在一次替换调用中替换多个字符 [英] Replace multiple characters in one replace call

问题描述

很简单的小问题，但是我不太明白怎么做.

Very simple little question, but I don't quite understand how to do it.

我需要用空格替换每个 '_' 实例，用空/空替换每个 '#' 实例.

I need to replace every instance of '_' with a space, and every instance of '#' with nothing/empty.

var string = '#请发送_an_information_pack_to_the_following_address:';

我已经试过了:

string.replace('#','').replace('_',' ');

我真的不喜欢像这样链接命令.有没有另一种方法可以做到这一点?

I don't really like chaining commands like this. Is there another way to do it in one?

推荐答案

使用 OR 运算符 (|):

Use the OR operator (|):

var str = '#this #is__ __#a test###__';
str.replace(/#|_/g,''); // result: "this is a test"

你也可以使用字符类:

str.replace(/[#_]/g,'');

小提琴

如果你想用一个东西替换散列，用另一个替换下划线，那么你只需要链接.但是，您可以添加原型:

Fiddle

If you want to replace the hash with one thing and the underscore with another, then you will just have to chain. However, you could add a prototype:

String.prototype.allReplace = function(obj) {
    var retStr = this;
    for (var x in obj) {
        retStr = retStr.replace(new RegExp(x, 'g'), obj[x]);
    }
    return retStr;
};

console.log('aabbaabbcc'.allReplace({'a': 'h', 'b': 'o'}));
// console.log 'hhoohhoocc';

为什么不连锁呢?我看不出有什么问题.

Why not chain, though? I see nothing wrong with that.

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