javascript - 删除括号外字符串的所有点

Question

我有字符串

var string =  .row-4 .col-2.grid-unit+.grid-unit+.grid-unit,.row-4 .col-3 .grid-unit .row-4 .grid-unit:nth-of-type(2n+3) .show-grid+.show-grid-reportdiv

我需要使用javascript从字符串中删除所有加号，将加号留在括号内

Answer 1

我会沿着这些思路去做：

var i, splits, string = ".row-4 .col-2.grid-unit+.grid-unit+.grid-unit,.row-4 .col-3 .grid-unit .row-4 .grid-unit:nth-of-type(2n+3) .show-grid+.show-grid-reportdiv";
splits = string.split(/(\([^)]+\))/);
for (i = 0; i< splits.length; i++) {
    if (splits[i].charAt(0) !== "(") {
        splits[i] = splits[i].replace("+"," ");
    }
}
string = splits.join();

另一种解决方法（不知道是否性能更好）将使用以下内容：

var string = ".row-4 .col-2.grid-unit+.grid-unit+.grid-unit,.row-4 .col-3 .grid-unit .row-4 .grid-unit:nth-of-type(2n+3) .show-grid+.show-grid-reportdiv";
function replacer (match, offset, string) { 
    var posOpen = string.indexOf("(",offset);
    var posClose = string.indexOf(")",offset);
    // we replace it if there are no more closing parenthesis or if there is one that is located after an opening one.
    if (posClose === -1 || (posClose > posOpen && posOpen !== -1)) {
        return " ";
    } else {
        return "+";
    }
};
string.replace(/\+/g, replacer);

编辑：添加了 bergi 建议，以便在循环内更快地检查。

EDIT2：第二种解决方案

Answer 2

使用以下代码，让我知道它是否有效:)

var myString =  ".row-4 .col-2.grid-unit+.grid-unit+.grid-unit,.row-4:nth-of-type(2n+3) .col-3 .grid-unit .row-4 .grid-unit:nth-of-type(2n+3) .show-grid+.show-grid-reportdiv";

var myArray = myString.split(/\(.[\(\)A-Za-z0-9-.+]*\)/);

for(var i = 0; i < myArray.length; i++) { 
  myString = myString.replace(myArray[i], myArray[i].replace(/[+]/g,' ')); 
}