I am trying to write a regular expression to select all characters between underscores. I end up with _([^_]+)_ but it does not match all the groups,
String : abc_bca_vag_hag_bag
output : bca vag hag
can someone help with this?
I used split function and select every elements except first one and last one:
st = 'abc_bca_vag_hag_bag'
lis = st.split('_')[1:-1]
# output ['bca', 'vag', 'hag']
Your regex will match the underscores, and things that are matched once will not be matched again. So after matching _bca_, it does not see that the last underscore in _bca_ is actually the same underscore that is before vag. It thinks that vag is not preceded by an underscore because it has already matched the underscore preceding it in the previous match.
You need to use lookaheads and lookbehinds:
(?<=_)[^_]+(?=_)
These will not match the underscore. They will only "look" and see whether there is an underscore.
Demo
Related
I have the following String
""00001","Open","2020-10-23 12:45","2022-10-20 15:48","Error not found","Alex Smith","NO ERROR","ErrorApp","FOB","CNSHA","GB","Plane","MODEL","2020-10-24 00:00","New","1","","2022-10-20 15:48""
I need to replace the 4th and the last words (dates) with an empty String "".
I have tried several regex expressions but couldn't get them to work.
The result should be:
""00001","Open","2020-10-23 12:45","","Error not found","Alex Smith","NO ERROR","ErrorApp","FOB","CNSHA","GB","Plane","MODEL","2020-10-24 00:00","New","1","","""
This regex:
actualLines[1].replaceAll("^(\".+?\",\".+?\",\".+?\",\")([0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2})(.*)$", "$1$3}");
It replaces the 4th, but I need both 4th and the last element to be replaced.
Does anyone know how this can be done? Thanks
An idea to use an alternation for matching both, the fourth and last double-quoted part:
^((?:\"[^\"]*\",){3})\"[^\"]*\"|\"[^\"]*\"$
See this demo at regex101 or a Java demo at tio.run
Replace with $1\"\" where $1 is a reference to what's captured by the first group. It holds the first three comma separated quoted parts (?:"[^"]*",){3} where [^"] matches characters other than quotes. On the right side of the alternation the part at the $ end gets matched (and $1 is empty).
I have a string from which I would like to caputre all after and including colon until (excluding) white space or paranthesis.
Why does the following regex include the paranthesis in the string match?
:(.*?)[\(\)\s] or also :(.+?)[\)\s] (non-greedy) does not work.
Example input: WHERE t.operator_id = :operatorID AND (t.merchant_id = :merchantID) AND t.readerApplication_id = :readerApplicationID AND t.accountType in :accountTypes
Should exctract :operatorID, :merchantID, :readerApplicationID, :accountTypes.
But my regexes extract for the second match :marchantID)
What is wrong and why?
Even if I use an exacter mapping condition in the capture, it does not work: :([a-zA-z0-9_]+?)[\)\(\s]
Put your conditional "followed by space or paren" as a lookahead, so that it sees but doesn't match. Right now you are explicitly matching parentheses with [\(\)\s]:
:(.+?)(?=[\s\(\)])
https://regex101.com/r/im8KWF/1/
Or, use the built-in \b "word boundary", which is also a "zero-width" assertion meaning the same thing*:
:(.+?)\b
https://regex101.com/r/FnnzGM/3/
*Definition of word boundary from regular-expressions.info:
There are three different positions that qualify as word boundaries:
Before the first character in the string, if the first character is a
word character. After the last character in the string, if the last
character is a word character. Between two characters in the string,
where one is a word character and the other is not a word character.
I have text that looks like something like this:
1. Must have experience in Java 2. Team leader...
I want to render this in HTML as an ordered list. Now adding the </li> tag to the end is simple enough:
s = replace(s, ". ", "</li>");
But how do I go about replacing the 1., 2. etc with <li>?
I have the regular expression \d*\.$ which matches a number with a period, but the problem is is that is a substring so matching 1. Must have experience in Java 2. Team leader with \d*\.$ returns false.
Code
See regex in use here
\d+\.\s+(.*?)\s*(?=\d+\.\s+|$)
Replace
<li>$1</li>\n
Results
Input
Must have experience in Java 2. Team leader...
Output
<li>Must have experience in Java</li>
<li>Team leader...</li>
Explanation
\d+ Match one or more digits
\. Match the dot character . literally
\s+ Match one or more whitespace characters
(.*?) Capture any character any number of times, but as few as possible, into capture group 1
\s* Match any number of whitespace characters
(?=\d+\.\s+|$) Positive lookahead ensuring either of the following doesn't match
\d+\.\s+
\d+ Match one or more digits
\. Match the dot character . literally
\s+ Match one or more whitespace characters
$ Assert position at the end of the line
But how do I go about replacing the 1., 2. etc with <li>?
You can use String#replaceAll which can allow regex instead of replace :
s = s.replaceAll("\\d+\\.\\s", "</li>");
Note
You don't need to use $ in the end of your regex.
You have to escape dot . because it's mean any character in regex
You can use \s for one space or \s* for zero or more spaces or \s+ for one or more space
We want
<ol>
<li>one</li>
<li>two<li>
</ol>
This can be done as:
s = s.replaceAll("(?s)(\\d+\\.)\\s+(.*\\.)\\s*", "<li>$2</li></ol>");
s = s.replaceFirst("<li>", "<ol><li>");
s = s.replaceAll("(?s)</li></ol><li>", "</li>\n<li>");
The trick is to first add </li></ol> with a spurious </ol> that should only remain after the last list item.
(?s) is the DOTALL notation, causing . to also match line breaks.
In case of more than one numbered list this will not do. Also it assumes one single sentence per list item.
I have this requirement - for an input string such as the one shown below
8This8 is &reallly& a #test# of %repl%acing% %mul%tiple 9matched9 9pairs
I would like to strip the matched word boundaries (where the matching pair is 8 or & or % etc) and will result in the following
This is really a test of repl%acing %mul%tiple matched 9pairs
This list of characters that is used for the pairs can vary e.g. 8,9,%,# etc and only the words matching the start and end with each type will be stripped of those characters, with the same character embedded in the word remaining where it is.
Using Java I can do a pattern as \\b8([^\\s]*)8\\b and replacement as $1, to capture and replace all occurrences of 8...8, but how do I do this for all the types of pairs?
I can provide a pattern such as \\b8([^\\s]*)8\\b|\\b9([^\\s]*)9\\b .. and so on that will match all types of matching pairs *8,9,..), but how do I specify a 'variable' replacement group -
e.g. if the match is 9...9, the the replacement should be $2.
I can of course run it through multiple of these, each replacing a specific type of pair, but I am wondering if there is a more elegant way.
Or is there a completely different way of approaching this problem?
Thanks.
You could use the below regex and then replace the matched characters by the characters present inside the group index 2.
(?<!\S)(\S)(\S+)\1(?=\s|$)
OR
(?<!\S)(\S)(\S*)\1(?=\s|$)
Java regex would be,
(?<!\\S)(\\S)(\\S+)\\1(?=\\s|$)
DEMO
String s1 = "8This8 is &reallly& a #test# of %repl%acing% %mul%tiple 9matched9 9pairs";
System.out.println(s1.replaceAll("(?<!\\S)(\\S)(\\S+)\\1(?=\\s|$)", "$2"));
Output:
This is reallly a test of repl%acing %mul%tiple matched 9pairs
Explanation:
(?<!\\S) Negative lookbehind, asserts that the match wouldn't be preceded by a non-space character.
(\\S) Captures the first non-space character and stores it into group index 1.
(\\S+) Captures one or more non-space characters.
\\1 Refers to the character inside first captured group.
(?=\\s|$) And the match must be followed by a space or end of the line anchor.
This makes sure that the first character and last character of the string must be the same. If so, then it replaces the whole match by the characters which are present inside the group index 2.
For this specific case, you could modify the above regex as,
String s1 = "8This8 is &reallly& a #test# of %repl%acing% %mul%tiple 9matched9 9pairs";
System.out.println(s1.replaceAll("(?<!\\S)([89&#%])(\\S+)\\1(?=\\s|$)", "$2"));
DEMO
(?<![a-zA-Z])[8&#%9](?=[a-zA-Z])([^\s]*?)(?<=[a-zA-Z])[8&#%9](?![a-zA-Z])
Try this.Replace with $1 or \1.See demo.
https://regex101.com/r/qB0jV1/15
(?<![a-zA-Z])[^a-zA-Z](?=[a-zA-Z])([^\s]*?)(?<=[a-zA-Z])[^a-zA-Z](?![a-zA-Z])
Use this if you have many delimiters.
I'm trying to compare following strings with regex:
#[xyz="1","2"'"4"] ------- valid
#[xyz] ------------- valid
#[xyz="a5","4r"'"8dsa"] -- valid
#[xyz="asd"] -- invalid
#[xyz"asd"] --- invalid
#[xyz="8s"'"4"] - invalid
The valid pattern should be:
#[xyz then = sign then some chars then , then some chars then ' then some chars and finally ]. This means if there is characters after xyz then they must be in format ="XXX","XXX"'"XXX".
Or only #[xyz]. No character after xyz.
I have tried following regex, but it did not worked:
String regex = "#[xyz=\"[a-zA-z][0-9]\",\"[a-zA-z][0-9]\"'\"[a-zA-z][0-9]\"]";
Here the quotations (in part after xyz) are optional and number of characters between quotes are also not fixed and there could also be some characters before and after this pattern like asdadad #[xyz] adadad.
You can use the regex:
#\[xyz(?:="[a-zA-z0-9]+","[a-zA-z0-9]+"'"[a-zA-z0-9]+")?\]
See it
Expressed as Java string it'll be:
String regex = "#\\[xyz=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\"\\]";
What was wrong with your regex?
[...] defines a character class. When you want to match literal [ and ] you need to escape it by preceding with a \.
[a-zA-z][0-9] match a single letter followed by a single digit. But you want one or more alphanumeric characters. So you need [a-zA-Z0-9]+
Use this:
String regex = "#\\[xyz(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")?\\]";
When you write [a-zA-z][0-9] it expects a letter character and a digit after it. And you also have to escape first and last square braces because square braces have special meaning in regexes.
Explanation:
[a-zA-z0-9]+ means alphanumeric character (but not an underline) one or more times.
(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")? means that expression in parentheses can be one time or not at all.
Since square brackets have a special meaning in regex, you used it by yourself, they define character classes, you need to escape them if you want to match them literally.
String regex = "#\\[xyz=\"[a-zA-z][0-9]\",\"[a-zA-z][0-9]\"'\"[a-zA-z][0-9]\"\\]";
The next problem is with '"[a-zA-z][0-9]' you define "first a letter, second a digit", you need to join those classes and add a quantifier:
String regex = "#\\[xyz=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\"\\]";
See it here on Regexr
there could also be some characters before and after this pattern like
asdadad #[xyz] adadad.
Regex should be:
String regex = "(.)*#\\[xyz(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")?\\](.)*";
The First and last (.)* will allow any string before the pattern as you have mentioned in your edit. As said by #ademiban this (=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")? will come one time or not at all. Other mistakes are also very well explained by Others +1 to all other.