How can I verify in Javascript that a string contains at least one set of letters and numbers using regex?

Question

How can I verify in Javascript that a string contains at least one set of letters and numbers using regex?

I am currently trying to decipher the appropriate regex for identifying a pattern similar to 2d1h30m10s, accepting any valid variation such as 1h, 1h30m, 30m, 10s, or any combination of these. Could regex be the right solution in this scenario?

I have been attempting to comprehend it, yet despite my efforts, all I receive is false when running various tests:

/^(0?[1-9]|1[0-2][h])([1-6][0-9][m])([1-6][0-9][s])\d$/.test('2d1h10m10s')
/^(0?[1-9]|1[0-2][h])([1-6][0-9][m])([1-6][0-9][s])\d$/.test('10m10s')
/^(0?[1-9]|1[0-2][h])([1-6][0-9][m])([1-6][0-9][s])\d$/.test('10s')

What elements am I overlooking here?

javascript regex

Answer 1

Answer №1

To make each section of the regular expression optional, you should exclude that unit.

Remove [h] from one of the alternatives to match both 12h and 01h.

Avoid using \d at the end.

The pattern does not account for single-digit minutes or seconds. Additionally, there is no need for 60s or 60m as they can be represented as 1m and 1h respectively.

/^((0?[1-9]|1[0-2])h)?([1-5]?[0-9]m)?([1-5]?[0-9]s)?$/

DEMO

It is unnecessary to enclose h, m, and s in square brackets since they are single characters.

Note that with each unit being optional, the pattern will also match an empty string. It's recommended to validate this separately from the regular expression.

Answer 2

To make each section of the regular expression optional, you should exclude that unit.

Remove [h] from one of the alternatives to match both 12h and 01h.

Avoid using \d at the end.

The pattern does not account for single-digit minutes or seconds. Additionally, there is no need for 60s or 60m as they can be represented as 1m and 1h respectively.

/^((0?[1-9]|1[0-2])h)?([1-5]?[0-9]m)?([1-5]?[0-9]s)?$/

DEMO

It is unnecessary to enclose h, m, and s in square brackets since they are single characters.

Note that with each unit being optional, the pattern will also match an empty string. It's recommended to validate this separately from the regular expression.

Answer 3

Answer №2

Perhaps this will be the solution

\d+\w+

If my comprehension is accurate, you are attempting to verify if a string includes characters that come after numbers.

Answer 4

Perhaps this will be the solution

\d+\w+

If my comprehension is accurate, you are attempting to verify if a string includes characters that come after numbers.

Answer 5

Answer №3

Give this a shot:

^(\d{0,2}[hms]){0,3}$

See some examples here: https://regex101.com/r/NV6oEu/1

Answer 6

Give this a shot:

^(\d{0,2}[hms]){0,3}$

See some examples here: https://regex101.com/r/NV6oEu/1

Answer 7

Answer №4

Here is the correct format for the expression you are trying to use:

/^((0?[1-9]|1[0-2])h)?((([0-5]?[0-9])|60)m)?((([0-5]?[0-9])|60)s)?$/

The original expression you had provided has a couple of issues: 1. the use of \d at the end, and 2. the capturing scope (extra parentheses in my corrected expression).

With this new expression, it will also be able to capture formats like 1h13s.

Answer 8

Here is the correct format for the expression you are trying to use:

/^((0?[1-9]|1[0-2])h)?((([0-5]?[0-9])|60)m)?((([0-5]?[0-9])|60)s)?$/

The original expression you had provided has a couple of issues: 1. the use of \d at the end, and 2. the capturing scope (extra parentheses in my corrected expression).

With this new expression, it will also be able to capture formats like 1h13s.

How can I verify in Javascript that a string contains at least one set of letters and numbers using regex?

Answer №1

Answer №2

Answer №3

Answer №4

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