Need regex for a string having characters followed by even number of digits
Can anyone tell how I can write regex for a string that take one or more alphanumeric character followed by an even number of digits?
Valid:
a11a1121
bbbb11a1121
Invalid:
a11a1
I have tried ^[a-zA-Z*20-9]*$ but it is always giving true.
Can you please help in this regard?
java regex
edited Nov 12 at 11:53
Biffen
4,16352129
asked Nov 12 at 11:47
Icche Guri
96211130
add a comment |
Can anyone tell how I can write regex for a string that take one or more alphanumeric character followed by an even number of digits?
Valid:
a11a1121
bbbb11a1121
Invalid:
a11a1
I have tried ^[a-zA-Z*20-9]*$ but it is always giving true.
Can you please help in this regard?
java regex
edited Nov 12 at 11:53
Biffen
4,16352129
asked Nov 12 at 11:47
Icche Guri
96211130
2
Possible duplicate of Learning Regular Expressions
– Biffen
Nov 12 at 11:52
add a comment |
Can anyone tell how I can write regex for a string that take one or more alphanumeric character followed by an even number of digits?
Valid:
a11a1121
bbbb11a1121
Invalid:
a11a1
I have tried ^[a-zA-Z*20-9]*$ but it is always giving true.
Can you please help in this regard?
java regex
edited Nov 12 at 11:53
Biffen
4,16352129
asked Nov 12 at 11:47
Icche Guri
96211130
Can anyone tell how I can write regex for a string that take one or more alphanumeric character followed by an even number of digits?
Valid:
a11a1121
bbbb11a1121
Invalid:
a11a1
I have tried ^[a-zA-Z*20-9]*$ but it is always giving true.
Can you please help in this regard?
java regex
java regex
edited Nov 12 at 11:53
Biffen
4,16352129
asked Nov 12 at 11:47
Icche Guri
96211130
edited Nov 12 at 11:53
Biffen
4,16352129
asked Nov 12 at 11:47
Icche Guri
96211130
edited Nov 12 at 11:53
Biffen
4,16352129
edited Nov 12 at 11:53
Biffen
4,16352129
edited Nov 12 at 11:53
Biffen
4,16352129
4,16352129
asked Nov 12 at 11:47
Icche Guri
96211130
asked Nov 12 at 11:47
Icche Guri
96211130
asked Nov 12 at 11:47
Icche Guri
96211130
96211130
2
Possible duplicate of Learning Regular Expressions
– Biffen
Nov 12 at 11:52
add a comment |
2
Possible duplicate of Learning Regular Expressions
– Biffen
Nov 12 at 11:52
2
2
Possible duplicate of Learning Regular Expressions
– Biffen
Nov 12 at 11:52
Possible duplicate of Learning Regular Expressions
– Biffen
Nov 12 at 11:52
add a comment |
2 Answers
2
active
oldest
votes
The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]
You can break down your requirement to groups and then handle it accordingly.
Like you require at least one character. so you start with ^([a-zA-Z]+)$
Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$
Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$
You can use online tools like regex101 for these purpose. The provided regex in action here
answered Nov 12 at 12:01
suvartheec
1,121917
add a comment |
You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$
Explanation :
[a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one
([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use([0-9]{2})+instead.
answered Nov 12 at 11:54
Amessihel
1,9631723
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]
You can break down your requirement to groups and then handle it accordingly.
Like you require at least one character. so you start with ^([a-zA-Z]+)$
Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$
Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$
You can use online tools like regex101 for these purpose. The provided regex in action here
answered Nov 12 at 12:01
suvartheec
1,121917
add a comment |
The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]
You can break down your requirement to groups and then handle it accordingly.
Like you require at least one character. so you start with ^([a-zA-Z]+)$
Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$
Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$
You can use online tools like regex101 for these purpose. The provided regex in action here
answered Nov 12 at 12:01
suvartheec
1,121917
add a comment |
The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]
You can break down your requirement to groups and then handle it accordingly.
Like you require at least one character. so you start with ^([a-zA-Z]+)$
Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$
Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$
You can use online tools like regex101 for these purpose. The provided regex in action here
answered Nov 12 at 12:01
suvartheec
1,121917
The regex that you have mentioned will search for any number of [either a-z, or A-Z or 2 or 0-9]
You can break down your requirement to groups and then handle it accordingly.
Like you require at least one character. so you start with ^([a-zA-Z]+)$
Then you need numbers in the multiple of 2. so you add ^([a-zA-Z]+(dd)+)$
Now you need any number of combination of these. So the exp becomes: ^([a-zA-Z]+(dd)+)*$
You can use online tools like regex101 for these purpose. The provided regex in action here
answered Nov 12 at 12:01
suvartheec
1,121917
answered Nov 12 at 12:01
suvartheec
1,121917
answered Nov 12 at 12:01
suvartheec
1,121917
answered Nov 12 at 12:01
suvartheec
1,121917
1,121917
add a comment |
add a comment |
You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$
Explanation :
[a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one
([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use([0-9]{2})+instead.
answered Nov 12 at 11:54
Amessihel
1,9631723
add a comment |
You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$
Explanation :
[a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one
([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use([0-9]{2})+instead.
answered Nov 12 at 11:54
Amessihel
1,9631723
add a comment |
You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$
Explanation :
[a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one
([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use([0-9]{2})+instead.
answered Nov 12 at 11:54
Amessihel
1,9631723
You can achieve it with this regexp: ^[a-z0-9]*[a-z]+([0-9]{2})*$
Explanation :
[a-z0-9]*[a-z]+: a string of at least one character terminated by a non digit one
([0-9]{2})*: an odd sequence of digits (0 or 2*n digits). If the even sequence cannot be null, use([0-9]{2})+instead.
answered Nov 12 at 11:54
Amessihel
1,9631723
edited Nov 12 at 12:29
answered Nov 12 at 11:54
Amessihel
1,9631723
answered Nov 12 at 11:54
Amessihel
1,9631723
answered Nov 12 at 11:54
Amessihel
1,9631723
1,9631723
add a comment |
add a comment |
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2
Possible duplicate of Learning Regular Expressions
– Biffen
Nov 12 at 11:52