仅当 % 出现在字符串的第一个或最后一个时，如何从字符串中删除 %

Question

I want to remove the % character from my string if the % character present in the string, then it should check whether it is in the beginning or end then it should trim the value then the provide the result.

Eg: var str = "Value%" or "%Value" or "%Value%"
The result should be = Value.
Eg: var str="Va%ue"
The result should be =Va%ue.
Eg: var str= "Value"
The result should be = Value.

Thanks in Advance

Answer 1

str = (str[0] == '%' || str.endsWith('%')? str.replace(/%/g, '') : str);

Check if the string starts or ends with % before replacing

Answer 2

Your regex basically needs to have two alternatives combined with an 'or' sign. You use ^ to signal beginning and $ to signal ending, then combine them with | .

The regex for percent lookup: ^%|%$ .

If you put that in to the replace() function and add the global lookup flag g , you can easily achieve what you're looking for:

const percentLookupRegex = /^%|%$/g;

str.replace(percentLookupRegex, '');

Here's a live example: https://regex101.com/r/nAK64n/2

Answer 3

If you want to use regex then you should look into the anchors ^ and $

Eg.

str.replace(/^%/, '');

Will replace % in the beginning of the line with nothing.

An alternative approach is to use startsWith and endsWith and then slice the string appropriately:

if (str.startsWith('%') {
   str = str.slice(1);
}

Removing a % at the end of the string is left as an exercise for the reader

Answer 4

str = (str.indexOf('%') === 0) ? str.substring(1) : str; // check for first character
str = (str.lastIndexOf('%') === (str.length - 1)) ? str.substring(0, str.length - 1) : str; //check for last character

This will only check for first and last character and remove only those