"{length=1,size=2}"的string类型如何转为map

对于字符串"{length=1,size=2}"，可以使用如下代码将其转换为Map:

String str = "{length=1,size=2}";
Map<String, String> map = new HashMap<>();
String[] pairs = str.substring(1, str.length() - 1).split(",");
for (String pair : pairs) {
    String[] keyValue = pair.split("=");
    map.put(keyValue[0], keyValue[1]);
}

以上代码将字符串首尾的"{"和"}"去掉，然后按照","进行拆分，再按照"="进行拆分，最终将每个键值对存储到Map中。注意，这里的Map的键和值都是字符串类型。

相关问题

string 转map

假设字符串的格式为 "key1=value1&key2=value2&key3=value3"，可以按照以下方式转换为 map：

#include <iostream>
#include <string>
#include <map>

int main() {
    std::string str = "key1=value1&key2=value2&key3=value3";
    std::map<std::string, std::string> kv_map;

    size_t pos = 0;
    while (pos < str.length()) {
        size_t equal_pos = str.find('=', pos);
        size_t amp_pos = str.find('&', pos);
        if (equal_pos == std::string::npos) {
            break;
        }
        std::string key = str.substr(pos, equal_pos - pos);
        std::string value;
        if (amp_pos == std::string::npos) {
            value = str.substr(equal_pos + 1);
        } else {
            value = str.substr(equal_pos + 1, amp_pos - equal_pos - 1);
        }
        kv_map[key] = value;
        if (amp_pos == std::string::npos) {
            break;
        }
        pos = amp_pos + 1;
    }

    for (const auto& kv : kv_map) {
        std::cout << kv.first << " => " << kv.second << std::endl;
    }

    return 0;
}

输出结果为：

key1 => value1
key2 => value2
key3 => value3

public static List<Map<String,Object>> hour(List<Map<String,Object>> list){ if(list != null && list.size()>0){ String item=""; String [] s1 =null; for(Map<String,Object> m : list){ item = m.get("item_info").toString(); s1 = item.split(";"); for(String s : s1){ String [] s2 = null; s2=s.split(","); m.put(s2[0],s2[2]); m.put(s2[0]+"code",s2[0]); m.put(s2[0]+"detname",s2[3]); if(s2.length>=5) { m.put(s2[0] + "bzz", s2[4]); } if(s2.length>=6) { m.put(s2[0] + "yj", s2[5]); } if(s2.length>=7) { m.put(s2[0] + "bj", s2[6]); } if(s2.length>=8) { m.put(s2[0] + "dw", s2[7]); } } } } return list; } 会报空指针异常 如何改正

可以在代码中加入对 `s2` 是否为 `null` 的判断，如下所示：

public static List<Map<String, Object>> hour(List<Map<String, Object>> list) {
    if (list != null && list.size() > 0) {
        String item = "";
        String[] s1 = null;
        for (Map<String, Object> m : list) {
            item = m.get("item_info").toString();
            s1 = item.split(";");
            for (String s : s1) {
                String[] s2 = null;
                s2 = s.split(",");
                if (s2 != null && s2.length > 2) {
                    m.put(s2[0], s2[2]);
                    m.put(s2[0] + "code", s2[0]);
                    m.put(s2[0] + "detname", s2[3]);
                    if (s2.length >= 5) {
                        m.put(s2[0] + "bzz", s2[4]);
                    }
                    if (s2.length >= 6) {
                        m.put(s2[0] + "yj", s2[5]);
                    }
                    if (s2.length >= 7) {
                        m.put(s2[0] + "bj", s2[6]);
                    }
                    if (s2.length >= 8) {
                        m.put(s2[0] + "dw", s2[7]);
                    }
                }
            }
        }
    }
    return list;
}

在对 `s2` 进行操作前，加入了 `s2 != null && s2.length > 2` 的判断，确保了 `s2` 不为 `null` 且至少有 3 个元素，避免了空指针异常的发生。