Question

I am trying to find a link which contains http or // or \ and surround with a href tag once its found,does anyone have any ideas on how this can be done

 INput:-http://pastebin.com/p9H8GQt4

sanity_results = sanity_results.replace('\n','<br>\n')
return sanity_results

def main ():
resultslis=[]
xmlfile = open('results.xml','r')
contents = xmlfile.read()
testresults=getsanityresults(contents)
#print testresults
for line in testresults:
    #print line
    line = line.strip()
    #print line
    line = re.sub(r'(http://[^\s]+|//[^\s]+|\\\\[^\s]+)', r'<a href="\1">\1</a>', line)
    print line       
    resultslis.append(line)
print resultslis

if __name__ == '__main__':
main()

Answer 1

You might want to use regular expressions for this:

line = re.sub(r'(http://[^\s]+)', r'<a href>\1</a>', line)

That just handles the http:// case. To handle all three, just do this:

line = re.sub(r'(http://[^\s]+|//[^\s]+|\\\\[^\s]+)', r'<a href>\1</a>', line)

Play with that regexp in the console to make sure it does what you want, but it seems to do what you asked for with your posted input data. As I mentioned in a comment, in general, you need to figure out what delimiters can end a link if you want to auto-link text.

Meanwhile, are you sure the problem spec is correct? Normally, you don’t want this:

<a href>http://foo/bar</a>

… but this:

<a href="http://foo/bar">http://foo/bar</a>

To get that, just change the sub replacement expression to r'<a href="\1">\1</a>'.

You could also write the whole thing with string functions, but for anything but simple cases, that actually turns out to be a lot harder than learning regexps. For example, the equivalent of the above one-liner is something like this:

index = 0
while index is not None:
    index = min(line.find(pattern, index) for pattern in ('http:', '//', '\\\\'))
    if index == -1:
        break
    space = line.find(' ', index)
    if space == -1:
        space = None
    line = line[:index] + '<a href>' + line[index:space] + '</a>' + line[space:]
    index = space

Except that I’m willing to bet I’ve got at least one obvious fencepost error in there, and likely at least one subtle bug with possible overlapping patterns, and so on.