节点打字稿：正则表达式允许至少一个带有 joi 验证的字母字符的空格

Question

I am using joi and regex to validate user input, I would like to accept only alpha num caracters where whitespace can be allowed. I've done the following :

const schema = Joi.object({
  value: Joi.string()
    .regex(
      /^[a-zA-Z0-9 ]*$/,
      'message',
    )
    .max(100)
    .required()
...

The problem is that now I allow user to fill a full string composed only of whitespaces, and that's not what I want, I accept whitespaces, only if they are between alphanum caracters. I am using this to test : https://www.regextester.com/104025

Answer 1

Try this one:

/^[a-zA-Z0-9]+[a-zA-Z0-9 ]*[a-zA-Z0-9]+$/

I dont know if this is the easiest or most elegant way but [a-zA-Z0-9]+ matches one or more of [a-zA-Z0-9] using the +

I added this to the front and end of your regex.

Test and explanation: https://regex101.com/r/79uEWb/2/

Answer 2

Specify at the beginning that space is not allowed [^ ]

 const withSpaceInFront = " hello"; const text = "hello there" const textAndNumbers = "hello 5" const textWithSpecialChars = "hello#" const numeric = 123 const textWithSpecialCharsAllowed = "general Kenobi..?" const regexNoSpace = /^[^ ][a-zA-Z0-9 ]*$/; console.log("no space in front") console.log(regexNoSpace.test(withSpaceInFront)); console.log(regexNoSpace.test(text)); console.log(regexNoSpace.test(textAndNumbers)); console.log(regexNoSpace.test(textWithSpecialChars)); console.log(regexNoSpace.test(numeric)); //if you want to accept special chars like ? or . just add them in your list of accepted chars const regexAllowedSpecialChars = /^[^ ][a-zA-Z0-9 ?.]*$/; console.log("no space in front & specific chars allowed") console.log(regexAllowedSpecialChars.test(textWithSpecialCharsAllowed));

Regex tests: https://regex101.com/r/8Ulpu9/1/