I'd like to transform what the user inputs into an textarea on a html page into a <p>-tagged output where each <p> is replacing new lines.
I'm trying with regular expressions but I can't get it to work. Will someone correct my expression?
String = "Hey, this is paragraph 1 \n and this is paragraph 2 \n and this will be paragraph 3"
Regex = r'(.+?)$'
It just results in Hey, this is paragraph 1 \n and this is paragraph 2 \n<p>and this will be paragraph 3</p>
I wouldn't use regular expressions for this, simply because you do not need it. Check this out:
text = "Hey, this is paragraph 1 \n and this is paragraph 2 \n and this will be paragraph 3"
html = ''
for line in text.split('\n'):
html += '<p>' + line + '</p>'
print html
To make it one line, because shorter is better, and clearer:
html = ''.join('<p>'+L+'</p>' for L in text.split('\n'))
I would do it this way:
s = "Hey, this is paragraph 1 \n and this is paragraph 2 \n and this will be paragraph 3"
"".join("<p>{0}</p>".format(row) for row in s.split('\n'))
You basically split your string into a list of lines. Then wrap each line with paragraph tags. In the end just join your lines.
Above answers relying on identifying '\n' do not work reliably. You need to use .splitlines(). I don't have enough rep to comment on the chosen answer, and when I edited the wiki, someone just reverted it. So can someone with more rep please fix it.
Text from a textarea may use '\r\n' as a new line character.
>> "1\r\n2".split('\n')
['1\r', '2']
'\r' alone is invalid inside a webpage, so using any of the above solutions produce ill formed web pages.
Luckily python provides a function to solve this. The answer that works reliably is:
html = ''.join('<p>'+L+'</p>' for L in text.splitlines())
You need to get rid of the anchor, $. Your regex is trying to match one or more of any non-newline characters, followed by the end of the string. You could use MULTILINE mode to make the anchors match at line boundaries, like so:
s1 = re.sub(r'(?m)^.+$', r'<p>\g<0></p>', s0)
...but this works just as well:
s1 = re.sub(r'.+', r'<p>\g<0></p>', s0)
The reluctant quantifier ( .+? ) wasn't doing anything useful either, but it didn't mess up the output like the anchor did.
Pretty easy >>
html='<p>'+s.replace("\n",'</p><p>')+'</p>'
Related
I'm trying to parse a webpage:However, I want to only focus on text within the div tag labelled "class='body conbody'". I want my program to look inside of this tag and output the text exactly like how they appear on the webpage.
Here is my code so far:
pres_file = directory + "\\" + pres_number + ".html"
with open(pres_file) as html_file:
soup = BeautifulSoup(html_file, 'html.parser')
desiredText = soup.find('div', class_='body conbody')
for para in desiredText.find_all('p'):
print(para.get_text())
The problem with my current code is that whenever I try to print the paragraphs, (a), (1), (2), (b), and (c) are always formatted with a lot of unnecessary newlines and additional spaces after it. However, I would like for it to output text that is equivalent to how it looks on the webpage. How can I change my code to accomplish this?
I want my program to look inside of this tag and output the text exactly like how they appear on the webpage.
The browser does a lot of processing to display a web page. This includes removing extra spaces. Additionally, the browser developer tools show a parsed version of the HTML as well as potential additions from dynamic JavaScript code.
On the other hand, you are opening a raw text file and get the text as it is, including any formatting such as indentation and line breaks. You will need to process this yourself to format it the way you want when you output it.
There are at least two things to look for:
Is the indentation tab or space characters? By default print() represents a tab as 8 spaces. You can either replace the tabs with spaces to reduce the indentation or you can use another output method that allows you to configure specify how to show tabs.
The strings themselves will include a newline character. But then print() also adds a line break. So either remove the newline character from each string or do print(para.get_text(), end='') to disable print adding another newline.
You can use strip() on strings, like para.get_text().strip(). This will remove any whitespaces before and after the string.
You can use either lstrip() and rstrip() to remove only the exceeding whitespaces from the left or right side of the string.
s = " \t \n\n something \t \n "
print(s.strip()) # 'something'
print(s.lstrip()) # 'something \t \n '
print(s.rstrip()) # ' \t \n\n something'
Would something like this work:
Strip left and right of the p
Indent the paragraph with 1em (so 1 times the font size)
Newline each paragraph
font_size = 16 # get the font size
for para in desiredText.find_all('p'):
print(font_size * " " + para.get_text().strip(' \t\n\r') + "\n")
When I print the string (in Python) coming from a website I scraped it from, it looks like this:
"His this
is
a sample
String"
It does not show the \n breaks. this is what I see in a Python interpreter.
And I want to convert it to HTML that will add in the line breaks. I was looking around and didn't see any libraries that do this out of the box.
I was thinking BeautifulSoup, but wasn't quite sure.
If you have a String that you have readed it from a file you can just replace \n to <br>, which is a line break in html, by doing:
my_string.replace('\n', '<br>')
You can use the python replace(...) method to replace all line breaks with the html version <br> and possibly surround the string in a paragraph tag <p>...</p>. Let's say the name of the variable with the text is text:
html = "<p>" + text.replace("\n", "<br>") + "</p>"
searching for this answer in found this, witch is likely better because it encodes all characters, at least for python 3
Python – Convert HTML Characters To Strings
# import html
import html
# Create Text
text = 'Γeeks for Γeeks'
# It Converts given text To String
print(html.unescape(text))
# It Converts given text to HTML Entities
print(html.escape(text))
I believe this will work
for line in text:
for char in line:
if char == "/n":
text.replace(char, "<br>")
Consider the text on this page. If you look at the source code, you'll see that the main text is presented exactly as in the page -- no HTML divisions or any other way to obviously find paragraphs/tabbed in sections.
Is there a way to automatically identify and remove sections that are tabbed in from the raw text?
One thing I notice is that when I encode the text as text = unicode(raw_text).encode("utf-8"), I can then see a bunch of \n's for line skips. But no \t's. (This might be not a useful direction to think, but just an idea).
Edit: The following works
text = unicode(raw_text).encode("utf-8")
y = [x for x in text.split("\n") if " " not in x]
final = " ".join(y)
Well, after looking at the page, they are 'tabbed' in with spaces rather than the tab character; looking for tabs would not be useful. It looks like the section is tabbed in with 5 spaces.
raw_text.replace(' ','')
To replace all occurances of 5 spaces...
from re import sub
...
raw_text = sub(r' .*\n', '', raw_text)
I have a Python script that generates some HTML. It does so using the Python markdown library. I'd like to stick the original Markdown text in a comment at the end of the HTML, where it will occasionally be useful for debugging purposes. I've tried just plunking the Markdown text after the end of the HTML, and it doesn't work for me (Firefox). So the way I imagine this working is that I run Markdown and then simply append the Markdown source, marked as a comment, after the HTML. However, HTML is apparently somewhat finicky about what it will allow in comments. The site htmlhelp.com gives the following advice after some discussion:
For this reason, use the following simple rule to compose valid and accepted [portable] comments:
An HTML comment begins with "" and does not contain "--" or ">" anywhere in the comment.
(source)
So it looks like I need to do some escaping or something to get my bunch of markdown text into a form that HTML will accept as a comment. Is there an existing tool that will help me do this?
According to the w3:
Comments consist of the following parts, in exactly the following order:
- the comment start delimiter "<!--"
- text
- the comment end delimiter "-->"
The text part of comments has the following restrictions:
1. must not start with a ">" character
2. must not start with the string "->"
3. must not contain the string "--"
4. must not end with a "-" character
These are very simple rules. You could regex-enforce them, but they are so simple you don't even need that!
3 of the 4 conditions can be met with concatenation, and the other one with a simple replace(). All in all, it's a one-liner:
def html_comment(text):
return '<!-- ' + text.replace('--', '- - ') + ' -->'
Note the spaces.
Can't you just .replace it? Ultimately, you could replace those characters with anything, but substituting with escape codes probably won't make your comment any more readible than substituting with nothing.
commented = '<!-- %s -->' % markdown_text.replace('--', '').replace('>', '')
I read this thread about extracting url's from a string. https://stackoverflow.com/a/840014/326905
Really nice, i got all url's from a XML document containing http://www.blabla.com with
>>> s = '<link href="http://www.blabla.com/blah" />
<link href="http://www.blabla.com" />'
>>> re.findall(r'(https?://\S+)', s)
['http://www.blabla.com/blah"', 'http://www.blabla.com"']
But i can't figure out, how to customize the regex to omit the double qoute at the end of the url.
First i thought that this is the clue
re.findall(r'(https?://\S+\")', s)
or this
re.findall(r'(https?://\S+\Z")', s)
but it isn't.
Can somebody help me out and tell me how to omit the double quote at the end?
Btw. the questionmark after the "s" of https means "s" can occur or can not occur. Am i right?
>>>from lxml import html
>>>ht = html.fromstring(s)
>>>ht.xpath('//a/#href')
['http://www.blabla.com/blah', 'http://www.blabla.com']
You're already using a character class (albeit a shorthand version). I might suggest modifying the character class a bit, that way you don't need a lookahead. Simply add the quote as part of the character class:
re.findall(r'(https?://[^\s"]+)', s)
This still says "one or more characters not a whitespace," but has the addition of not including double quotes either. So the overall expression is "one or more character not a whitespace and not a double quote."
You want the double quotes to appear as a look-ahead:
re.findall(r'(https?://\S+)(?=\")', s)
This way they won't appear as part of the match. Also, yes the ? means the character is optional.
See example here: http://regexr.com?347nk
I used to extract URLs from text through this piece of code:
url_rgx = re.compile(ur'(?i)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:\'".,<>?\xab\xbb\u201c\u201d\u2018\u2019]))')
# convert string to lower case
text = text.lower()
matches = re.findall(url_rgx, text)
# patch the 'http://' part if it is missed
urls = ['http://%s'%url[0] if not url[0].startswith('http') else url[0] for url in matches]
print urls
It works great!
Thanks. I just read this https://stackoverflow.com/a/13057368/326905
and checked out this which is also working.
re.findall(r'"(https?://\S+)"', urls)