I need to capture numbers and dots between brackets on lines containing the string 0020,000d, for example:
I: (0020,000d) UI [1.2.410.200001.1104.20160720104648421 ] # 38, 1 StudyInstanceUID
Using this regexp 0020,000d.*\[([\.0-9]+)\] I can match the needed value only if it doesn't have a space inside the brackets. How can I match the needed value ignoring any other character?.
Edit
If I use this regexp 0020,000d.*\[([\.0-9(\s|^\s))]+)\] I can capture numbers and dots and/or spaces, now if the string contains a space how can I capture in a group everything but the space?.
To clarify, I want to extract the 1.2.410.200001.1104.20160720104648421 string.
Codifying my (apparently helpful) answer from the comments:
You just need to allow zero or more spaces after the numbers-and-dots sequence before the closing bracket:
0020,000d.*\[([.0-9]+) *\]
Also, please note that you don't need to escape a dot in a character class.
Try this
let regex = /(?!\[)[.\d]+(?=[(\s)*\]])/g
let str = 'I: (0020,000d) UI [1.2.410.200001.1104.20160720104648421 ]'
let result = str.match(regex);
console.log(result);
Related
I need to extract a string from an URL. Here are some examples:
Input: https://www.example.net/eur_en/bas-026-009-basic-baby-hat-beige.html – Output: bas-026-009
Input: https://www.example.net/eur_en/aw18-245-b86-big-cherries-snow-jacket-plum-red.html – Output: aw18-245-b86
Input: https://www.example.net/eur_en/ss20-028-e70-hearts-tee-off-white-yellow.html – Output: ss20-028-e70
I want to be able to extract the string that goes from the first character after the "/eur_en/" until the third dash. Can someone help me? Thanks
You're looking for regexp: \/eur_en\/([^-]+-[^-]+-[^-]+)
Play & test it at regex101: https://regex101.com/r/RvGROG/1
You need something like this:
const urls = [
"https://www.example.net/eur_en/bas-026-009-basic-baby-hat-beige.html",
"https://www.example.net/eur_en/aw18-245-b86-big-cherries-snow-jacket-plum-red.html",
"https://www.example.net/eur_en/ss20-028-e70-hearts-tee-off-white-yellow.html",
]
const rg = new RegExp(`\/eur_en\/([^-]+-[^-]+-[^-]+)`)
const strs = urls.map(url => url.match(rg)[1])
console.log(strs)
// Output:
// [
// "bas-026-009",
// "aw18-245-b86",
// "ss20-028-e70"
// ]
Of course, it's a simple example. In real cases don't forget to check that .match returned array with length greater than 1.
So, the first element is full captured string and the second (as third and next) it's a sub-strings, which is captured by parentheses.
We can improve and complicate our regex like so:
\/((?:[^-\/]+-){2}[^-\/]+)
It'll allow us to not to use a specific anchor /eur_en/ and control the number of dash divided parts.
The expression you're looking for is the following:
/(?<=eur_en\/)[^-]*-[^-]*-[^-]*/
Here is how it works:
(?<=eur_en\/): will look behind for eur_env/ but will not use it in the output
[^-]*: it will match any character that is not a dash. So it will get everything up to the first dash (not including the dash)
[^-]*: it will match any character that is not a dash. So it will get everything up to the second dash (not including the dash)
[^-]*: it will match any character that is not a dash. So it will get everything up to the third dash (not including the dash).
/(?<=\/eur_en\/)\w+-\w+-\w+/g
Tolkens
Description
(?<=\/eur_en\/)
Look behind - If /eur_en/ is found, match whatever proceeds it.
\w+-\w+-\w+
One or more Word character = [A-Za-z0-9] and a literal hyphen three consecutive times.
Review: https://regex101.com/r/Ge0zA3/1
I'm using the following regular expression pattern:
.*(?<line>^\s*Extends\s+#(?<extends>[_A-Za-z0-9]+)\s*$)?.*
And the following text:
Name #asdf
Extends #extendedClass
Origin #id
What I don't understand is that both of the caught group results (line and extends) are empty, but when I remove the last question mark from the expression the groups are caught.
The line group must be optional since the Extends line is not always present.
I created a fiddle using this expression, which can be accessed at https://regexr.com/4rekk
EDIT
I forgot to mention that I'm using the multiline and dotall flags along with the expression.
It's already been mentioned that the leading .* is capturing everything when you make your (?<line>) group optional. The following is not directly related to your question but it may be useful information (if not, just ignore):
You need to be careful elsewhere. You are using ^ and $ to match the start and end of lines as well as the start and end of the string. But the $ character will not consume the newline character that marks the end of a line. So:
'Line 1\nLine 2'.match(/^Line 1$^Line 2/m) returns null
while
'Line 1\nLine 2'.match(/^Line 1\n^Line 2/m) returns a match
So in your case if you were trying to capture all three lines, any of which were optional, you would write the regex for one of the lines as follows to make sure you consume the newline:
/(?<line>^\s*Extends\s+#(?<extends>[_A-Za-z0-9]+)[^\S\n]*\n)?/ms
Where you had specified \s*$, I have [^\S\n]*\n. [^\S\n]* is a double negative that says one or more non non-white space character excluding the newline character. So it will consume all white space characters except the newline character. If you wanted to look for any of the three lines in your example (any or all are optional), then the following code snippet should do it. I have used the RegExp function to create the regex so that it can be split across multiple lines. Unfortunately, it takes a string as its argument and so some backslash characters have to be doubled up:
let s = ` Name #asdf
Extends #extendedClass
Origin #id
`;
let regex = new RegExp(
"(?<line0>^\\s*Name\\s+#(?<name>[_A-Za-z0-9]+)[^\\S\\n]*\\n)?" +
"(?<line>^\\s*Extends\\s+#(?<extends>[_A-Za-z0-9]+)[^\\S\\n]*\\n)?" +
"(?<line2>^\\s*Origin\\s+#(?<id>[_A-Za-z0-9]+)[^\\S\\n]*\\n)?",
'm'
);
let m = s.match(regex);
console.log(m.groups);
The above code snippet seems to have a problem under Firefox (an invalid regex flag, 's', is flagged on a line that doesn't exist in the above snippet). See the following regex demo.
And without named capture groups:
let s = ` Name #asdf
Extends #extendedClass
Origin #id
`;
let regex = new RegExp(
"(^\\s*Name\\s+#([_A-Za-z0-9]+)[^\\S\\n]*\\n)?" +
"(^\\s*Extends\\s+#([_A-Za-z0-9]+)[^\\S\\n]*\\n)?" +
"(^\\s*Origin\\s+#([_A-Za-z0-9]+)[^\\S\\n]*\\n)?",
'm'
);
let m = s.match(regex);
console.log(m);
I am trying to parse a file that contains parameter attributes. The attributes are setup like this:
w=(nf*40e-9)*ng
but also like this:
par_nf=(1) * (ng)
The issue is, all of these parameter definitions are on a single line in the source file, and they are separated by spaces. So you might have a situation like this:
pd=2.0*(84e-9+(1.0*nf)*40e-9) nf=ng m=1 par=(1) par_nf=(1) * (ng) plorient=0
The current algorithm just splits the line on spaces and then for each token, the name is extracted from the LHS of the = and the value from the RHS. My thought is if I can create a Regex match based on spaces within parameter declarations, I can then remove just those spaces before feeding the line to the splitter/parser. I am having a tough time coming up with the appropriate Regex, however. Is it possible to create a regex that matches only spaces within parameter declarations, but ignores the spaces between parameter declarations?
Try this RegEx:
(?<=^|\s) # Start of each formula (start of line OR [space])
(?:.*?) # Attribute Name
= # =
(?: # Formula
(?!\s\w+=) # DO NOT Match [space] Word Characters = (Attr. Name)
[^=] # Any Character except =
)* # Formula Characters repeated any number of times
When checking formula characters, it uses a negative lookahead to check for a Space, followed by Word Characters (Attribute Name) and an =. If this is found, it will stop the match. The fact that the negative lookahead checks for a space means that it will stop without a trailing space at the end of the formula.
Live Demo on Regex101
Thanks to #Andy for the tip:
In this case I'll probably just match on the parameter name and equals, but replace the preceding whitespace with some other "parse-able" character to split on, like so:
(\s*)\w+[a-zA-Z_]=
Now my first capturing group can be used to insert something like a colon, semicolon, or line-break.
You need to add Perl tag. :-( Maybe this will help:
I ended up using this in C#. The idea was to break it into name value pairs, using a negative lookahead specified as the key to stop a match and start a new one. If this helps
var data = #"pd=2.0*(84e-9+(1.0*nf)*40e-9) nf=ng m=1 par=(1) par_nf=(1) * (ng) plorient=0";
var pattern = #"
(?<Key>[a-zA-Z_\s\d]+) # Key is any alpha, digit and _
= # = is a hard anchor
(?<Value>[.*+\-\\\/()\w\s]+) # Value is any combinations of text with space(s)
(\s|$) # Soft anchor of either a \s or EOB
((?!\s[a-zA-Z_\d\s]+\=)|$) # Negative lookahead to stop matching if a space then key then equal found or EOB
";
Regex.Matches(data, pattern, RegexOptions.IgnorePatternWhitespace | RegexOptions.ExplicitCapture)
.OfType<Match>()
.Select(mt => new
{
LHS = mt.Groups["Key"].Value,
RHS = mt.Groups["Value"].Value
});
Results:
Input:
1.\frac{[a+b]}{xjch}
2.\frac{pqz}{xjch}
Wanted output is
1.[a+b]/(xjch)
2.(pqz)/(xjch)
My regex is:
\\frac\{(.{2,})\}\{(.{2,})\}
if i apply this regex,
the output will be,
1.([a+b])/(xjch)
2.(pqz)/(xjch)
But i dont want () in [a+b]. ie if any special character inside the {...}, the round bracket should not come. otherwise, (Without special characters) ,the round bracket should come like (pqz),(xjch).
I want two regex for both 1. and 2. then only i will get wanted output.
Could anyone help me?
you can write a Regex that contain within the bracket and replace the group 1 and 2 with a condition
if(nextchar == "[")
TypeOfYourInstuction = 1;
else
TypeOfYourInstuction = 2;`
and this regex is
\\frac\{\[?([a-zA-Z1-9\+]{2,})\]?\}\{\[?([a-zA-Z1-9\+]{2,})\]?\}
http://regex101.com/r/dN8sA5/18
but as you mention it, you can write two regex for first type and the second one:
the first regex: \[[^\]]{2,}\] // Demo = http://regex101.com/r/dN8sA5/20
the second regex: \{[^\[^\}]*\} // Demo = http://regex101.com/r/dN8sA5/19
you have to replace the second type with parenthesis
I'm very new to regular expression. I want to extract the following string
"109_Admin_RegistrationResponse_20130103.txt"
from this file content, the contents is selected per line:
01-10-13 10:44AM 47 107_Admin_RegistrationDetail_20130111.txt
01-10-13 10:40AM 11 107_Admin_RegistrationResponse_20130111.txt
The regular expression should not pick the second line, only the first line should return a true.
Your Regex has a lot of different mistakes...
Your line does not start with your required filename but you put an ^ there
missing + in your character group [a-zA-Z], hence only able to match a single character
does not include _ in your character group, hence it won't match Admin_RegistrationResponse
missing \ and d{2} would match dd only.
As per M42's answer (which I left out), you also need to escape your dot . too, or it would match 123_abc_12345678atxt too (notice the a before txt)
Your regex should be
\d+_[a-zA-Z_]+_\d{4}\d{2}\d{2}\.txt$
which can be simplified as
\d+_[a-zA-Z_]+_\d{8}\.txt$
as \d{2}\d{2} really look redundant -- unless you want to do with capturing groups, then you would do:
\d+_[a-zA-Z_]+_(\d{4})(\d{2})(\d{2})\.txt$
Remove the anchors and escape the dot:
\d+[a-zA-Z_]+\d{8}\.txt
I'm a newbie in php but i think you can use explode() function in php or any equivalent in your language.
$string = "01-09-13 10:17AM 11 109_Admin_RegistrationResponse_20130103.txt";
$pieces = explode("_", $string);
$stringout = "";
foreach($i = 0;$i<count($pieces);i++){
$stringout = $stringout.$pieces[$i];
}