Question

我的字符串中包括括号：

我想大写这个字符串（除了这个）和（this）

我想将其转换为：

我想要超过这个字符串（除了这个）和（这个）

在JavaScript中执行此操作的最佳方法是什么？如果我必须使用正则表达式，请清楚解释因为我不擅长正则表达式。

致以最诚挚的问候，

Answer 1

可能更简单的是大写字母然后是小写之间的小写

var str = "I want to uppercase this string (except this) and (this)";

str = str.toUpperCase().replace(/(\(.+?)\)/g, m => m.toLowerCase());

保留案例：

var str = "I want to uppercase this string (eXcEpT ThIs) and (ThIS)";

str = str.toUpperCase().replace(/\((.+?\))/g, (m, c, o) => str.substr(o, m.length));

> I WANT TO UPPERCASE THIS STRING (eXcEpT ThIs) AND (ThIS)

Answer 2

您可以使用正则表达式，然后使用回调替换

＆＃13;

var str = "I want to uppercase this string (except this) and (this)";
var res = str.replace(/[^()](?=([^()]*\([^()]*\))*[^()]*$)/g, t => t.toUpperCase());

console.log(res)

＆＃13;

正则表达式执行以下操作

  NODE                     EXPLANATION
-----------------------------------------
  [^()]                    any character except: '(', ')'
--------------------------------------------------------------------------------
  (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
    (                        group and capture to \1 (0 or more times
                             (matching the most amount possible)):
--------------------------------------------------------------------------------
      [^()]*                   any character except: '(', ')' (0 or
                               more times (matching the most amount
                               possible))
--------------------------------------------------------------------------------
      \(                       '('
--------------------------------------------------------------------------------
      [^()]*                   any character except: '(', ')' (0 or
                               more times (matching the most amount
                               possible))
--------------------------------------------------------------------------------
      \)                       ')'
--------------------------------------------------------------------------------
    )*                       end of \1 (NOTE: because you are using a
                             quantifier on this capture, only the
                             LAST repetition of the captured pattern
                             will be stored in \1)
--------------------------------------------------------------------------------
    [^()]*                   any character except: '(', ')' (0 or
                             more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
    $                        before an optional \n, and the end of
                             the string
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
  /g                       global

没有任何正则表达式

＆＃13;

var str = "I want to uppercase this string (except this) and (this)";
var par = false;
var res = str.split('').map(function(c) {
  par = c == '(' ? true : c == ')' ? false : par;
  return par ? c : c.toUpperCase();
}).join('')

console.log(res)

＆＃13;

Answer 3

这可能更简单：

var str = 'I want to uppercase this string (except this) and (this)';

var repl = str.replace(/[a-z]+(?![^()]*\))/g, m => m.toUpperCase())

console.log(repl);

//=> "I WANT TO UPPERCASE THIS STRING (except this) AND (this)"

这假设(和)在输入中均衡且未转义。

否定前瞻(?![^()]*\))断言我们匹配的字符串后面没有)，中间没有(或)。

Answer 4

可能是一种更好的方法，但使用的方式却快速而肮脏

＆＃13;

var x = 'I want to uppercase this string (except this) and (this)'.replace( /(^[^(]+)|(\)[^(]+)/g, function (a,b) { return (a || b).toUpperCase()})
console.log(x)

＆＃13;

大写括号外的每个字符

4 个答案: