Regex for Alternating Numbers
Question:
I am trying to write a regex pattern for phone numbers consisting of 9 fixed digits.
I want to identify numbers that have two numbers alternating for four times such as 5XYXYXYXY
I used the below sample
number = 561616161
I tried the below pattern but it is not accurate
^5(d)(?=d1).+
can someone point out what i am doing wrong?
Answers:
I would use:
^(?=d{9}$)d*(d)(d)(?:12){3}d*$
Demo
Here is an explanation of the pattern:
^ from the start of the number(?=d{9}$) assert exactly 9 digitsd* match optional leading digits(d) capture a digit in 1(d) capture another digit in 2(?:12){3} match the XY combination 3 more timesd* more optional digits$ end of the number
If you want to repeat 4 times 2 of the same pairs and matching 9 digits in total:
^(?=d{9}$)d*(dd)1{3}d*$
Explanation
^ Start of string(?=d{9}$) Positive lookahead, assert 9 digits till the end of the stringd* Match optional digits(dd)1{3} Capture group 1, match 2 digits and then repeat what is captured in group 1 3 timesd* Match optional digits$ End of string
If you want to match a pattern repeating 4 times 2 digits where the 2 digits are not the same:
^(?=d{9}$)d*((d)(?!2)d)1{3}d*$
Explanation
^ Start of string(?=d{9}$) Positive lookahead, assert 9 digits till the end of the stringd* Match optional digits( Capture group 1
(d) Capture group 2, match a single digit(?!2)d Negative lookahead, assert not the same char as captured in group 2. If that is the case, then match a single digit
) Close group 11{3} Repeat the captured value of capture group 1 3 timesd* Match optional digits$ End of string
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My first guess from OP’s self tried regex ^5(d)(?=d1).+ without any own additions was a regex is needed to verify numbers starting with 5 and followed by 4 pairs of same two digits.
^5(dd)1{3}$
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The same idea with the "added guess" to disallow all same digits like e.g. 511111111
^5((d)(?!2)d)1{3}$
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Guessing further that 5 is a variable value and assuming if one variable at start/end with the idea of taking out invalid values early – already having seen the other nice provided answers.
^(?=d?(dd)1{3})d{9}$
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Solution 3 with solution 2’s assumption of two different digits in first pairing.
^(?=d?((d)(?!2)d)1{3})d{9}$
Solutions 3 and 4 are most obvious playings with @4thBird’s nice answer in changed order.
I am trying to write a regex pattern for phone numbers consisting of 9 fixed digits.
I want to identify numbers that have two numbers alternating for four times such as 5XYXYXYXY
I used the below sample
number = 561616161
I tried the below pattern but it is not accurate
^5(d)(?=d1).+
can someone point out what i am doing wrong?
I would use:
^(?=d{9}$)d*(d)(d)(?:12){3}d*$
Demo
Here is an explanation of the pattern:
^from the start of the number(?=d{9}$)assert exactly 9 digitsd*match optional leading digits(d)capture a digit in1(d)capture another digit in2(?:12){3}match the XY combination 3 more timesd*more optional digits$end of the number
If you want to repeat 4 times 2 of the same pairs and matching 9 digits in total:
^(?=d{9}$)d*(dd)1{3}d*$
Explanation
^Start of string(?=d{9}$)Positive lookahead, assert 9 digits till the end of the stringd*Match optional digits(dd)1{3}Capture group 1, match 2 digits and then repeat what is captured in group 1 3 timesd*Match optional digits$End of string
If you want to match a pattern repeating 4 times 2 digits where the 2 digits are not the same:
^(?=d{9}$)d*((d)(?!2)d)1{3}d*$
Explanation
^Start of string(?=d{9}$)Positive lookahead, assert 9 digits till the end of the stringd*Match optional digits(Capture group 1(d)Capture group 2, match a single digit(?!2)dNegative lookahead, assert not the same char as captured in group 2. If that is the case, then match a single digit
)Close group 11{3}Repeat the captured value of capture group 1 3 timesd*Match optional digits$End of string
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My first guess from OP’s self tried regex
^5(d)(?=d1).+without any own additions was a regex is needed to verify numbers starting with5and followed by 4 pairs of same two digits.^5(dd)1{3}$ -
The same idea with the "added guess" to disallow all same digits like e.g.
511111111^5((d)(?!2)d)1{3}$ -
Guessing further that
5is a variable value and assuming if one variable at start/end with the idea of taking out invalid values early – already having seen the other nice provided answers.^(?=d?(dd)1{3})d{9}$ -
Solution 3 with solution 2’s assumption of two different digits in first pairing.
^(?=d?((d)(?!2)d)1{3})d{9}$
Solutions 3 and 4 are most obvious playings with @4thBird’s nice answer in changed order.