time, csv
driver = webdriver.Chrome("C:\\Users\\Ashraf%20Misran\\Installer\\chromedriver.exe")
driver.get("file:///C:/Users/Ashraf%20Misran/Devs/project-html/learning-html/selenium sandbox.html")
driver.maximize_window()
username = driver.find_element_by_xpath(".//input")
username.click()
username.send_keys("000200020002")
The page I opened is coded as below:
<!DOCTYPE html>
<html>
<head>
<title>Sandbox</title>
</head>
<body>
<form>
<input type="text" name="username">
</form>
</body>
</html>
I think the problem is with web-page, you are trying to open. Would suggest to try first with simple test, like Open Google page, enter something in search field. With this you will be able to verify, if you correctly implemented driver initialization.
Update: try to use this css selector: input[name='username'], if page is loaded correctly, then you have a problem with your web element selector.
I think, there is a problem with using relative xpath locator. Please try that one:
username = driver.findElement(By.xpath("//input"))
I need a url to be opened in a IE browser specifically.
I know the following code is wrong but I don't know what else to try.
How can I achieve that through python?
template
Open in IE
urls.py
url(r'^open-ie$', views.open_in_ie, name='ie'),
views.py
import webbrowser
def open_in_ie(request):
ie = webbrowser.get(webbrowser.iexplore)
return ie.open('https://some-link.com')
Again, I know this is wrong and it tries to open the ie browser at a server level. Any advices? Thank you!
Shot Answer: You can't.
Long Answer: If user is using IE to view your website you can open links in other browsers. But if user is using any other browser(firefox, chrome, etc.) all links will open in same browser, you can't access other browsers. So in your case answer is no, because you are trying to open IE from some other browser.
Here is the code to open another browser from IE if you are interested:
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html lang="en">
<head>
<title>HTA Test</title>
<hta:application applicationname="HTA Test" scroll="yes" singleinstance="yes">
<script type="text/javascript">
function openURL()
{
var shell = new ActiveXObject("WScript.Shell");
shell.run("http://www.google.com");
}
</script>
</head>
<body>
<input type="button" onclick="openURL()" value="Open Google">
</body>
</html>
Code from here