I have a variable:
var str = "@devtest11 @devtest1";
I use this way to replace @devtest1 with another string:
str.replace(new RegExp('@devtest1', 'g'), "aaaa")
However, its result (aaaa1 aaaa) is not what I expect. The expected result is: @devtest11 aaaa. I just want to replace the whole word @devtest1.
How can I do that?
Use the \b zero-width word-boundary assertion.
var str = "@devtest11 @devtest1";
str.replace(/@devtest1\b/g, "aaaa");
// => @devtest11 aaaa
If you need to also prevent matching the cases like hello@devtest1, you can do this:
var str = "@devtest1 @devtest11 @devtest1 hello@devtest1";
str.replace(/( |^)@devtest1\b/g, "$1aaaa");
// => @devtest11 aaaa
"!!@devtest1" and a pattern of /\b@devtest1\b/. There is no word-boundary between "!" and "@". (That is, the definition of word boundary is specific; but not always appropriate.)
hello@devtest1 which is not what op would expectUse word boundary \b for limiting the search to words.
Because @ is special character, you need to match it outside of the word.
\b assert position at a word boundary (^\w|\w$|\W\w|\w\W), since \b does not include special characters.
var str = "@devtest11 @devtest1";
str = str.replace(/@devtest1\b/g, "aaaa");
document.write(str);If your string always starts with @ and you don't want other characters to match
var str = "@devtest11 @devtest1";
str = str.replace(/(\s*)@devtest1\b/g, "$1aaaa");
// ^^^^^ ^^
document.write(str);\b between @ and devtest1 is useless, as it will always match (the neighbours being constant).@\bd is a bit useless; it is equivalent to @d. Putting a \b before @ is also problematic .. but it can be omitted for the sown data.hello@devtest1
\b won't work properly if the words are surrounded by non space characters..I suggest the below method
var output=str.replace('(\s|^)@devtest1(?=\s|$)','$1aaaa');