I'm trying to figure out how to write a regex that will match every charter up to, but not including the first number in the character sequence if there is one.
Ex:
Input: abc123
Output: abc
Input: #$%@#<>@<123
Output: #$%@#<>@<
Input: abc
Output: abc
Input: abc @#@#@-122
Output: abc @#@#@-
[Update] Try this regex:
([^0-9\n]+)[0-9]?.*
Regex explains:
( capturing group starts
[^0-9\n] match a single character other than numbers and new line
+ match one or more times
) capturing group ends
[0-9] match a single digit number (0-9)
? match zero or more times
.* if any, match all other than new line
Thanks @Robbie Averill for clarifying OP's requirement. Here is the demo.
You can use:
/^([^\d\n]+)\d*.*$/gm
This will also handle scenarios where you have multiple sets of numbers in a string. Example here.
Explanation:
^ # define the start of the stirng
( # open capture group
[^\d\n]+ # match anything that isn't a digit or a newline that occurs once or more
) # close capture group
\d* # zero or more digits
.* # anything zero or more times
$ # define the end of the string
g # global
m # multi line
The greedy matching will mean that by default you will match the capture group and stop capturing as soon as either a digit or anything that isn't matched in the capture group or the end of the string it encountered.
I did not select a correct answer because the correct answer was left in the comments. "^\D+"
I am working in java, so putting it all together I got:
Pattern p = Pattern.compile("^\\D+");
Matcher m = p.matcher("Testing123Testing");
String extracted = m.group(1);
Use the character class feature: [...]
Identify numbers: [0-9]
Negate the class: [^0-9]
Allow as many as you like: [^0-9]*
