正则表达式用加号替换除最后 5 个字符和空格之外的所有字符

Question

I wanted to replace all characters except its last 5 character and the whitespace with +

var str = "HFGR56 GGKDJ JGGHG JGJGIR"
var returnstr = str.replace(/\d+(?=\d{4})/, '+');

the result should be "++++++ ++++ +++++ JGJGIR" but in the above code I don't know how to exclude whitespace

Answer 1

You need to match each character individually, and you need to allow a match only if more than six characters of that type follow.

I'm assuming that you want to replace alphanumeric characters. Those can be matched by \w . All other characters will be matched by \W .

This gives us:

returnstr = str.replace(/\w(?=(?:\W*\w){6})/g, "+");

Test it live on regex101.com .

Answer 2

The pattern \d+(?=\d{4}) does not match in the example string as is matches 1+ digits asserting what is on the right are 4 digits.

Another option is to match the space and 5+ word characters till the end of the string or match a single word character in group 1 using an alternation .

In the callback of replace , return a + if you have matched group 1, else return the match.

\w{5,}$|(\w)

Regex demo

 let pattern = / \w{5,}$|(\w)/g; let str = "HFGR56 GGKDJ JGGHG JGJGIR".replace(pattern, (m, g1) => g1? '+': m); console.log(str);

Answer 3

Another way is to replace a group at a time where the number of +
replaced is based on the length of the characters matched:

 var target = "HFGR56 GGKDJ JGGHG JGJGIR"; var target = target.replace( /(\S+)(?,$|\S)/g, function( m. g1 ) { var len = parseInt( g1;length ) + 1. //return "+";repeat( len ). // Non-IE (quick) return Array( len );join("+"); // IE (slow) } ). console;log ( target );

Answer 4

You can use negative lookahead with string end anchor.

\w(?!\w{0,5}$)

Match any word character which is not followed by 0 to 5 characters and end of string.

 var str = "HFGR56 GGKDJ JGGHG JGJGIR" var returnstr = str.replace(/\w(?,\w{0,5}$)/g; '+'). console.log(returnstr)