I am looking for a regex to test if a file has any extension. I define it as: file has an extension if there is no slashes present after the last ".". The slashes are always backslashes.
I started with this regex
.*\..*[^\\]
Which translates to
.* Any char, any number of repetitions
\. Literal .
.* Any char, any number of repetitions
[^\\] Any char that is NOT in a class of [single slash]
This is my test data (excluding ##, which is my comments)
\path\foo.txt ## I only want to capture this line
\pa.th\foo ## But my regex also captures this line <-- PROBLEM HERE
\path\foo ## This line is correctly filtered out
What would be a regex to do this?
Your solution is almost correct. Use this:
^.*\.[^\\]+$
Sample at rubular.
I wouldn't use a regular expression here. I'd split on / and ..
var path = '\some\path\foo\bar.htm',
hasExtension = path.split('\').pop().split('.').length > 1;
if (hasExtension) console.log('Weee!');
Here goes a more simple function to check it.
const hasExtension = path => {
const lastDotIndex = path.lastIndexOf('.')
return lastDotIndex > 1 && path.length - 1 > lastDotIndex
}
if (hasExtension(path)) console.log('Sweet')
You can also try even more simpler approach:
(\.[^\\]+)$
Details:
$ = Look from the end of string
[^\\]+ = Any character except path separator one or more time
\. = looks for <dot> character before extension
Live Demo
Related
The following should be matched:
AAA123
ABCDEFGH123
XXXX123
can I do: ".*123" ?
Yes, you can. That should work.
. = any char except newline
\. = the actual dot character
.? = .{0,1} = match any char except newline zero or one times
.* = .{0,} = match any char except newline zero or more times
.+ = .{1,} = match any char except newline one or more times
Yes that will work, though note that . will not match newlines unless you pass the DOTALL flag when compiling the expression:
Pattern pattern = Pattern.compile(".*123", Pattern.DOTALL);
Matcher matcher = pattern.matcher(inputStr);
boolean matchFound = matcher.matches();
Use the pattern . to match any character once, .* to match any character zero or more times, .+ to match any character one or more times.
The most common way I have seen to encode this is with a character class whose members form a partition of the set of all possible characters.
Usually people write that as [\s\S] (whitespace or non-whitespace), though [\w\W], [\d\D], etc. would all work.
.* and .+ are for any chars except for new lines.
Double Escaping
Just in case, you would wanted to include new lines, the following expressions might also work for those languages that double escaping is required such as Java or C++:
[\\s\\S]*
[\\d\\D]*
[\\w\\W]*
for zero or more times, or
[\\s\\S]+
[\\d\\D]+
[\\w\\W]+
for one or more times.
Single Escaping:
Double escaping is not required for some languages such as, C#, PHP, Ruby, PERL, Python, JavaScript:
[\s\S]*
[\d\D]*
[\w\W]*
[\s\S]+
[\d\D]+
[\w\W]+
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegularExpression{
public static void main(String[] args){
final String regex_1 = "[\\s\\S]*";
final String regex_2 = "[\\d\\D]*";
final String regex_3 = "[\\w\\W]*";
final String string = "AAA123\n\t"
+ "ABCDEFGH123\n\t"
+ "XXXX123\n\t";
final Pattern pattern_1 = Pattern.compile(regex_1);
final Pattern pattern_2 = Pattern.compile(regex_2);
final Pattern pattern_3 = Pattern.compile(regex_3);
final Matcher matcher_1 = pattern_1.matcher(string);
final Matcher matcher_2 = pattern_2.matcher(string);
final Matcher matcher_3 = pattern_3.matcher(string);
if (matcher_1.find()) {
System.out.println("Full Match for Expression 1: " + matcher_1.group(0));
}
if (matcher_2.find()) {
System.out.println("Full Match for Expression 2: " + matcher_2.group(0));
}
if (matcher_3.find()) {
System.out.println("Full Match for Expression 3: " + matcher_3.group(0));
}
}
}
Output
Full Match for Expression 1: AAA123
ABCDEFGH123
XXXX123
Full Match for Expression 2: AAA123
ABCDEFGH123
XXXX123
Full Match for Expression 3: AAA123
ABCDEFGH123
XXXX123
If you wish to explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:
There are lots of sophisticated regex testing and development tools, but if you just want a simple test harness in Java, here's one for you to play with:
String[] tests = {
"AAA123",
"ABCDEFGH123",
"XXXX123",
"XYZ123ABC",
"123123",
"X123",
"123",
};
for (String test : tests) {
System.out.println(test + " " +test.matches(".+123"));
}
Now you can easily add new testcases and try new patterns. Have fun exploring regex.
See also
regular-expressions.info/Tutorial
No, * will match zero-or-more characters. You should use +, which matches one-or-more instead.
This expression might work better for you: [A-Z]+123
Specific Solution to the example problem:-
Try [A-Z]*123$ will match 123, AAA123, ASDFRRF123. In case you need at least a character before 123 use [A-Z]+123$.
General Solution to the question (How to match "any character" in the regular expression):
If you are looking for anything including whitespace you can try [\w|\W]{min_char_to_match,}.
If you are trying to match anything except whitespace you can try [\S]{min_char_to_match,}.
Try the regex .{3,}. This will match all characters except a new line.
[^] should match any character, including newline. [^CHARS] matches all characters except for those in CHARS. If CHARS is empty, it matches all characters.
JavaScript example:
/a[^]*Z/.test("abcxyz \0\r\n\t012789ABCXYZ") // Returns ‘true’.
I like the following:
[!-~]
This matches all char codes including special characters and the normal A-Z, a-z, 0-9
https://www.w3schools.com/charsets/ref_html_ascii.asp
E.g. faker.internet.password(20, false, /[!-~]/)
Will generate a password like this: 0+>8*nZ\\*-mB7Ybbx,b>
I work this Not always dot is means any char. Exception when single line mode. \p{all} should be
String value = "|°¬<>!\"#$%&/()=?'\\¡¿/*-+_#[]^^{}";
String expression = "[a-zA-Z0-9\\p{all}]{0,50}";
if(value.matches(expression)){
System.out.println("true");
} else {
System.out.println("false");
}
Given string:
some_function(inputId = "select_something"),
(...)
some_other_function(inputId = "some_other_label")
I would like to arrive at:
some_function(inputId = ns("select_something")),
(...)
some_other_function(inputId = ns("some_other_label"))
The key change here is the element ns( ... ) that surrounds the string available in the "" after the inputId
Regex
So far, I have came up with this regex:
:%substitute/\(inputId\s=\s\)\(\"[a-zA-Z]"\)/\1ns(/2/cgI
However, when deployed, it produces an error:
E488: Trailing characters
A simpler version of that regex works, the syntax:
:%substitute/\(inputId\s=\s\)/\1ns(/cgI
would correctly inser ns( after finding inputId = and create string
some_other_function(inputId = ns("some_other_label")
Challenge
I'm struggling to match the remaining part of the string, ex. "select_something") and return it as:
"select_something")).
You have many problems with your regex.
[a-zA-Z] will only match one letter. Presumably you want to match everything up to the next ", so you'll need a \+ and you'll also need to match underscores too. I would recommend \w\+. Unless more than [a-zA-Z_] might be in the string, in which case I would do .\{-}.
You have a /2 instead of \2. This is why you're getting E488.
I would do this:
:%s/\(inputId = \)\(".\{-}\)"/\1ns(\2)/cgI
Or use the start match atom: (that is, \zs)
:%s/inputId = \zs\".\{-}"/ns(&)/cgI
You can use a negated character class "[^"]*" to match a quoted string:
%s/\(inputId\s*=\s*\)\("[^"]*"\)/\1ns(\2)/g
I am trying to work on regular expressions. I have a mainframe file which has several fields. I have a flat file parser which distinguishes several types of records based on the first three letters of every line. How do I write a regular expression where the first three letters are 'CTR'.
Beginning of line or beginning of string?
Start and end of string
/^CTR.*$/
/ = delimiter
^ = start of string
CTR = literal CTR
$ = end of string
.* = zero or more of any character except newline
Start and end of line
/^CTR.*$/m
/ = delimiter
^ = start of line
CTR = literal CTR
$ = end of line
.* = zero or more of any character except newline
m = enables multi-line mode, this sets regex to treat every line as a string, so ^ and $ will match start and end of line
While in multi-line mode you can still match the start and end of the string with \A\Z permanent anchors
/\ACTR.*\Z/m
\A = means start of string
CTR = literal CTR
.* = zero or more of any character except newline
\Z = end of string
m = enables multi-line mode
As such, another way to match the start of the line would be like this:
/(\A|\r|\n|\r\n)CTR.*/
or
/(^|\r|\n|\r\n)CTR.*/
\r = carriage return / old Mac OS newline
\n = line-feed / Unix/Mac OS X newline
\r\n = windows newline
Note, if you are going to use the backslash \ in some program string that supports escaping, like the php double quotation marks "" then you need to escape them first
so to run \r\nCTR.* you would use it as "\\r\\nCTR.*"
^CTR
or
^CTR.*
edit:
To be more clear: ^CTR will match start of line and those chars. If all you want to do is match for a line itself (and already have the line to use), then that is all you really need. But if this is the case, you may be better off using a prefab substr() type function. I don't know, what language are you are using. But if you are trying to match and grab the line, you will need something like .* or .*$ or whatever, depending on what language/regex function you are using.
Regex symbol to match at beginning of a line:
^
Add the string you're searching for (CTR) to the regex like this:
^CTR
Example: regex
That should be enough!
However, if you need to get the text from the whole line in your language of choice, add a "match anything" pattern .*:
^CTR.*
Example: more regex
If you want to get crazy, use the end of line matcher
$
Add that to the growing regex pattern:
^CTR.*$
Example: lets get crazy
Note: Depending on how and where you're using regex, you might have to use a multi-line modifier to get it to match multiple lines. There could be a whole discussion on the best strategy for picking lines out of a file to process them, and some of the strategies would require this:
Multi-line flag m (this is specified in various ways in various languages/contexts)
/^CTR.*/gm
Example: we had to use m on regex101
Try ^CTR.\*, which literally means start of line, CTR, anything.
This will be case-sensitive, and setting non-case-sensitivity will depend on your programming language, or use ^[Cc][Tt][Rr].\* if cross-environment case-insensitivity matters.
^CTR.*$
matches a line starting with CTR.
Not sure how to apply that to your file on your server, but typically, the regex to match the beginning of a string would be :
^CTR
The ^ means beginning of string / line
There's are ambiguities in the question.
What is your input string? Is it the entire file? Or is it 1 line at a time? Some of the answers are assuming the latter. I want to answer the former.
What would you like to return from your regular expression? The fact that you want a true / false on whether a match was made? Or do you want to extract the entire line whose start begins with CTR? I'll answer you only want a true / false match.
To do this, we just need to determine if the CTR occurs at either the start of a file, or immediately following a new line.
/(?:^|\n)CTR/
(?i)^[ \r\n]*CTR
(?i) -- case insensitive -- Remove if case sensitive.
[ \r\n] -- ignore space and new lines
* -- 0 or more times the same
CTR - your starts with string.
This is not correct use of wildcards ? I'm attempting to match String that contains a date. I don't want to include the date in the returned String or the String value that prepends the matched String.
object FindText extends App{
val toFind = "find1"
val line = "this is find1 the line 1 \n 21/03/2015"
val find = (toFind+".*\\d{2}/\\d{2}/\\d{4}").r
println(find.findFirstIn(line))
}
Output should be : "find1 the line 1 \n "
but String is not found.
Dot does not match newline characters by default. You can set a DOTALL flag to make it happen (I have also added a "positive look-ahead - the (?=...) thingy - since you did not want the date to be included in the match": val find = (toFind+"""(?s).*(?=\d{2}/\d{2}/\d{4})""").r
(Note also, that in scala you do not need to escape special characters in strings, enclosed in a triple-quote pairs ... pretty neat).
The problem lies with the newline in the test string. A .* does not match newlines apparently. Replacing this with .*\\n?.* should fix it. One could also use a multiline flag in the regex such as:
val find = ("(?s)"+toFind+".*\\d{2}/\\d{2}/\\d{4}").r
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];
}