在trim（）之后获取字符串的最后一个字符不起作用

Question

How to remove the white spaces from string, in order to get the last character?

const email = "abc@";
const emailWhiteSpace = "abc@ ";
console.log(email.trim()[email.length - 1]) //==> @
console.log(emailWhiteSpace.trim()[emailWhiteSpace.length - 1]) //==> undefinied

Any idea how to solve this issue?

Answer 1

It might be easier to use a regular expression - match a non-space character, followed by space characters, followed by the end of the line ( $ ):

 const lastChar = str => str.match(/(\\S)(?=\\s*$)/)[1]; console.log(lastChar("abc@")); console.log(lastChar("abc@ ")); 

Of course, you can also save the trimmed text in a variable:

 const lastChar = str => { const trimmed = str.trim(); return trimmed[trimmed.length - 1]; }; console.log(lastChar("abc@")); console.log(lastChar("abc@ ")); 

Answer 2

You need to also refer to the trimmed string when accessing the length:

 const email = "abc@"; const emailWhiteSpace = "abc@ "; console.log(email.trim()[email.length - 1]) //==> @ console.log(emailWhiteSpace.trim()[emailWhiteSpace.trim().length - 1]) // ^^^^^^^^ 

But a better approach, which would have avoided your error in the first place, would be to just assign the trimmed string to an actual variable:

 var emailWhiteSpace = "abc@ "; var emailWhiteSpaceTrimmed = emailWhiteSpace.trim(); console.log(emailWhiteSpaceTrimmed[emailWhiteSpaceTrimmed.length - 1]) 

Answer 3

这是因为emailWhiteSpace.trim()没有改变字符串，它返回一个新的，这意味着emailWhiteSpace.trim().length与emailWhiteSpace.length 。

Answer 4

You can use trimEnd() to trim the last space.

 const emailWhiteSpace = "abc@ "; console.log(emailWhiteSpace.trimEnd()[emailWhiteSpace.trimEnd().length-1]) /* You can also use a separate variable */ const trimmedEmail = emailWhiteSpace.trimEnd(); console.log(trimmedEmail[trimmedEmail.length-1]); 

Answer 5

You've put trim on the wrong place. Should be

emailWhiteSpace[emailWhiteSpace.trim().length - 1] //-> "@"

This is because of the emailWhiteSpace.length returns an untrimmed string (with the whitespace). Thus you have either assign the trim result to some variable or to put it in the indexer as shown. This would depend on what are you're trying to achieve.

Answer 6

mine would be ..

const email = "abc@";
const emailWhiteSpace = "abc@ ";

const f = function( v) { v = String(v); return v[v.length-1]};

console.log( f( email.trim())) //==> @
console.log( f( emailWhiteSpace.trim())) //==> @