以符号开头的字符串替换n次

I am trying to replace a string starting with a specific symbol '@' with the symbol '%', but the condition is that the symbol should be at the start of the string.

我试图用符号“%”替换以特定符号“@”开头的字符串，但条件是符号应该在字符串的开头。

For eg.

如。

@@@hello@hi@@

@@@hello@hi@@

should be replaced by

应该被

%%%hello@hi@@

% % % hello@hi@@

I have come up with the regex that matches the starting '@' symbols, but I am able to replace it only once, instead of replacing it with the number of times it matched.

我想到了与开始的“@”符号匹配的regex，但是我只能替换它一次，而不能用它匹配的次数来替换它。

The code is

代码是

var str = "@@@hello@hi@@";
var exp = new RegExp('^@+', 'g');
var mystr = str.replace(exp, '%');

But, it outputs

但是,它输出

%hello@hi@@

% hello@hi@@

But, the intended output is

但是，预期的输出是

%%%hello@hi@@

% % % hello@hi@@

My current solution is something like this:

我目前的解决方案是这样的:

var str = "@@@hello@hi@@";
var match = str.match(/^@+/g)[0];

var new_str = str.replace(match, "");

var diff_count = str.length-new_str.length;
var new_sub_str = Array(diff_count+1).join("%")

var mystr = new_sub_str + new_str;

This solution does give me the intended output, but I am worried about the performance.

这个解决方案确实提供了预期的输出，但是我担心性能。

Is there any better way to achieve this ?

有更好的方法来实现这一点吗?

3 个解决方案

#1

You can use a callback function:

您可以使用回调函数:

var mystr = '@@@hello@hi@@'.replace(/^@+/g, function(match) {
  return Array(match.length + 1).join('%');
});
document.write(mystr);

The Array(n).join(s) construction is simply a shorthand way of repeating the string s n-1 times.

数组(n).join(s)结构只是简单地重复字符串s n-1次的一种简写方式。

#2

An interesting solution without regexp:

没有regexp的有趣解决方案:

var mystr = '@@@@@hello@hi@@'.split('').map(function(item) {
    if (item == '@' && !this.stop) {
      return '%';
    } else {
      this.stop = true;
      return item;
    }
  }, {}).join('');

  console.log(mystr);

And an alternative:

和另一个:

var mystr = Array.prototype.map.call('@@@@@hello@hi@@', function(item) {
    if (item == '@' && !this.stop) {
      return '%';
    } else {
      this.stop = true;
      return item;
    }
  }, {}).join('');

  console.log(mystr);

#3

You can do it without a callback function as replacement with this pattern:

您可以不使用回调函数来替换此模式:

if (mystr.charAt(0)=='@')
    mystr = mystr.replace(/@((?=@)|.*)/g, '%%$1');

Obviously, if you already know that the first character is always a @, remove the if condition.

显然，如果您已经知道第一个字符总是@，那么删除if条件。

If your string has newlines replace the dot with [^] or [\s\S].

如果你的字符串有换行点替换[^]或[\ s \ s]。

#1