I'm trying to replace part of a string with the same number of dummy characters in JavaScript, for example: '==Hello==' with '==~~~~~=='.
This question has been answered using Perl and PHP, but I can't get it to work in JavaScript. I've been trying this:
txt=txt.replace(/(==)([^=]+)(==)/g, "$1"+Array("$2".length + 1).join('~')+"$3");
The pattern match works fine, but the replacement does not - the second part adds '~~' instead of the length of the pattern match. Putting the "$2" inside th开发者_运维问答e parentheses doesn't work. What can I do to make it insert the right number of characters?
Use a function for replacement instead:
var txt = "==Hello==";
txt = txt.replace(/(==)([^=]+)(==)/g, function ($0, $1, $2, $3) {
return $1 + (new Array($2.length + 1).join("~")) + $3;
});
alert(txt);
//-> "==~~~~~=="
The issue with the expression
txt.replace(/(==)([^=]+)(==)/g, "$1"+Array("$2".length + 1).join('~')+"$3")
is that "$2".length
forces $2
to be taken as a string literal, namely the string "$2"
, that has length 2.
From the MDN docs:
Because we want to further transform the result of the match before the final substitution is made, we must use a function.
This forces evaluation of the match before the transformation.
With an inline function as parameter (and repeat
) -- here $1, $2, $3
are local variables:
txt.replace(/(==)([^=]+)(==)/g, (_,$1,$2,$3) => $1+'~'.repeat($2.length)+$3);
txt = '==Hello==';
//inline function
console.log(
txt.replace(/(==)([^=]+)(==)/g, (_, g1, g2, g3) => g1 + '~'.repeat(g2.length) + g3)
);
The length attribute is being evaluated before the $2 substitution so replace() won't work. The function call suggested by Augustus should work, another approach would be using match() instead of replace().
Using match() without the /g, returns an array of match results which can be joined as you expect.
txt="==Hello==";
mat=txt.match(/(==)([^=]+)(==)/); // mat is now ["==Hello==","==","Hello","=="]
txt=mat[1]+Array(mat[2].length+1).join("~")+mat[3]; // txt is now "==~~~~~=="
You excluded the leading/trailing character from the middle expression, but if you want more flexibility you could use this and handle anything bracketed by the leading/trailing literals.
mat=txt.match(/(^==)(.+)(==$)/)
A working sample uses the following fragment:
var processed = original.replace(/(==)([^=]+)(==)/g, function(all, before, gone, after){
return before+Array(gone.length+1).join('~')+after;
});
The problem in your code was that you always measured the length of "$2"
(always a string with two characters). By having the function you can measure the length of the matched part. See the documentation on replace for further examples.
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