NSRegularExpression to match a certain pattern - ios

Question

Backgroud:
I have a response like so (please note the new lines):
HTTP/1.1 200 OK
CACHE-CONTROL: max-age = 180
EXT:
LOCATION: http://172.16.16.16:80/upnp.jsp
SERVER: Linux/2.6.32.24 UPNP/1.0 ZD3025/9.5.1.0
ST: upnp:rootdevice
USN: uuid:6e4bb543-fff6-4384-a4be-::upnp:rootdevice
I would like to match the Server line which is:
Linux/2.6.32.24 UPNP/1.0 ZD3025/9.5.1.0
Implementation:
I don't care about any character in the above given response EXCEPT for the characters Linux, UPNP and ZD (in that order).
So I am using .* to match any number of characters, numbers, special chars.
-(void) regexCompareForUPnP:(NSString *) string {
NSError * err = nil;
//Building expression
NSString *expression = #"^.*Linux.*UPNP.*ZD.*$";
NSRegularExpression *regex = [[NSRegularExpression alloc] initWithPattern:expression options:0 error:&err];
NSUInteger numberOfMatches = [regex numberOfMatchesInString:string
options:0
range:NSMakeRange(0, [string length])];
NSLog(#"Matches found: %d", numberOfMatches);
}
So far, Matches found: 0.
Question:
So, in short does this:
NSString *expression = #"^.*Linux.*UPNP.*ZD.*$";
Match this:
Linux/2.6.32.24 UPNP/1.0 ZD3025/9.5.1.0
EDIT:
Thank you omz that worked perfectly.
Now I am trying to match the upnp.jsp (in the LOCATION line) alongwith what I was trying to match before. So I tried this:
NSString *expression = #"^.*upnp.jsp\n.*Linux.*UPNP.ZD.$";
That does not work. Appreciate your help..

Answer 1

I'm not familiar with this regex library. Based on what you're trying to match, however, it may be to do with newlines between OS's and individual programs.
Try this: *expression = #"upnp.jsp\r?\n[^\n]*?Linux[^\n]*?UPNP[^\n]*?ZD"
A few other changes made from what you're trying as well;
I've replaced .* with [^\n]*? for two reasons:
I'm not sure how this library deals with .*, but using that is typically inefficient in comparison to something more speicific such as looking for a non-greedy set of characters that aren't a new line. The difference between .* and .*? depends on the regex engine. In most regex engines I work with at least, .* is by default greedy; i.e. it will capture as much data as it can while still having the expression find a match. Using the non-greedy version, .*?, means that the regex will capture only as much as it needs to to satisfy the whole expression. This is particularly evident in an example I'll expand on at the end.
I removed the surrounding ^$ because when they are appended and prepended with .* they only work towards increasing processing time (unless you were using matching groups, which you aren't). If you were doing ^(.*)upnp or ^(.*upnp) and then using the contents of the matched group, there could be purpose in having the initial ^.*.
As for greedy/non-greedy, the difference becomes evident when trying to match data between a set of double quotes ". For the sake of simplicity, I won't deal with having escaped double quotes in the middle of a string you also want to capture.
Given the string: I said, "Hi." She responded with, "Hello!"
Using the greedy regex "(.*)", the matched group would be Hi." She responded with, "Hello!
Using the non-greedy regex "(.*?)", the matched group would be Hi.