4 回答
TA贡献1808条经验 获得超4个赞
您可以创建自己的解析器,并跟踪“堆栈”以检测之前是否打开了括号。下面的示例适用于()、[]、{}或您想要的任何内容。它们可以相互嵌套。
你可以像这样使用它:
const mySplit = customSplitFactory({
delimiter: ',',
escapedPairs: {
'(': ')',
'{': '}',
'[': ']'
}
});
mySplit('one, two, start (a, b) end'); // ["one"," two"," start (a, b) end"]
代码和演示:
// Generic factory function
function customSplitFactory({ delimiter, escapedPairs }) {
const escapedStartChars = Object.keys(escapedPairs);
return (str) => {
const result = str.split('')
// For each character
.reduce((res, char) => {
// If it's a start escape char `(`, `[`, ...
if (escapedStartChars.includes(char)) {
// Add the corresponding end char to the stack
res.escapeStack.push(escapedPairs[char]);
// Add the char to the current group
res.currentGroup.push(char);
// If it's the end escape char we were waiting for `)`, `]`, ...
} else if (
res.escapeStack.length &&
char === res.escapeStack[res.escapeStack.length - 1]
) {
// Remove it from the stack
res.escapeStack.pop();
// Add the char to the current group
res.currentGroup.push(char);
// If it's a delimiter and the escape stack is empty
} else if (char === delimiter && !res.escapeStack.length) {
if (res.currentGroup.length) {
// Push the current group into the results
res.groups.push(res.currentGroup.join(''));
}
// Reset it
res.currentGroup = [];
} else {
// Otherwise, just push the char into the current group
res.currentGroup.push(char);
}
return res;
}, {
groups: [],
currentGroup: [],
escapeStack: []
});
// If the current group was not added to the results yet
if (result.currentGroup.length) {
result.groups.push(result.currentGroup.join(''));
}
return result.groups;
};
}
// Usage
const mySplit = customSplitFactory({
delimiter: ',',
escapedPairs: {
'(': ')',
'{': '}',
'[': ']'
}
});
function demo(s) { // Just for this demo
const res = mySplit(s);
console.log([s, res].map(JSON.stringify).join(' // '));
}
demo('one, two, start (a, b) end,'); // ["one"," two"," start (a, b) end"]
demo('one, two, start {a, b} end'); // ["one"," two"," start {a, b} end"]
demo('one, two, start [{a, b}] end,'); // ["one"," two"," start [{a, b}] end"]
demo('one, two, start ((a, b)) end,'); // ["one"," two"," start ((a, b)) end"]
TA贡献2011条经验 获得超2个赞
您需要首先考虑特殊情况,即括号,首先处理它:
var str, mtc;
str = "one, two, start (a, b) end, hello";
mtc = str.match(/[^,]*\([^\)]+\)[^,]+|[^,]+/g);
console.log(mtc);
//Expected output: ["one","two", " start (a, b) end", " hello"]
首先,处理括号:
patt = /[^,]*\([^\)]+\)[^,]+/g
//That will match any character after ,
//Then match character "(" and then match any charecter with no ")" then ends with )
//Now is easy things, we just matches character withno colon
patt = /[^,]+/g
TA贡献1829条经验 获得超7个赞
正如一些评论所建议的,您可以使用 split 功能。例子:
let str = "one, two, start (a, b) end,"; let matches = str.split(/(?<!(\"|\{|\()[a-zA-Z0-9]*),(?![a-zA-Z0-9]*\)|\}|\")/);
match 将是一个包含 ["one", "two", "start (a, b) end", "" ] 的数组;
文档:https ://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/split
TA贡献1793条经验 获得超6个赞
如果不需要处理不匹配的大括号,则可以将其简化为天真的平衡大括号计数器。
目前使用默认正常文本尽力而为:
如果检测到右大括号,它将尝试找到起始大括号并将其括起来,将封闭的段视为文本
如果没有找到起始大括号,则将其视为普通文本
const braces = {'{':'}','[':']','(':')'}
// create object map of ending braces to starting braces
const inv_braces = Object.fromEntries(Object.entries(braces).map(x=>x.reverse()))
const red = new RegExp(`(,)|` +
`([${Object.keys(braces).join('')}])|` +
`([${Object.values(braces).map(x=>`\\${x}`).join('')}])` , 'g')
// pre-build break-point scanning regexes
// group1 comma detection, group2 start braces, group3 end braces
element_extract= str => {
let res = []
let stack = [], next, last = -1
// search until no more break-points found
while(next = red.exec(str)) {
const [,comma,begin,end] = next, {index} = next
if(begin) stack.push(begin) // beginning brace, push to stack
else if(end){ //ending brace, pop off stack to starting brace
const start = stack.lastIndexOf(inv_braces[end])
if(start!==-1) stack.length = start
}
else if(!stack.length && comma) res.push(str.slice(last+1,last=index))
// empty stack and comma, slice string and push to results
}
if(last<str.length) res.push(str.slice(last+1)) // final element
return res
}
data = [
"one, two, start (a, b) end",
"one, two, start ((a, (b][,c)]) ((d,e),f)) end, two",
"one, two ((a, (b,c)) ((d,e),f)) three, start (a, (b,c)) ((d,e),f) end, four",
"(a, (b,c)) ((d,e)],f))"
]
for(const x of data)
console.log(element_extract(x))
笔记:
可以通过为 \ 添加另一个匹配组并增加索引以跳过来添加转义
可以添加正则表达式字符串清理器以允许匹配特殊字符
可以添加第二个正则表达式以跳过逗号进行优化(请参阅编辑历史记录)
可以通过替换逗号匹配器并在计算中包括定界符长度来添加对可变长度定界符的支持。大括号也是如此。
例如,我可以使用 (\s*,\s*) 而不是 (,) 来去除空格,或者通过调整正则表达式生成器以使用 '|' 来使用 '{{':'}}' 作为大括号 而不是字符类
为简单起见,我省略了这些
添加回答
举报