Is there a possibility to use Regex for finding matching string between two dots?
I have strings with direcotries and I need to find string between two dots. Eg:
$string = '/Folder/file.co.txt';
and regex will return ONLY co between two dots.
I've tried following pattern: '/..../', but it returned .co. with dots.
Is there a possibility to do this with regex or all I can do is splice returned string?
You could use lookaround:
$string = '/Folder/file.co.txt';
preg_match('/(?<=\.)..(?=\.)/', $string, $matches);
echo $matches[0];
Output:
co
If you use preg_match you can set braces like () to define a group.
Your Statement can look like
"~\.(..)\.~g"
Related
I need to extract from a string a set of characters which are included between two delimiters, without returning the delimiters themselves.
A simple example should be helpful:
Target: extract the substring between square brackets, without returning the brackets themselves.
Base string: This is a test string [more or less]
If I use the following reg. ex.
\[.*?\]
The match is [more or less]. I need to get only more or less (without the brackets).
Is it possible to do it?
Easy done:
(?<=\[)(.*?)(?=\])
Technically that's using lookaheads and lookbehinds. See Lookahead and Lookbehind Zero-Width Assertions. The pattern consists of:
is preceded by a [ that is not captured (lookbehind);
a non-greedy captured group. It's non-greedy to stop at the first ]; and
is followed by a ] that is not captured (lookahead).
Alternatively you can just capture what's between the square brackets:
\[(.*?)\]
and return the first captured group instead of the entire match.
If you are using JavaScript, the solution provided by cletus, (?<=\[)(.*?)(?=\]) won't work because JavaScript doesn't support the lookbehind operator.
Edit: actually, now (ES2018) it's possible to use the lookbehind operator. Just add / to define the regex string, like this:
var regex = /(?<=\[)(.*?)(?=\])/;
Old answer:
Solution:
var regex = /\[(.*?)\]/;
var strToMatch = "This is a test string [more or less]";
var matched = regex.exec(strToMatch);
It will return:
["[more or less]", "more or less"]
So, what you need is the second value. Use:
var matched = regex.exec(strToMatch)[1];
To return:
"more or less"
Here's a general example with obvious delimiters (X and Y):
(?<=X)(.*?)(?=Y)
Here it's used to find the string between X and Y. Rubular example here, or see image:
You just need to 'capture' the bit between the brackets.
\[(.*?)\]
To capture you put it inside parentheses. You do not say which language this is using. In Perl for example, you would access this using the $1 variable.
my $string ='This is the match [more or less]';
$string =~ /\[(.*?)\]/;
print "match:$1\n";
Other languages will have different mechanisms. C#, for example, uses the Match collection class, I believe.
[^\[] Match any character that is not [.
+ Match 1 or more of the anything that is not [. Creates groups of these matches.
(?=\]) Positive lookahead ]. Matches a group ending with ] without including it in the result.
Done.
[^\[]+(?=\])
Proof.
http://regexr.com/3gobr
Similar to the solution proposed by null. But the additional \] is not required. As an additional note, it appears \ is not required to escape the [ after the ^. For readability, I would leave it in.
Does not work in the situation in which the delimiters are identical. "more or less" for example.
Most updated solution
If you are using Javascript, the best solution that I came up with is using match instead of exec method.
Then, iterate matches and remove the delimiters with the result of the first group using $1
const text = "This is a test string [more or less], [more] and [less]";
const regex = /\[(.*?)\]/gi;
const resultMatchGroup = text.match(regex); // [ '[more or less]', '[more]', '[less]' ]
const desiredRes = resultMatchGroup.map(match => match.replace(regex, "$1"))
console.log("desiredRes", desiredRes); // [ 'more or less', 'more', 'less' ]
As you can see, this is useful for multiple delimiters in the text as well
PHP:
$string ='This is the match [more or less]';
preg_match('#\[(.*)\]#', $string, $match);
var_dump($match[1]);
This one specifically works for javascript's regular expression parser /[^[\]]+(?=])/g
just run this in the console
var regex = /[^[\]]+(?=])/g;
var str = "This is a test string [more or less]";
var match = regex.exec(str);
match;
To remove also the [] use:
\[.+\]
I had the same problem using regex with bash scripting.
I used a 2-step solution using pipes with grep -o applying
'\[(.*?)\]'
first, then
'\b.*\b'
Obviously not as efficient at the other answers, but an alternative.
I wanted to find a string between / and #, but # is sometimes optional. Here is the regex I use:
(?<=\/)([^#]+)(?=#*)
Here is how I got without '[' and ']' in C#:
var text = "This is a test string [more or less]";
// Getting only string between '[' and ']'
Regex regex = new Regex(#"\[(.+?)\]");
var matchGroups = regex.Matches(text);
for (int i = 0; i < matchGroups.Count; i++)
{
Console.WriteLine(matchGroups[i].Groups[1]);
}
The output is:
more or less
If you need extract the text without the brackets, you can use bash awk
echo " [hola mundo] " | awk -F'[][]' '{print $2}'
result:
hola mundo
I'm trying to run a test to see if a long string with multiple lines has multiple occurrences of the same pattern, or lets say 5 or 10 occurrences.
So a string like:
$string = "this is a test pattern1 more of a test pattern1
and so on and so on pattern1";
So in this case I was trying in PHP:
if (preg_match('/(pattern1)\1{2,}/m',$string)) print "Found 2+ occurrences of pattern1\n";
Of course this does not work.
And I cannot use preg_match_all.
Can someone correct my regex please?
If I understand well, you was not far from the good pattern (for three occurrences here):
/(pattern1)(?:.*?\1){2,}/s
where the s modifier allows the dot to match newlines.
You can use the following..
if (preg_match('/(pattern1)(?:((?!\1).)*\1){2,}/s', $string)) {
Working Demo
Check this pattern
/(pattern1)/g
g - modifier finds all matches instead of returning first match.
why dont you simply search for the word/pattern using preg_match_all and count the number of occurences :
<?php
$message="this is a test pattern1 more of a test pattern1 and one more pattern1";
echo preg_match_all('/pattern1/i', $message, $matches);
will return 3.
or rather in your case:
<?php
$message="this is a test pattern1 more of a test pattern1 and one more pattern1";
if(preg_match_all('/pattern1/i', $message, $matches) >= 2 ) print "Found 2+ occurences of pattern1\n";
For example i have this string.
$string = 'test***bas';
How can I display text before the stars with Regex?
You could use a regular expression which makes use of Capture Groups. Once that you have matched your input, you could then access the captured group and print the output.
The following pattern
^(.+?)\*\*\*
will create a group match using the parenthesis operators. See http://gskinner.com/RegExr/ for testing your regular expressions (there are many ways of testing online)
The language you use around your regular expression will have different ways of capturing groups so you will need to better explain what language you are using for any further advice.
Example for before and after asterix
^(.+?)\*\*\*(.+)$
If tou also want what is located after the ***, you can use the following:
$string = 'test***bas';
$pattern = '/(.+)\*{3}(.+)/';
preg_match($pattern, $string, $matches);
$matches will contain the results:
$matches[1] will be "test"
$matches[2] will be "bas"
I have some strings like below
my-name-is-2547
this-is-stack-2012
hllo-how-2011
Now I want the above strings to be changed to something like the ones below using regex.
my-name-is-(2547)
this-is-stack-(2012)
hllo-how-(2011)
I don't want to use substr or other, only regex replace.
$pattern = '/(\d+)$/';
$replacement = '($1)';
echo preg_replace($pattern, $replacement, $string);
If you are sure that a numbers are only at the end:
regular expression:
(\d+)
using 1 capturing group. Replaced by: ($1).
so the outpu will be:
my-name-is-(2547)
this-is-stack-(2012)
hllo-how-(2011)
I need to extract from a string a set of characters which are included between two delimiters, without returning the delimiters themselves.
A simple example should be helpful:
Target: extract the substring between square brackets, without returning the brackets themselves.
Base string: This is a test string [more or less]
If I use the following reg. ex.
\[.*?\]
The match is [more or less]. I need to get only more or less (without the brackets).
Is it possible to do it?
Easy done:
(?<=\[)(.*?)(?=\])
Technically that's using lookaheads and lookbehinds. See Lookahead and Lookbehind Zero-Width Assertions. The pattern consists of:
is preceded by a [ that is not captured (lookbehind);
a non-greedy captured group. It's non-greedy to stop at the first ]; and
is followed by a ] that is not captured (lookahead).
Alternatively you can just capture what's between the square brackets:
\[(.*?)\]
and return the first captured group instead of the entire match.
If you are using JavaScript, the solution provided by cletus, (?<=\[)(.*?)(?=\]) won't work because JavaScript doesn't support the lookbehind operator.
Edit: actually, now (ES2018) it's possible to use the lookbehind operator. Just add / to define the regex string, like this:
var regex = /(?<=\[)(.*?)(?=\])/;
Old answer:
Solution:
var regex = /\[(.*?)\]/;
var strToMatch = "This is a test string [more or less]";
var matched = regex.exec(strToMatch);
It will return:
["[more or less]", "more or less"]
So, what you need is the second value. Use:
var matched = regex.exec(strToMatch)[1];
To return:
"more or less"
Here's a general example with obvious delimiters (X and Y):
(?<=X)(.*?)(?=Y)
Here it's used to find the string between X and Y. Rubular example here, or see image:
You just need to 'capture' the bit between the brackets.
\[(.*?)\]
To capture you put it inside parentheses. You do not say which language this is using. In Perl for example, you would access this using the $1 variable.
my $string ='This is the match [more or less]';
$string =~ /\[(.*?)\]/;
print "match:$1\n";
Other languages will have different mechanisms. C#, for example, uses the Match collection class, I believe.
[^\[] Match any character that is not [.
+ Match 1 or more of the anything that is not [. Creates groups of these matches.
(?=\]) Positive lookahead ]. Matches a group ending with ] without including it in the result.
Done.
[^\[]+(?=\])
Proof.
http://regexr.com/3gobr
Similar to the solution proposed by null. But the additional \] is not required. As an additional note, it appears \ is not required to escape the [ after the ^. For readability, I would leave it in.
Does not work in the situation in which the delimiters are identical. "more or less" for example.
Most updated solution
If you are using Javascript, the best solution that I came up with is using match instead of exec method.
Then, iterate matches and remove the delimiters with the result of the first group using $1
const text = "This is a test string [more or less], [more] and [less]";
const regex = /\[(.*?)\]/gi;
const resultMatchGroup = text.match(regex); // [ '[more or less]', '[more]', '[less]' ]
const desiredRes = resultMatchGroup.map(match => match.replace(regex, "$1"))
console.log("desiredRes", desiredRes); // [ 'more or less', 'more', 'less' ]
As you can see, this is useful for multiple delimiters in the text as well
PHP:
$string ='This is the match [more or less]';
preg_match('#\[(.*)\]#', $string, $match);
var_dump($match[1]);
This one specifically works for javascript's regular expression parser /[^[\]]+(?=])/g
just run this in the console
var regex = /[^[\]]+(?=])/g;
var str = "This is a test string [more or less]";
var match = regex.exec(str);
match;
To remove also the [] use:
\[.+\]
I had the same problem using regex with bash scripting.
I used a 2-step solution using pipes with grep -o applying
'\[(.*?)\]'
first, then
'\b.*\b'
Obviously not as efficient at the other answers, but an alternative.
I wanted to find a string between / and #, but # is sometimes optional. Here is the regex I use:
(?<=\/)([^#]+)(?=#*)
Here is how I got without '[' and ']' in C#:
var text = "This is a test string [more or less]";
// Getting only string between '[' and ']'
Regex regex = new Regex(#"\[(.+?)\]");
var matchGroups = regex.Matches(text);
for (int i = 0; i < matchGroups.Count; i++)
{
Console.WriteLine(matchGroups[i].Groups[1]);
}
The output is:
more or less
If you need extract the text without the brackets, you can use bash awk
echo " [hola mundo] " | awk -F'[][]' '{print $2}'
result:
hola mundo