Question

我有一个字符串需要根据“，”（逗号）的出现进行拆分，但需要忽略一对括号中出现的任何字符串。例如，B2B,(A2C,AMM),(BNC,1NF),(106,A01),AAA,AX3 应该分成

B2B,
(A2C,AMM),
(BNC,1NF),
(106,A01),
AAA,
AX3

Answer 1

FOR NON NESTED

,(?![^\(]*\))

FOR NESTED （括号内的括号）

(?<!\([^\)]*),(?![^\(]*\))

Answer 2

尝试以下：

var str = 'B2B,(A2C,AMM),(BNC,1NF),(106,A01),AAA,AX3';
console.log(str.match(/\([^)]*\)|[A-Z\d]+/g));
// gives you ["B2B", "(A2C,AMM)", "(BNC,1NF)", "(106,A01)", "AAA", "AX3"]

Java版：

String str = "B2B,(A2C,AMM),(BNC,1NF),(106,A01),AAA,AX3";
Pattern p = Pattern.compile("\\([^)]*\\)|[A-Z\\d]+");
Matcher m = p.matcher(str);
List<String> matches = new ArrayList<String>();
while(m.find()){
    matches.add(m.group());
}

for (String val : matches) {
    System.out.println(val);
}

Answer 3

一个简单的迭代可能比任何正则表达式都更好，特别是如果你的数据在括号内有圆括号。例如：

String data="Some,(data,(that),needs),to (be, splited) by, comma";
StringBuilder buffer=new StringBuilder();
int parenthesesCounter=0;
for (char c:data.toCharArray()){
    if (c=='(') parenthesesCounter++;
    if (c==')') parenthesesCounter--;
    if (c==',' && parenthesesCounter==0){
        //lets do something with this token inside buffer
        System.out.println(buffer);
        //now we need to clear buffer  
        buffer.delete(0, buffer.length());
    }
    else 
        buffer.append(c);
}
//lets not forget about part after last comma
System.out.println(buffer);

输出

Some
(data,(that),needs)
to (be, splited) by
 comma

Answer 4

试试这个

\w{3}(?=,)|(?<=,)\(\w{3},\w{3}\)(?=,)|(?<=,)\w{3}

说明：有三个部分由OR (|)

分隔

\w{3}(?=,) - 匹配3个任何字母数字字符（包括下划线），并为逗号做好积极展望
(?<=,)\(\w{3},\w{3}\)(?=,) - 匹配此模式(ABC,E4R)，也会做出积极的向前看，并留意逗号
(?<=,)\w{3} - 匹配3个任何字母数字字符（包括下划线），并为逗号做正面看法

根据正则表达式拆分字符串

4 个答案: