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How to remove alpha characters from timestamp using RegEx? [duplicate]

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

Split a mathematical function containing using signs in java [duplicate]

This question already has answers here:
How to Split a mathematical expression on operators as delimiters, while keeping them in the result?
(5 answers)
Closed 4 years ago.
I want to split a mathematical function by the sign of the variables in it like this :
input--> x-5y+3z=10
output--> [x,-5y,+3z,=10]
this code does not work in the way i want :
String function = "x-5y+3z=10";
String split = function.split("=|-|\\+");
the output of the array is :
[x,5y,3z,10]
so what is the correct regex for this ?

The "problem" using split is that the delimiter used will be removed, because it'll takt the parts that are between this delimiter, you need a pattern that is non-capturing or with a simple lookahead : match something wich is before something else
The pattern (?=[-+=]) would work, it'll take the part that starts with a -+= symbol without removing it :
String function = "x-5y+3z=10";
String[] split = function.split("(?=[-+=])");
System.out.println(Arrays.toString(split)); //[x, -5y, +3z, =10]
Some doc on Lookahead

Why is the beginning of a line not recognized by a regex? [duplicate]

This question already has answers here:
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.

Regex, trim multiple characters? [duplicate]

This question already has answers here:
Removing repeated characters in String
(4 answers)
Closed 8 years ago.
Lets say I have a string:
tttteeeeeeessssssttttttt
Using the power of regex, how can that string be turned into:
test
At first look it seems easy to do, but the current code (not regex) I have for it is not behaving well and im pretty sure regex is the way to go.

You can use:
str = str.replaceAll("([A-Za-z])\\1+", "$1");
RegEx Demo

Use string.replaceAll function.
strng.replaceAll("(.)\\1+", "$1");
The above regex captures the first character in the sequence of same characters and matches all the following one or more characters (which must be same as the one inside the capturing group) . Replacing those characters with the character inside group index 1 will give you the desired output.
Example:
System.out.println("tttteeeeeeessssssttttttt".replaceAll("(.)\\1+","$1" ));
Output:
test

(.)(?=\1)
Try this.Replace by empty string.See demo.
https://regex101.com/r/tX2bH4/41
str = str.replaceAll("(.)(?=\\1)", "");

Replace all with a string having regex wild chars [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
java String.replaceAll without regex
I have a string and I need to replace some parts of it.
The replacement text contains regex wild chars though. Example:
String target = "Something * to do in ('AAA', 'BBB')";
String replacement = "Hello";
String originalText = "ABCDEFHGIJKLMN" + target + "ABCDEFHGIJKLMN";
System.out.println(originalText.replaceAll(target, replacement));
I get:
ABCDEFHGIJKLMNSomething * to do in ('AAA', 'BBB')ABCDEFHGIJKLMN
Why doesn't the replacement occur?

Because *, ( and ) are all meta-characters in regular expressions. Hence all of them need to be escaped. It looks like Java has a convenient method for this:
java.util.regex.Pattern.quote(target)
However, the better option might be, to just not use the regex-using replaceAll function but simply replace. Then you do not need to escape anything.

String.replaceAll() takes a regular expression and so it's trying to expand these metacharacters.
One approach is to escape these chars (e.g. \*).
Another would be to do the replacement yourself by using String.indexOf() and finding the start of the contained string. indexOf() doesn't take a regexp but rather a normal string.