Question

我有以下响应我想解析它以获取使用正则表达式用等号（=）分隔的名称和值：

anyType{
  IsFound=true;
  SubscriberId=7;
  FormMode=Update;
  MobileNumber=0785347032;
  SimNumber=78987530047;
  FirstName=Heba;
  SecondName=Ali;
  ThirdName=Ali;
  LastName=Ali;
  DateOfBirth=2001-01-01T00:00:00;
  Sex=1;
  Address=Zarqa;
  ReigonId=null;
  DistrictId=null;
  Email=test@gmail.com;
  IdentityNo=12312;
  IdentityTypeId=3;
  ContactNo1=0789876576;
  ContactNo2=0789876456;
}
anyType{
  IsFound=true;
  SubscriberId=7;
  FormMode=Update;
  MobileNumber=0785347032;
  SimNumber=78987530047;
  FirstName=Heba;
  SecondName=Ali;
  ThirdName=Ali;
  LastName=Ali;
  DateOfBirth=2001-01-01T00:00:00;
  Sex=1;
  Address=Zarqa;
  ReigonId=null;
  DistrictId=null;
  Email=test@gmail.com;
  IdentityNo=12312;
  IdentityTypeId=3;
  ContactNo1=0789876576;
  ContactNo2=0789876456;
}

我正在执行以下操作:(我想用正则表达式替换它）

public static ArrayList<ParsedPair> parse(String string)
{
  ArrayList<ParsedPair> parsedPairs = new ArrayList<ParsedPair>();

  String[] parts = string.split("anyType\\{");
  if (parts != null && parts.length != 0)
      parts[parts.length-1].replace("}", "");

  for (int i=0; i<parts.length; i++)
  {
    String[] pairs = parts[i].split(";");
    for (int j=0; j<pairs.length; j++)
    {
      ParsedPair parsedPair = new ParsedPair();
      parsedPair.setName(pairs[j].substring(0, pairs[j].indexOf("=")).trim());
      parsedPair.setValue(pairs[j].substring(pairs[j].indexOf("=")+1).trim());
      parsedPairs.add(parsedPair);
    }
  }

  return parsedPairs;
}

另一个问题：任何想法这种形式的数据格式化叫什么？是否有解析器？

Answer 1

我不确定你的真正目标是什么，但也许这个简单的解析器会帮助你一点

public static List<Map<String, String>> parse(String input) {
    List<Map<String, String>> result = new ArrayList<>();

    boolean insideBraces = false;
    boolean appendToKeys = true;
    StringBuilder keyBiulder = new StringBuilder();
    StringBuilder valueBiulder = new StringBuilder();
    Map<String, String> map = null;

    for (char ch : input.toCharArray()){
        if (ch == '{'){
            insideBraces = true;
            map = new LinkedHashMap<>();//I assume that order of keys does matter
        }else if (ch == '}'){
            insideBraces = false;
            result.add(map);
        }else if (!Character.isWhitespace(ch) && insideBraces){//based on your example we ignore spaces
            if (ch == '=') appendToKeys = false;//from now we need to append to valueBuilder
            else if (ch == ';'){//no more data for this pair
                map.put(keyBiulder.toString(), valueBiulder.toString());
                //lets reset settings for next pair
                appendToKeys = true;//we will start appending to keyBuilder again
                keyBiulder.delete(0, keyBiulder.length());
                valueBiulder.delete(0, valueBiulder.length());
            }else{
                if (appendToKeys) keyBiulder.append(ch);
                else valueBiulder.append(ch);
            }
        }
    }

    return result;
}

用法示例

public static void main(String... args) {
    String str="anyType{IsFound=true; SubscriberId=7; FormMode=Update; MobileNumber=0785347032 ; SimNumber=78987530047 ; FirstName=Heba; SecondName=Ali ; ThirdName=Ali\n" + 
            "; LastName=Ali ; DateOfBirth=2001-01-01T00:00:00; Sex=1; Address=Zarqa ; ReigonId=null; DistrictId=null; Email=test@gmail.com ; IdentityNo=12312; IdentityTypeId=3; ContactNo1=0789876576 ; ContactNo2=0789876456 ; } anyType{IsFound=true; SubscriberId=7; FormMode=Update; MobileNumber=0785347032 ; SimNumber=78987530047 ; FirstName=Heba; SecondName=Ali ; ThirdName=Ali\n" + 
            "; LastName=Ali ; DateOfBirth=2001-01-01T00:00:00; Sex=1; Address=Zarqa ; ReigonId=null; DistrictId=null; Email=test@gmail.com ; IdentityNo=12312; IdentityTypeId=3; ContactNo1=0789876576 ; ContactNo2=0789876456 ; }";
    for (Map<String, String> map : parse(str)){
        for (Map.Entry<String, String> entry : map.entrySet())
            System.out.println(entry);
        System.out.println("--------");
    }
}

Ouptut：

IsFound=true
SubscriberId=7
FormMode=Update
MobileNumber=0785347032
SimNumber=78987530047
FirstName=Heba
SecondName=Ali
ThirdName=Ali
LastName=Ali
DateOfBirth=2001-01-01T00:00:00
Sex=1
Address=Zarqa
ReigonId=null
DistrictId=null
Email=test@gmail.com
IdentityNo=12312
IdentityTypeId=3
ContactNo1=0789876576
ContactNo2=0789876456
--------
IsFound=true
SubscriberId=7
FormMode=Update
MobileNumber=0785347032
SimNumber=78987530047
FirstName=Heba
SecondName=Ali
ThirdName=Ali
LastName=Ali
DateOfBirth=2001-01-01T00:00:00
Sex=1
Address=Zarqa
ReigonId=null
DistrictId=null
Email=test@gmail.com
IdentityNo=12312
IdentityTypeId=3
ContactNo1=0789876576
ContactNo2=0789876456
--------

Answer 2

这应该有效，

    String str="anyType{IsFound=true; SubscriberId=7; FormMode=Update; MobileNumber=0785347032 ; SimNumber=78987530047 ; FirstName=Heba; SecondName=Ali ; ThirdName=Ali\n" + 
    "; LastName=Ali ; DateOfBirth=2001-01-01T00:00:00; Sex=1; Address=Zarqa ; ReigonId=null; DistrictId=null; Email=test@gmail.com ; IdentityNo=12312; IdentityTypeId=3; ContactNo1=0789876576 ; ContactNo2=0789876456 ; } anyType{IsFound=true; SubscriberId=7; FormMode=Update; MobileNumber=0785347032 ; SimNumber=78987530047 ; FirstName=Heba; SecondName=Ali ; ThirdName=Ali\n" + 
    "; LastName=Ali ; DateOfBirth=2001-01-01T00:00:00; Sex=1; Address=Zarqa ; ReigonId=null; DistrictId=null; Email=test@gmail.com ; IdentityNo=12312; IdentityTypeId=3; ContactNo1=0789876576 ; ContactNo2=0789876456 ; }";

    Pattern p=Pattern.compile("(\\w+)=(\\w+)\\s*;");
    Matcher m=p.matcher(str1);
    while(m.find()){
        System.out.println(m.group(1)+"\t\t"+m.group(2));
    }

使用Java Regular Expression解析服务响应

2 个答案: