JavaScriptTypeScript: Concatenate string variables and string arrays with accordingly set spaces - Stack Overflow

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Situation

I want to concatenate string variables and string arrays in TypeScript like in this simple example:

// var1, var2: string
// arr1: string[]
const myString = 'result:' + var1 + ' ' + arr1.join(' ') + ' ' var2;

I don't want multiple spaces concatenated together and there shouldn't be any spaces at the end.

Problem

It is possible that those variables are set to '' or that the array is empty. These leads to multiple spaces concatenated together which I want to avoid.

Question

Is there a more elegant way to concatenate only the set variables separated with exactly one space?

Conditions

• The solution should work with more variables/arrays
• The solution should be a one-liner

Situation

I want to concatenate string variables and string arrays in TypeScript like in this simple example:

// var1, var2: string
// arr1: string[]
const myString = 'result:' + var1 + ' ' + arr1.join(' ') + ' ' var2;

I don't want multiple spaces concatenated together and there shouldn't be any spaces at the end.

Problem

It is possible that those variables are set to '' or that the array is empty. These leads to multiple spaces concatenated together which I want to avoid.

Question

Is there a more elegant way to concatenate only the set variables separated with exactly one space?

Conditions

• The solution should work with more variables/arrays
• The solution should be a one-liner

• javascript
• string
• typescript

Share Improve this question Follow asked Sep 9, 2020 at 8:10 winklerrrwinklerrr 14.9k99 gold badges8181 silver badges9696 bronze badges 8

• 1 const myString = 'result:' + [var1, ...arr1, var2].filter(s => s?.length ?? 0 > 0).join(' '); – Aluan Haddad Commented Sep 9, 2020 at 8:32
• 1 @AluanHaddad Your solution looks really interesting but I don't understand it pletely. Could you please write it as an answer and explain why you use s? and ?? and how they work? Thank you in advance. – winklerrr Commented Sep 9, 2020 at 8:48
• 1 It's called Optional chaining. My ment is almost exactly the same as @SpoonMeiser's answer, which I hadn't noticed. s?.length evaluates to undefined if s` is undefined or null, where s.length would throw. ?? a fallback – Aluan Haddad Commented Sep 9, 2020 at 8:52
• 1 @AluanHaddad and that 0 > 0 part? – winklerrr Commented Sep 9, 2020 at 8:57
• 1 So, if s is null or undefined then s?.length will evaluate to undefined And undefined > 0 evaluates to false (as does undefined < 0) so s?.length ?? 0 specifies that 0 should be used as a fallback. The thought here is to treat null or undefined values as if they were empty strings, filtering them out. – Aluan Haddad Commented Sep 9, 2020 at 9:02

 |  Show 3 more ments

3 Answers 3

Sorted by: Reset to default 5

You can use destructuring to concatenate everything into a temporary array, and then use filter to remove any empty elements before calling join on the lot.

let var1 = "var1";
let arr1 = ["one", "", "two"];
let var2 = ""

const myString = "result:" + [var1, ...arr1, var2].filter(s => s).join(' ');

console.log(myString);

Does this satisfy your requirements?

const strArr = ['result:' + var1, ...arr1, var2].map((a) => a.trim()).join(' ')

Update

As suggested, following handles

1. empty string
2. string with multiple spaces
3. trim additional spaces

const strArr = ['result:' + var1, ...arr1, var2].map((a) => a.trim()).filter(Boolean).join(' ')

Update 2

If you are using ES2019, you could do this in single pass:

const strArr = 
['result:' + var1, ...arr1, var2]
         .flatMap((el) => Boolean(el.trim()) ? [el.trim()] :[]).join(' ')

You could try this: const cleanedResult = result.replace(/\s\s+/g, ' ').trim()

This would remove multiple spaces in your variables as well though.

If you do not want that you can try out this solution:

const result = 'result:' + (var1 ? var1 + ' ' : '')  + (arr1 ? arr1.join(' ') : '') + (var2 ? var2 + ' ' : '')

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