According to the docs of Zod I can use trim, but trim will only remove the whitespace characters at the end and beginning. I am looking for a way to remove all of the whitespace characters. To give you an idea of the setup I am using (removed irrelevant code):

const stp = zodString({
    invalid_type_error: '',
    required_error: '',
}).trim()
    .min(1, { message: '' })
    .max(10, { message: '' })
    .regex(reg, { message: '', });

Is there a way to implement some sort of trim to remove all white space with zodString?

According to the docs of Zod I can use trim, but trim will only remove the whitespace characters at the end and beginning. I am looking for a way to remove all of the whitespace characters. To give you an idea of the setup I am using (removed irrelevant code):

const stp = zodString({
    invalid_type_error: '',
    required_error: '',
}).trim()
    .min(1, { message: '' })
    .max(10, { message: '' })
    .regex(reg, { message: '', });

Is there a way to implement some sort of trim to remove all white space with zodString?

• javascript
• typescript
• zod

• 2 "Trimming space" in the middle of a string is odd. Could you provide sample data demonstrating this? I can see compacting multiple spaces into one space. – Mr. Polywhirl Commented Sep 19, 2023 at 13:20
• Well for example, you just wrote "Trimming space". I wish to have the white character in between "Trimming space" removed. – I try so hard but I cry harder Commented Sep 19, 2023 at 13:21
• 1 Does this answer your question? Replace all whitespace characters – isherwood Commented Sep 19, 2023 at 13:21
• 1 Wouldn't you just need z.string().regex(/^\S+$/); – Mr. Polywhirl Commented Sep 19, 2023 at 13:23
• @isherwood No, because that would be done outside of the zodstring function. I am looking for a built-in way with zodString to resolve this. – I try so hard but I cry harder Commented Sep 19, 2023 at 13:24

3 Answers 3

Using a pipe, you can first remove the white spaces and then perform additional validations.

z.string().transform(value => value.replace(/\s+/g, ''))
.pipe(z.string().min(1, { message: 'This field is required' }))

zod supports the .trim() transformation

docs

const mySchema = z.object({ mykey: z.string().trim() })

The method .trim() you used in the code snippet will remove the whitespaces from both ends of the string.

If for some reason you want to filter all whitespaces use .transform()

z.string().transform(value => value.replaceAll(" ", ""))