I am using mapping template in AWS API Gateway methods Integration response.
I found that it adds unwanted whitespaces to the return string. I have read Velocity references and tried all trimming and replacing I could think of. However there still seems to be 3 whitespaces before text.
My actual code is more complicated but this is a short example of the problem.
#set ($test = "Foo bar")
##return that
$test.toString().trim()
Most probably, the extra spaces come from elsewhere in the template.
First, you can try to add some extra characters to understand where the spaces come from:
<#set ($test = "Foo bar")>
####return that
<$test.toString().trim()>
and look where the spaces do fall.
Also, you can try to comment end lines to see if it changes something:
#set ($test = "Foo bar")##
##return that
$test.toString().trim()##
Related
I'm using the following regex to find URLs in a text file:
/http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+/
It outputs the following:
http://rda.ucar.edu/datasets/ds117.0/.
http://rda.ucar.edu/datasets/ds111.1/.
http://www.discover-earth.org/index.html).
http://community.eosdis.nasa.gov/measures/).
Ideally they would print out this:
http://rda.ucar.edu/datasets/ds117.0/
http://rda.ucar.edu/datasets/ds111.1/
http://www.discover-earth.org/index.html
http://community.eosdis.nasa.gov/measures/
Any ideas on how I should tweak my regex?
Thank you in advance!
UPDATE - Example of the text would be:
this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/).
This will trim your output containing trail characters, ) .
import re
regx= re.compile(r'(?m)[\.\)]+$')
print(regx.sub('', your_output))
And this regex seems workable to extract URL from your original sample text.
https?:[\S]*\/(?:\w+(?:\.\w+)?)?
Demo,,, ( edited from https?:[\S]*\/)
Python script may be something like this
ss=""" this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/). """
regx= re.compile(r'https?:[\S]*\/(?:\w+(?:\.\w+)?)?')
for m in regx.findall(ss):
print(m)
So for the urls you have here:
https://regex101.com/r/uSlkcQ/4
Pattern explanation:
Protocols (e.g. https://)
^[A-Za-z]{3,9}:(?://)
Look for recurring .[-;:&=+\$,\w]+-class (www.sub.domain.com)
(?:[\-;:&=\+\$,\w]+\.?)+`
Look for recurring /[\-;:&=\+\$,\w\.]+ (/some.path/to/somewhere)
(?:\/[\-;:&=\+\$,\w\.]+)+
Now, for your special case: ensure that the last character is not a dot or a parenthesis, using negative lookahead
(?!\.|\)).
The full pattern is then
^[A-Za-z]{3,9}:(?://)(?:[\-;:&=\+\$,\w]+\.?)+(?:\/[\-;:&=\+\$,\w\.]+)+(?!\.|\)).
There are a few things to improve or change in your existing regex to allow this to work:
http[s]? can be changed to https?. They're identical. No use putting s in its own character class
[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),] You can shorten this entire thing and combine character classes instead of using | between them. This not only improves performance, but also allows you to combine certain ranges into existing character class tokens. Simplifying this, we get [a-zA-Z0-9$-_#.&+!*\(\),]
We can go one step further: a-zA-Z0-9_ is the same as \w. So we can replace those in the character class to get [\w$-#.&+!*\(\),]
In the original regex we have $-_. This creates a range so it actually inclues everything between $ and _ on the ASCII table. This will cause unwanted characters to be matched: $%&'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_. There are a few options to fix this:
[-\w$#.&+!*\(\),] Place - at the start of the character class
[\w$#.&+!*\(\),-] Place - at the end of the character class
[\w$\-#.&+!*\(\),] Escape - such that you have \- instead
You don't need to escape ( and ) in the character class: [\w$#.&+!*(),-]
[0-9a-fA-F][0-9a-fA-F] You don't need to specify [0-9a-fA-F] twice. Just use a quantifier like so: [0-9a-fA-F]{2}
(?:%[0-9a-fA-F][0-9a-fA-F]) The non-capture group isn't actually needed here, so we can drop it (it adds another step that the regex engine needs to perform, which is unnecessary)
So the result of just simplifying your existing regex is the following:
https?://(?:[$\w#.&+!*(),-]|%[0-9a-fA-F]{2})+
Now you'll notice it doesn't match / so we need to add that to the character class. Your regex was matching this originally because it has an improper range $-_.
https?://(?:[$\w#.&+!*(),/-]|%[0-9a-fA-F]{2})+
Unfortunately, even with this change, it'll still match ). at the end. That's because your regex isn't told to stop matching after /. Even implementing this will now cause it to not match file names like index.html. So a better solution is needed. If you give me a couple of days, I'm working on a fully functional RFC-compliant regex that matches URLs. I figured, in the meantime, I would at least explain why your regex isn't working as you'd expect it to.
Thanks all for the responses. A coworker ended up helping me with it. Here is the solution:
des_links = re.findall('http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+', des)
for i in des_links:
tmps = "/".join(i.split('/')[0:-1])
print(tmps)
I have to split URIs on the second portion:
/directory/this-part/blah
The issue I'm facing is that I have 2 URIs which logically need to be one
/directory/house-&-home/blah
/directory/house-%26-home/blah
This comes back as:
house-&-home and house-%26-home
So logically I need a regex to retrieve the second portion but also remove everything between the hyphens.
I have this, so far:
/[^(/;\?)]*/([^(/;\?)]*).*
(?<=directory\/)(.+?)(?=\/)
Does this solve your issue? This returns:
house-&-home and house-%26-home
Here is a demo
If you want to get the result:
house--home
then you should use a replace method. Because I am not sure what language you are using, I will give my example in java:
String regex = (?<=directory\/)(.+?)(?=\/);
String str = "/directory/house-&-home/blah"
Pattern.compile(regex).matcher(str).replaceAll("\&", "");
This replace method allows you to replace a certain pattern ( The & symbol ) with nothing ""
I'm trying to parse a gitolite.conf file, which is a whitespace-oriented conf file with a few regexes. The worst problem is that some options might appear anywhere:
#staff = dilbert alice # line 1
#projects = foo bar # line 2
repo #projects baz # line 3
RW+ = #staff # line 4
- master = ashok # line 5
RW = ashok # line 6
R = wally # line 7
config hooks.emailprefix = '[%GL_REPO] ' # line 8
Check the "master" attribute. Some repos have them, others do not. It's a real pain.
This answer assumes a goal of extracting key/value pairs into capturing groups, where key consists of contiguous non-whitespace before = and value includes everything after = but before #, trimmed of leading/trailing whitespace.
Basic version
([^\s]+)\s*=\s*((?:\s*[^\s#]+)*)
More advanced version
The regex above doesn't handle quoted strings very well (e.g. prefix = ' Quoted with # and leading/trailing whitespace '). Regex isn't great at this kind of thing but simple cases can be handled as follows:
([^\s]+)\s*=\s*('[^']*'|"[^"]*"|(?:(?:\s*[^\s#]+)*))
Here's the demo if you need to see what is captured and play around with it more: Debuggex Demo
First, you should know that this isn't entirely possible with Regex. Regex is a great tool for parsing regular languages (including some types of configuration files), but as soon as you get into "Well, this line is actually a header line and we need all lines under it, and some lines might have this token, and others might not", it gets quite messy. I'm not saying it's impossible, but you're going to waste a lot of time debugging your Regex pattern instead of just writing a parser in whatever language you're using this with.
Second, if you're going to ask a quesiton about Regex, it is always helpful to know what you want out of the expression. Do you want to tokenize everything, do you only want the configuration keys, do you only want the comments?
That being said, I took my best guess, here's an expression to get you started:
^(?:([^=#]+?)\s.?=?\s.?([^=#]+?)\s.?(?:#|$))
With this expression, please apply the g and m flags (global and multiline). In PCRE, this would look like:
/^(?:([^=#]+?)\s.?=?\s.?([^=#]+?)\s.?(?:#|$))/gm
There are two capture groups, one is whatever is before the = sign, and the other is whatever is after. If there is no = sign, the first capture group contains everything. Anything after "#" is ignored.
Here's a fiddle to demonstrate: http://www.rexfiddle.net/eQexbZU
I've searched around quite a bit now, but I can't get any suggestions to work in my situation. I've seen success with negative lookahead or lookaround, but I really don't understand it.
I wish to use RegExp to find URLs in blocks of text but ignore them when quoted. While not perfect yet I have the following to find URLs:
(https?\://)?(\w+\.)+\w{2,}(:[0-9])?\/?((/?\w+)+)?(\.\w+)?
I want it to match the following:
www.test.com:50/stuff
http://player.vimeo.com/video/63317960
odd.name.amazone.com/pizza
But not match:
"www.test.com:50/stuff
http://plAyerz.vimeo.com/video/63317960"
"odd.name.amazone.com/pizza"
Edit:
To clarify, I could be passing a full paragraph of text through the expression. Sample paragraph of what I'd like below:
I would like the following link to be found www.example.com. However this link should be ignored "www.example.com". It would be nice, but not required, to have "www.example.com and www.example.com" ignored as well.
A sample of a different one I have working below. language is php:
$articleEntry = "Hey guys! Check out this cool video on Vimeo: player.vimeo.com/video/63317960";
$pattern = array('/\n+/', '/(https?\:\/\/)?(player\.vimeo\.com\/video\/[0-9]+)/');
$replace = array('<br/><br/>',
'<iframe src="http://$2?color=40cc20" width="500" height="281" frameborder="0" webkitAllowFullScreen mozallowfullscreen allowFullScreen></iframe>');
$articleEntry = preg_replace($pattern,$replace,$articleEntry);
The result of the above will replace any new lines "\n" with a double break "" and will embed the Vimeo video by replacing the Vimeo address with an iframe and link.
I've found a solution!
(?=(([^"]+"){2})*[^"]*$)((https?:\/\/)?(\w+\.)+\w{2,}(:[0-9]+)?((\/\w+)+(\.\w+)?)?\/?)
The first part from (? to *$) what makes it work for me. I found this as an answer in java Regex - split but ignore text inside quotes? by https://stackoverflow.com/users/548225/anubhava
While I had read that question before, I had overlooked his answer because it wasn't the one that "solved" the question. I just changed the single quote to double quote and it works out for me.
add ^ and $ to your regex
^(https?\://)?(\w+\.)+\w{2,}(:[0-9])?\/?((/?\w+)+)?(\.\w+)?$
please notice you might need to escape the slashes after http (meaning https?\:\/\/)
update
if you want it to be case sensitive, you shouldn't use \w but [a-z]. the \w contains all letters and numbers, so you should be careful while using it.
I'm trying to search a field in a database to extract URLs. Sometimes there will be more than 1 URL in a field and I would like to extract those in to separate variables (or an array).
I know my regex isn't going to cover all possibilities. As long as I flag on anything that starts with http and ends with a space I'm ok.
The problem I'm having is that my efforts either seem to get only 1 URL per record or they get only 1 the last letter from each URL. I've tried a couple different techniques based on solutions other have posted but I haven't found a solution that works for me.
Sample input line:
Testing http://marko.co http://tester.net Just about anything else you'd like.
Output goal
$var[0] = http://marko.co
$var[1] = http://tester.net
First try:
if ( $status =~ m/http:(\S)+/g ) {
print "$&\n";
}
Output:
http://marko.co
Second try:
#statusurls = ($status =~ m/http:(\S)+/g);
print "#statusurls\n";
Output:
o t
I'm new to regex, but since I'm using the same regex for each attempt, I don't understand why it's returning such different results.
Thanks for any help you can offer.
I've looked at these posts and either didn't find what I was looking for or didn't understand how to implement it:
This one seemed the most promising (and it's where I got the 2nd attempt from, but it didn't return the whole URL, just the letter: How can I store regex captures in an array in Perl?
This has some great stuff in it. I'm curious if I need to look at the URL as a word since it's bookended by spaces: Regex Group in Perl: how to capture elements into array from regex group that matches unknown number of/multiple/variable occurrences from a string?
This one offers similar suggestions as the first two. How can I store captures from a Perl regular expression into separate variables?
Solution:
#statusurls = ($status =~ m/(http:\S+)/g);
print "#statusurls\n";
Thanks!
I think that you need to capture more than just one character. Try this regex instead:
m/http:(\S+)/g