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What is the most efficient way with regex to match:
test
test case
tester
The last one should not match.
Use a word boundary:
/test\b/
This means that the word must end after test.
\btest\b
\b is the word boundary anchor.
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what is the regular expression to validate a string. the string should contain
all lower case
_ is allowed in the string.
Thanks in advance.
You need to use regex pattern
^[a-z_]*$
You can use a POSIX class as an alternative to a simple character class to achieve this:
[[:lower:]_]*
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I am writing regex to match the pattern like this.
abc:123-12-4
abc: It should be exact match including colon
123 Number match any length
- Exact Match
12 Number Match any length
- Exact match
4 Exact match
Any ideas how it can be done in a simpler way.
You can use this regex:
^abc:[0-9]+-[0-9]+-4$
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I have the following line of XML:
<indexentry><secondaryie>definition, 3/2/4</secondaryie></indexentry>
And I need a regex that matches the above and converts it as below:
ABC3(the first number)/P-2(second number)-4(third number)
How can I do this?
Use this regex:
([0-9]+)/([0-9]+)/([0-9]+)
And from captured groups #1, #2, #3 make your string:
ABC3\1/P-2\2-4\3
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I need a regular expression to only match rising numbers.
ex: 22335566 66678
but not: 444663 33997777666664
the length of the number is not fixed nor is the starting digit.
Any help please?
The best way to implement this using regex is:
^(?=\d)1*2*3*4*5*6*7*8*9*$
The regex matches:
0 or more 1's, followed by,
0 or more 2's, followed by,
0 or more 3's... and so on.
(?=\d) ensures that there is atleast 1 digit in your string.
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I currently have this Regex:
.*[a-z]$|.*[A-Z]$|.*[0-9]$
I need to add the following to the existing - /:.#[]()
Does anyone know how to do it?
You don't need multiple OR conditions. Have only one and include special characters also:
[0-9A-Za-z\/#:.()\[\]]
Per your comments, this should do it for you:
^[a-zA-Z0-9/:.#\[\]()]+$
This will match a string containing only the characters you've specified.