A string have data with semicolons now i want to remove all the data within the 2 semicolons and leave the rest as it is. I am using perl regex to remove the unwanted data from the string:
String :
$val="Data;test is here ;&data=1dffvdviofv;&dt&;&data=343";
Now we want to remove all the data between each semicolons ,throughout the string :
$val=~s/(.*)(\;.*\;)(.*)$/$1$3/g;
But this is not working for me. Final out should be like below :
Data &data=1dffvdviofv&data=343
One of the problems is that .* is greedy, that is, it will consume as much as it can. You can make it non-greedy by writing .*?, but that alone won't fix your regex since you've anchored it to the end of the string with $. Personally I don't think there is a need for the capture groups, you can just write
$val =~ s/;.*?;//g;
I'm assuming that the extra space in your expected output (Data &data...) is a typo.
You might also want to consider using a proper parser for whatever data format this is.
Related
I have $stringF. Contained within $stringF is the following (the string is all one line, not word-wrapped as below):
http://news.google.com/news/url?sa=t&fd=R&ct2=us&usg=
AFQjCNHWQk0M4bZi9xYO4OY4ZiDqYVt2SA&clid=
c3a7d30bb8a4878e06b80cf16b898331&cid=52779892300270&ei=
H4IAW6CbK5WGhQH7s5SQAg&url=https://abcnews.
go.com/Lifestyle/wireStory/latest-royal-wedding-thousands-streets-windsor-55280649
I want to locate that string and make it look like this:
https://abcnews.go.com/Lifestyle/wireStory/latest-royal-
wedding-thousands-streets-windsor-55280649
Basically I need to use preg_replace to find the following string:
http://news.google.com/news/url?sa= ***SOME UNKNOWN CONTENT*** &url=http
and replace it with the following string:
http
I'm a little rusty with my php, and even rustier with regular expressions, so I'm struggling to figure this one out. My code looks like this:
$stringG = preg_replace('http://news.google.com/news/url?sa=*&url=http','http',$stringH);
except I know I can't use wildcards and I know I need to specially deal with the special characters (colon, forward slash, question mark, and sign, etc). Hoping someone can help me out here.
Also of note is that my $stringF contains multiple instances of such strings, so I need the preg_replace to be not greedy - otherwise it will replace a huge chunk of my string unnecessarily.
PHP has tools for that, no need to use a regex. parse_url to get the components of an url (scheme, host, path, anchor, query, ...) and parse_str to get the keys/values of the query part.
$url = 'http://news.google.com/news/url?sa=t&fd=R&ct2=us&usg=AFQjCNHWQk0M4bZi9xYO4OY4ZiDqYVt2SA&clid=c3a7d30bb8a4878e06b80cf16b898331&ci=52779892300270&ei=H4IAW6CbK5WGhQH7s5SQAg&url=https://abcnews.go.com/Lifestyle/wireStory/latest-royal-wedding-thousands-streets-windsor-55280649';
parse_str(parse_url($url, PHP_URL_QUERY), $arr);
echo $arr['url'];
I am still learning perl and have all most got a program written. My question, as simple as it may be, is if I want to hardcode a string to a field would the below do that? Thank you :).
$out[45]="VUS";
In the other lines I use the below to define the values that are passed into the `$[out], but the one in question is hardcoded and the others come from a split.
my #vals = split/\t/; # this splits the line at tabs
my #mutations=split/,/,$vals[9]; # splits on comma to create an array of mutations
my ($gene,$transcript,$exon,$coding,$aa);
for (#mutations)
{
($gene,$transcript,$exon,$coding,$aa) = split/\:/; # this takes col AB and splits it at colons
grep {$transcript eq $_} keys %nms or next;
}
my #out=($.,#colsleft,$_,#colsright);
$out[2]=$gene;
$out[3]=$nms{$transcript};
$out[4]=$transcript;
$out[15]=$coding;
$out[17]=$aa;
Your line of code: $out[45]="VUS"; is correct in that it is defining that 46th element of the array #out to the string, "VUS". I am trying to understand from your code, however why you would want to do that? Usually, it is better practice to not hardcode if at all possible. You want to make it your goal to make your program as dynamic as possible.
I have an array that have some symbols that I want to remove and even thought I find a solution, I will like to know if this is the right way because I'm afraid if I use it with array will remove the character that I might need on future arrays.
Here is an example item on my array:
$string1='22 | logging monitor informational';
so I try the following:
$string1=~ s/\s{6}\|(?=\s{6})//;
So my output is:
22 logging monitor informational
Is the other way that best match "|". I just want to remove the pipe character.
Thanks in advance
"I want to remove just the pipe character."
OK, then do this:
$string1 =~ s/\|//;
This will remove the first pipe character in the string. (You said in another comment that you don't want to remove any additional pipe characters.) If that's not what you want, then I'd suggest telling us exactly what you do want. We can't read minds, you know.
In the mean time, I'd also strongly recommend reading the Perl regular expressions tutorial.
How to use matlab regexprep , for multiple expression and replacements?
file='http:xxx/sys/tags/Rel/total';
I want to replace 'sys' with sys1 and 'total' with 'total1'. For a single expression a replacement it works like this:
strrep(file,'sys', 'sys1')
and want to have like
strrep(file,'sys','sys1','total','total1') .
I know this doesn't work for strrep
Why not just issue the command twice?
file = 'http:xxx/sys/tags/Rel/total';
file = strrep(file,'sys','sys1')
strrep(file,'total','total1')
To solve it you need substitute functionality with regex, try to find in matlab's regexes something similar to this in php:
$string = 'http:xxx/sys/tags/Rel/total';
preg_replace('/http:(.*?)\//', 'http:${1}1/', $string);
${1} means 1st match group, that is what in parenthesis, (.*?).
http:(.*?)\/ - match pattern
http:${1}1/ - replace pattern with second 1 as you wish to add (first 1 is a group number)
http:xxx/sys/tags/Rel/total - input string
The secret is that whatever is matched by (.*?) (whether xxx or yyyy or 1234) will be inserted instead of ${1} in replace pattern, and then replace instead of old stuff into the input string. Welcome to see more examples on substitute functionality in php.
As documented in the help page for regexprep, you can specify pairs of patterns and replacements like this:
file='http:xxx/sys/tags/Rel/total';
regexprep(file, {'sys' 'total'}, {'sys1' 'total1'})
ans =
http:xxx/sys1/tags/Rel/total1
It is even possible to use tokens, should you be able to define a match pattern for everything you want to replace:
regexprep(file, '/([st][yo][^/$]*)', '/$11')
ans =
http:xxx/sys1/tags/Rel/total1
However, care must be taken with the first approach under certain circumstances, because MATLAB replaces the pairs one after another. That is to say if, say, the first pattern matches a string and replaces it with something that is subsequently matched by a later pattern, then that will also be replaced by the later replacement, even though it might not have matched the later pattern in the original string.
Example:
regexprep('This\is{not}LaTeX.', {'\\' '([{}])'}, {'\\textbackslash{}' '\\$1'})
ans =
This\textbackslash\{\}is\{not\}LaTeX.
=> This\{}is{not}LaTeX.
and
regexprep('This\is{not}LaTeX.', {'([{}])' '\\'}, {'\\$1' '\\textbackslash{}'})
ans =
This\textbackslash{}is\textbackslash{}{not\textbackslash{}}LaTeX.
=> This\is\not\LaTeX.
Both results are unintended, and there seems to be no way around this with consecutive replacements instead of simultaneous ones.
i am relatively new to NSRegularExpression and just can't come up with a pattern to find a string within a string....
here is the string...
##$294#001#[12345-678[123-456-7#15665#2
I want to extract the string..
#001#[12345-678[123-456-7#
for more info I know that there will be 3 digits(like 001) between two # 's and 20 characters between the last two # 's..
I have tried n number of combinations but nothing seem to work. any help is appreciated.
How about something like this:
#[0-9]{3}#.{20}#
If you know that the 20 characters will always consist of digits, [ and -, your pattern would become:
#[0-9]{3}#[0-9\[\-]{20}#
Be careful with the backslashes: When you use create the pattern with a string literal (#"..."), you need to add an extra backslash before each backslash.
You can test NSRegularExpression patterns without recompiling each time by using RegexTester https://github.com/liyanage/regextester