Xmlstarlet sel -t -v - # parse the stream with XPath expression Xmlstarlet format -H - 2>/dev/null | # convert broken HTML to HTML ^M is Control+v Enter xmlstarlet: curl -Ls | Or a xpath & network aware tool like xidel or saxon-lint: xidel -se 51K Announcement: We just launched Online Unicode Tools a collection of browser-based Unicode utilities. (Written by Andy Lester, the author of ack and more.) Just paste your text in the form below, press the Extract Links button, and you'll get a list of all links found in the text. This comes with the package www-mechanize-perl (Debian based distro). Instead, use a proper parser: mech-dump mech-dump -links -absolute -agent-alias='Linux Mozilla' Parsing HTML with regex is a regular discussion: this is a bad idea. I am in no way recommending regex for parsing HTML, unless you know what you're doing or have very limited needs (eg: only want links), like in this case. That could be done using webread to retrieve data from the webpage and regexp to extract all the hyperlinks in the page by parsing through the retrieved. P.S.(2): You should, if you want to make use of a better way to parse HTML, instead of using my solution above (which doesn't handle every corner case, given the lack of an HTML example/sample) use the better answer from Gilles Quenot that is more fit for general (eg: complete) and more optimized support of HTML syntaxes. P.S.: This may or may not be obvious, but this also doesn't take into account links/URLs that are generated dynamically (eg: PHP, JS, etc) since curl mostly works on static links. Given you can't/don't show an example of said structure or the webpage itself, it is difficult to make an answer that works on it, unless more HTML knowledge was involved. Thus certain knowledge of the webpage structure or HTML is required. Lastly, this does not take into account every possible way a link is displayed. If you want to remove that, make/use sed or something else like so:Ĭurl -f -L URL | grep -Eo "https?://\S+?\"" | sed 's/&.*//' This also doesn't "clean" whatever won't be part of the link (eg: a "&" character, etc).I don't recall where I saw this, but it should appear on certain sites under certain/particular HTML tags.ĮDIT: Gilles Quenot kindly provided a solution for what I wrongly described as "half-link" (the correct term being relative link):
#EXTRACT ALL LINKS FROM PAGE FULL#
This does not take into account links that aren't "full" or basically are what I call "half a link", where only a part of the full link is shown.Or curl -f -L URL | grep -Eo '"(http|https)://*"'
A nice tip is to add the &num100 parameter to the URL to force Google into showing 100 results per page. Step 1: Search for a Google term that you want to extract links. We want to extract all external links from a Google search result. This should do it: curl -f -L URL | grep -Eo "https?://\S+?\"" This is a step-by-step example with the Google results page. from import ByĮlems = driver.find_elements(by=By.XPATH, = Įlems2 = driver.find_elements(by=By.Warning: Using regex for parsing HTML in most cases (if not all) is bad, so proceed at your own discretion. If duplicates are OK, one liner list comprehension can be used. If (l not in href_links2) & (l is not None): from import ByĮlems = driver.find_elements(by=By.XPATH, = driver.find_elements(by=By.TAG_NAME, value="a") Regular expressions can only be used to extract the link from the text if it is displayed inside the javascript in clear text. Both are not needed.īy.XPATH IMO is the easiest as it does not return a seemingly useless None value like By.TAG_NAME does. One for By.XPATH and the other, By.TAG_NAME. The current method is to use find_elements() with the By class. All of the accepted answers using Selenium's driver.find_elements_by_*** no longer work with Selenium 4.
0 kommentar(er)
