根据引号将字符串设为小写 - Javascript

Question

我正在为 Domain Specific Language,（或 DSL）构建一个解析器，我正在尝试将字符串转换为全部小写。我知道 toLowerCase 可以轻松完成此任务，但我需要在其原始情况下保留用双引号或单引号（" 或 '）引用的字符串。例如，见下文：

输入： ThIs iS a teST "sTriNg Y'alL" aS yOu cAN sEE 'hEllO woRl\' o miNE'

输出： this is a test "sTriNg Y'alL" as you can see 'hEllO woRl\' o miNE'

编辑：添加反斜杠引号

Answer 1

我确定有一个正则表达式解决方案，但这里有另一个解决方案，它在将引用的字符串小写之前替换它：

String.prototype.toLowerCaseQuoted = function() {
   var str = this.valueOf();
   var replacements = [];
   var I = 0;
   str = str
      .replace(/((\".+\")|(\'.+\'))/g, function(s) {
         console.log(s)
         replacements.push(s);
         return "%s"+(I++)+"%"
      })
      .toLowerCase()
      .replace(/%s([0-9]+)%/g, function(s) {
         var k = parseInt(s.match(/([0-9])+/)[0]);
         console.log(k)
         return replacements[k];
      });
   return str;
}

例如：

"WILL BE LOWER CASE \"QUOTED\" \'MORE QUOTED\'".toLowerCaseQuoted()

Returns "will be lower case "QUOTED" 'MORE QUOTED'"

Answer 2

只是拼凑了一个快速解析器，不确定它的效果如何，但它应该处理无限制的反斜杠转义

function string_to_block(str) {
    var blocks = [],
        i, j, k;
    function isEscaped(str, i) {
        var escaped = false;
        while (str[--i] === '\') escaped = !escaped;
        return escaped;
    }
    start: for (i = 0; i < str.length; i = j + 1) {
        find: for (j = i; j < str.length; ++j) {
            if (str[j] === '"' && !isEscaped(str, j)) {
                if (j > i) {
                    blocks.push({type: 'regular', str: str.slice(i, j)});
                }
                end: for (k = j + 1; k < str.length; ++k) {
                    if (str[k] === '"' && !isEscaped(str, k)) {
                        // found a "str" block
                        blocks.push({type: 'quote', str: str.slice(j, k + 1)});
                        j = k;
                        break find;
                    }
                }
                throw new SyntaxError('unclosed "str... starting at index ' + j);
            }
            if (str[j] === "'" && !isEscaped(str, j)) {
                if (j > i) {
                    blocks.push({type: 'regular', str: str.slice(i, j)});
                }
                end: for (k = j + 1; k < str.length; ++k) {
                    if (str[k] === "'" && !isEscaped(str, k)) {
                        // found a 'str' block
                        blocks.push({type: 'quote', str: str.slice(j, k + 1)});
                        j = k;
                        break find;
                    }
                }
                throw new SyntaxError("unclosed 'str... starting at index " + j);
            }
        }
    }
    if (k + 1 < str.length) {
        blocks.push({type: 'regular', str: str.slice(k + 1)});
    }
    return blocks;
}

现在

var foo = string_to_block("ThIs iS a teST \"sTriNg Y'alL\" aS yOu cAN sEE 'hEllO woRl\' o miNE'");
/*
[
    {"type": "regular", "str": "ThIs iS a teST "},
    {"type": "quote"  , "str": "\"sTriNg Y'alL\""},
    {"type": "regular", "str": " aS yOu cAN sEE "},
    {"type": "quote"  , "str": "'hEllO woRl\' o miNE'"}
]
*/

因此我们可以根据需要重新构建您的字符串；

var i, str = '';
for (i = 0; i < foo.length; ++i) {
    if (foo[i].type === 'regular') str += foo[i].str.toLowerCase();
    else str += foo[i].str;
}
str; // this is a test "sTriNg Y'alL" as you can see 'hEllO woRl\' o miNE'

Answer 3

String.prototype.toLowerCaseQuoted = function() {
    var oldValue = this.valueOf();
    var newValue = '';
    var inside = 0;

    for (var i = 0; i < oldValue.length; i++) {
        if (oldValue[i] == '"') {
            if (inside == 0) {
                inside = 1;
            } else {
                inside = 0;
            }
        }

        if (inside == 1) {
            newValue += oldValue[i];
        } else {
            newValue += oldValue[i].toLowerCase();
        }
    }

    return newValue;
 }

Answer 4

这是@Paul S 的后续。这应该处理没有引号块的字符串...

function string_to_block(str) {
    var blocks = [],
        i, j, k;
    function isEscaped(str, i) {
        var escaped = false;
        while (str[--i] === '\') escaped = !escaped;
        return escaped;
    }
    start: for (i = 0; i < str.length; i = j + 1) {
        find: for (j = i; j <= str.length; ++j) {
            if (str[j] === '"' && !isEscaped(str, j)) {
                if (j > i) {
                    blocks.push({type: 'regular', str: str.slice(i, j)});
                }
                end: for (k = j + 1; k < str.length; ++k) {
                    if (str[k] === '"' && !isEscaped(str, k)) {
                        // found a "str" block
                        blocks.push({type: 'quote', str: str.slice(j, k + 1)});
                        j = k;
                        break find;
                    }
                }
                throw new SyntaxError('unclosed "str... starting at index ' + j);
            }
            if (str[j] === "'" && !isEscaped(str, j)) {
                if (j > i) {
                    blocks.push({type: 'regular', str: str.slice(i, j)});
                }
                end: for (k = j + 1; k < str.length; ++k) {
                    if (str[k] === "'" && !isEscaped(str, k)) {
                        // found a 'str' block
                        blocks.push({type: 'quote', str: str.slice(j, k + 1)});
                        j = k;
                        break find;
                    }
                }
                throw new SyntaxError("unclosed 'str... starting at index " + j);
            }
            if (j === str.length) {
                            // We reached the end without finding any quote blocks
              if (j > i) {
                blocks.push({type: 'regular', str: str.slice(i,j)});
              }
            }
        }
    }
    return blocks;
}

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