如何使用 Python 从远程 PHP 响应字符串中获取第一行 url？

Question

I want to extract a URL from a remote PHP response. My current code gets the link variable response, but how can I only a grab the first line of the response, which in this case is a URL?

URL to look in first line of PHP response in link variable:

http://test.awebsite.com/1.m3u8?token=454766879809809

Python code:

req = urllib2.Request('http://www.somesite.com/test.php')
req.add_header('User-Agent', 'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:19.0) Gecko/20100101 Firefox/19.0')
response = urllib2.urlopen(req)   
link = response.read()

Full PHP response using print link :

10:01:08 T:7688  NOTICE: 
    http://test.awebsite.com/1.m3u8?token=454766879809809 
    <!-- Start -->
    <script type="text/javascript" src="http://1.js"></script>
    <script type="text/javascript" src="http://1.js"></script>
    <script type="text/javascript" src="http://3.js"></script>
    <noscript><br><center><font color='#000000' face='Verdana' style='font-size: 11px; background-color:#FFFFFF'><a target='_blank' href='http://www.ads.com'><font color='#000000'>ads</font></a></font></center></noscript>
    <!-- End -->

Answer 1

You can try splitting the response by lines like this:

lines = link.split("\n")

And then get the first line:

anwser = lines[1] # 1st line is lines[0], 2nd is lines[1], 3rd is lines[2] etc...

Note that you should expect an exception to be raised if the substring you split at isn't in your string. One way you can check it like this:

first_line=""
if "\n" in link:
    first_line = link.split("\n")[1] # Gets the same result as before in one line...
else:
    print("Something went wrong...")
print(first_line)

Or with a try-except statement:

first_line=""
try:
    first_line = link.split("\n")[1]
except IndexError:
    print("Something went wrong...")
print(first_line)

I hope that answers your question!

Answer 2

_{I decided to clarify some things missed in ant0nisk's answer by editing it, but I ended up completely rewriting it, so I decided to post it as separate answer.}

You can try splitting the response by lines like this:

lines = link.splitlines()

And then get the first line:

anwser = lines[1] # 1st line is lines[0], 2nd is lines[1], 3rd is lines[2] etc...

Note that if link is made up from less than two lines, the line above raises raise IndexError exception, which (if unhandled) will terminate your program with message which will be meaningful for you, but might confuse non-programmers. If you for example want folks-friendly error message or have plan B what to do in case of one-line or empty request, you can handle the error with a try-except statement:

try:
    first_line = link.split("\n")[1]
except IndexError:
    print("Response is too short! (expected at least 2 lines)")
else: # Everything OK
    print(first_line)

Catching exception is expensive. Usually it is not a problem, because it happens when recovering from exceptional scenarios. But, if exception is raised often, so that handling it is not an exceptional scenario, using if instead of try-except might turn out to be faster.