Javascript 正则表达式用值替换所有逗号
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Regex replace all commas with value
提问by Jeff
I have a string that looks like this: "Doe, John, A" (lastname, firstname, middle initial).
我有一个看起来像这样的字符串：“Doe，John，A”（姓氏，名字，中间名首字母）。
I'm trying to write a regular expression that converts the string into "Doe*John*A".
我正在尝试编写一个将字符串转换为“Doe*John*A”的正则表达式。
However, I have to take into account all spaces for this string so "Doe , John , A" would still convert into "Doe*John*A".
但是，我必须考虑此字符串的所有空格，因此“Doe，John，A”仍会转换为“Doe*John*A”。
ALSO, the string "Doe John A" should convert into "Doe*John*A".
此外，字符串“Doe John A”应该转换为“Doe*John*A”。
I started writing this, but I think I'm stuck on the spaces & the possibility of the user not supplying the commas.
我开始写这个，但我想我被困在空格和用户不提供逗号的可能性上。
Here's what I have:
这是我所拥有的：
var myString = "John, Doe, A";
var myOtherString = "John Doe A";
var myFunction = function (aString) {
aString = aString.replace(", ", "*");
aString = aString.replace(", ", "*");
return aString;
};
These should both return "Doe*John*A".
这些都应该返回"Doe*John*A"。
I think I'm repeating myself too much in this function. I'm also not taking into account the possibility that no commas will be provided.
我想我在这个功能中重复了太多。我也没有考虑不提供逗号的可能性。
Is there a better way to do this?
有一个更好的方法吗？
回答by Martin Ender
Yes, there is. Use the replacefunction with a regex instead. That has a few advantages. Firstly, you don't have to call it twice anymore. Secondly it's really easy to account for an arbitrary amount of spaces and an optional comma:
就在这里。改用replace带有正则表达式的函数。这有几个优点。首先，你不必再调用它两次了。其次，考虑任意数量的空格和可选的逗号真的很容易：
aString = aString.replace(/[ ]*,[ ]*|[ ]+/g, '*');
Note that the square brackets around the spaces are optional, but I find they make the space characters more easily readable. If you want to allow/remove any kind of whitespace there (tabs and line breaks, too), use \sinstead:
请注意，空格周围的方括号是可选的，但我发现它们使空格字符更易于阅读。如果您想允许/删除任何类型的空格（也包括制表符和换行符），请\s改用：
aString = aString.replace(/\s*,\s*|\s+,/g, '*');
Note that in both cases we cannot simply make the comma optional, because that would allow zero-width matches, which would introduce a *at every single position in the string. (Thanks to CruorVult for pointing this out)
请注意，在这两种情况下，我们不能简单地将逗号设为可选，因为这将允许零宽度匹配，这将*在字符串中的每个位置引入 a 。（感谢 CruorVult 指出这一点）
回答by CruorVult
If you want to replace all non-word characters try this:
如果要替换所有非单词字符，请尝试以下操作：
str.replace(/\W+/g, '*');
回答by Philipp
String.replace only replaces the first occurence. To replace them all, add the "g" flag for "global". You can also use character groups and the +operator (one or more) to match chains of characters:
String.replace 只替换第一次出现。要全部替换它们，请为“global”添加“g”标志。您还可以使用字符组和 + 运算符（一个或多个）来匹配字符链：
aString.replace("[,\s]+", "*", "g");
This will replace all chains of commas and whitespaces with a *.
这会将所有逗号和空格链替换为*.
回答by bonCodigo
Try this out to remove all spaces and commas, then replace with *.
试试这个删除所有空格和逗号，然后用 * 替换。
Myname= myname.replace(/[,\s]/,"*")
Myname= myname.replace(/[,\s]/,"*")
Editted as removing 'at least two items' from the pattern. But to have at least on item.
编辑为从模式中删除“至少两个项目”。但至少要有项目。
Myname= myname.replace(/([,\s]{1,})/,"*")
Myname= myname.replace(/([,\s]{1,})/,"*")
Reference: on Rublar. However you are better off with regexpal as per m.buettner :)
参考：在 Rublar 上。但是，根据 m.buettner，您最好使用 regexpal :)
