Evaluate the following expression using meta character for regular expression:
‘[^Ale|ax.r$]’
Which two matches would be returned by this expression? (Choose two.)
A.
Alex
B.
Alax
C.
Alxer
D.
Alaxendar
E.
Alexender
Which two matches would be returned by this expression?
Evaluate the following expression using meta character for regular expression:
‘[^Ale|ax.r$]’
Which two matches would be returned by this expression? (Choose two.)
can anyone explain above question plz…
‘[Ale|ax.r$]’ is equal to ‘[Al(e|a)x.r$]’ am i correct?
I think that ‘^’ in this statement [^Ale|ax.r$] is a mistake. Because otherwise the answer is incorrect.
I think this function regexp_like respond the question :
select case when regexp_like (string,'[^Ale|ax.r$]’) then ‘MATCH’ else ‘NO MATCH’ end as output from dual;
string is A, B, C or D therefore D and E is response.
Thanks.
D and E
1) Pattern ‘[Aleaxr$]’ matches all characters in A to E, except characters ‘n’ and ‘d’ in D and E.
2) [^ (caret sign after opening square brackets), means “not equals”,
which INVERSES the matches found in 1), thus excluding A,B,and C.
‘[^Al(e|a)x.r$]
^ : start of string should be “Al”
| : after Al, character should be either “e or a”
$ : end of the string should be “r”
so answer D, E are correct.
I am sure that the question is defective. The [] makes it a set. But you cant use the “|”(or) Operation in a set.
CASE 1)
So lets say the [] is the fault. Then we search for that pattern: ‘^Ale|ax.r$’.
So we have an initial A followed by l, followed by either e or a, followed by x, followed by a random sign, followed by an r which has to be the last sign as well.
Non of A-E is matching this pattern.
CASE 2)
Lets say the | is the fault. Then we search for a word not containing of any of the following signs A l e a x . r $ (because ^ in a the beging of a “[]” negotiates). Any not empty word (NULL) wouldnt be matched by that because it would contain at least “.” as “any sign”.
To put that in a nutshell: The question is retarded.
It appears that answer is actually right. Check yourself, for example here: livesql.oracle.com
Since the string is given in brackets []. It is not a string it’s a matching list. which means that characters can be placed in any order.
[^] – negates expression. which equals to anything but…
| and $ within [] – mean nothing, so you can just ignore them
So basically expression equals to “anything that has more than just letters A l e a x r. And there just 2 values that qualify for that ‘Alaxendar’ and ‘Alexender’ since they have letters ‘n’ and ‘d’.
select * from
(SELECT ‘Alex’ a from dual
union all
SELECT ‘Alax’ from dual
union all
SELECT ‘Alxar’ from dual
union all
SELECT ‘Alaxendar’ from dual
union all
SELECT ‘Alexender’ from dual
)
where regexp_like(a,'[^Ale|ax.r$]’)
A
Alaxendar
Alexrende
2 rows selected.
I think, you have the right explanation. thanks
I choose