Why does this comparison return ? (Regular Expressions)

Question

1)mysql> SELECT 'aXbc' REGEXP '[a-dXYZ]';                 -> 1
2)mysql> SELECT 'aXbc' REGEXP '^[a-dXYZ]$';               -> 0    
3)mysql> SELECT 'aXbc' REGEXP '^[a-dXYZ]?$';              -> 0   // 0 or 1
4)mysql> SELECT 'aXbc' REGEXP '^[a-dXYZ]+$';              -> 1   // 1 or more
5)mysql> SELECT 'aXbc' REGEXP '^[a-dXYZ]*$';              -> 1   //0 or more

I'm confused about the second comparison. Doesn't it mean that a string which starts with [a-dXYZ] and ends with [a-dXYZ]?

Or does it mean that a string which starts with [a-dXYZ] and ends with [a-dXYZ] whose length is 1? If this is true, using ^ and $ symbols like '^....$' (leftmost and rightmost) eliminates all substrings and analyses only the string as a whole - is that right?

Note: This is not a how can I make it question. I want to understand a cause and effect relation.

Andreas Louv · Accepted Answer · 2016-04-17T18:39:21.663

2

^[a-dXYZ]$ will match exactly one character which it either: a, b, c, d, X, Y, Z and is the first character and the last. So your second observation is pretty straight on.

^ means start of input and $ means end of input.

You can use * to repeat 0 or more times or + to repeat 1 or more times:

^[a-dXYZ]+$

If you want to match a string starting and ending with [a-dXYZ], you could use the following:

^[a-dXYZ].*[a-dXYZ]$
#        ^ Match anything zero or more times

In some regex implementations you can use the m modifier which will make ^ and $ match beginning and end of lines, rater than input, consider the following Perl snippet:

s/^1+$/_/gm

It will replace all lines only consistent of ones:

% seq 1 11 | tac | perl -p0e 's/^1+$/_/gm'
_
10
9
8
7
6
5
4
3
2
_

edited Apr 17 '16 at 18:39

answered Apr 17 '16 at 18:26

Andreas Louv

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^[a-dXYZ]$ it looks for 1 length string as a whole, doesn't consider substrings ? – Burak Karasoy Apr 17 '16 at 18:28
It will match a input which is 1 character long. – Andreas Louv Apr 17 '16 at 18:31
am i allowed to use ^ $ symbols in the middle of string if so does '^[B](.*)(^[K]$)(.*)[B]$' return true with BAAKAAB. or it depends on where to use ?(oracle sql, mysql , java) – Burak Karasoy Apr 17 '16 at 18:59
@BurakKarasoy As I said before: `^ and $ will match beginning and end of input`. So using these metacharacters in the middle of a match will make the match invalid. – Andreas Louv Apr 17 '16 at 19:03
answer of this question confused me http://stackoverflow.com/questions/15467249/regular-expressions-middle-of-string and also thanks for your time. – Burak Karasoy Apr 17 '16 at 19:06
@BurakKarasoy That answer uses something called lookarounds: http://www.regular-expressions.info/lookaround.html – Andreas Louv Apr 17 '16 at 19:13
owner of that question should have mentioned that in the question.Thanks again for your effort – Burak Karasoy Apr 17 '16 at 19:19

Why does this comparison return ? (Regular Expressions)

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