This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed last month.
So, I wanted to know the meaning of ".$" and why does this command delete the last character of a String.
I got what I wanted but i just want an explanation for it.
Because is a regular expression and $ mean "last"
Related
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 3 years ago.
I have timestamps in a field and would like to remove the 'T' and the 'Z' in the value. An example value is 2019-11-01T14:47:43Z. I would like to use a RegEx to solve this problem. I plan to use this in Java.
You can use Java's String.replaceAll() function to remove values with regex. The regular expression [a-zA-Z] will match any one letter; replacing it with an empty string will remove it entirely.
String ts = "2019-11-01T14:47:43Z";
System.out.println(ts.replaceAll("[a-zA-Z]", ""));
2019-11-0114:47:43
Demo
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Closed 4 years ago.
I want to replace &sp; in the string below with Z.
Input text : ABCD&sp;EF&p;GHIJ&bsp;KL
Output text : ABCDZEFZGHIZKL
Can anyone tell me how to replace the every instance of &\D+; using java regular expression?
I am using /(&\D+;)?/ but it doesn't work.
Use String#replaceAll.
You also should use the ? modificator to +:
String str = "ABCD&sp;EF&p;GHIJ&bsp;KL";
String regex = "&\\D+?;";
System.out.println (str.replaceAll(regex,"Z"));
This should work
Match the initial &, then all characters that are not the tailing ;, then that tailing ; like so: &[^;]+; If not matching numbers (as suggested by your example with \D) is a requirement, add the numbers to the negated character set: [^;0-9] To make it replace all occurrences, add the global flag g. The site regexr.com is a handy tool to create regexes.
Edit: Sorry, I initially read your question wrong.
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How to use beginning and endline markers in regex for Java String?
(5 answers)
How to use java regex to match a line
(2 answers)
Closed 5 years ago.
given the following expression:
Pattern.compile("^Test.*\n").matcher("Test 123\nNothing\nTest 2\n").replaceAll("foo\n")
This yields:
"foo\nNothing\nTest 2\n"
for me. I expected that the last line is also replaced to foo\n since there is a linebreak immediately before Test 2 in the input string.
Why is doesn't the regex match there?
You have to add the multiline flag to the pattern: Pattern.MULTILINE.
Pattern.compile("^Test.*\n", Pattern.MULTILINE).matcher("Test 123\nNothing\nTest 2\n").replaceAll("foo\n")
By Default the match is only single line. For more Informations see the javadoc
At the beginning of your regex you have a ^ sign which normally anchors the regex to the beginning of a tested string. You need to specify multiline regex option (Oracle Documentation link) to make it apply to start of each line instead.
Try this (I have split the lines for legibility, feel free to oneline it back):
Pattern.compile("^Test.*\n", Pattern.MULTILINE)
.matcher("Test 123\nNothing\nTest 2\n")
.replaceAll("foo\n")
Unfortunately I do not have Java environment set up at the moment, so I'm unable to check this by myself.
This question already has answers here:
Regex to trim hyphens from start and end of a string
(2 answers)
Closed 7 years ago.
Given a word-string in Java, I want to strip off from beginning and from end, exactly these specified set of characters:
[?:!.,;'\"«»]
as many times as they appear.
For instance, «Be!!» should become just Be, "Here!!!" should become Here, «I should become I.
Can anyone provide a correct way to do this?
Use an anchored regex in string.replaceAll function.
string.replaceAll("^[?:!.,;'\"«»]+|[?:!.,;'\"«»]+$", "");
DEMO
This question already has answers here:
Regular expression to match a line that doesn't contain a word
(34 answers)
Closed 8 years ago.
I m trying regex to get the strings
starting with # and
not ending with a dot(.)
For that i tried the java code(link here) but this does not show any results -
#(\\w+)*(?<!.(.))*$
The string i m trying is -
This is a test\nAnother #pradyut#test ht#html.com\ntest\n#art\n#cool#paintings#collections
This should return
pradyut
test
cool
The result html ending with a .com should not return.
Regards
You can use this regex:
(?<=#)\w+\b(?!\.)
In Java you have to use:
(?<=#)\\w+\\b(?!\\.)
Regex Demo