c - 用 C 将 JSON 字符串放入数据库

Question

下面是我的代码:

#include <json/json.h>
#include <stdio.h>
#include <sys/types.h>
#include <string.h>
#include <stdlib.h>
#include <db.h>

#define  DATABASE "access.db"


/*printing the value corresponding to boolean, double, integer and strings*/
void print_json_value(json_object *jobj){
  enum json_type type;
  printf("type: ",type);
  type = json_object_get_type(jobj); /*Getting the type of the json object*/
  switch (type) {
    case json_type_boolean: printf("json_type_boolean\n");
                         printf("value: %s\n", json_object_get_boolean(jobj)? "true": "false");
                         break;
    case json_type_double: printf("json_type_double\n");
                        printf("          value: %lf\n", json_object_get_double(jobj));
                         break;
    case json_type_int: printf("json_type_int\n");
                        printf("          value: %d\n", json_object_get_int(jobj));
                         break;
    case json_type_string: printf("json_type_string\n");
                         printf("          value: %s\n", json_object_get_string(jobj));
                         break;
  }

}


/*Parsing the json object*/
void json_parse(json_object * jobj) {
  enum json_type type;
  json_object_object_foreach(jobj, key, val) { /*Passing through every array element*/
    printf("type: ",type);
    type = json_object_get_type(val);
    switch (type) {
      case json_type_boolean:
      case json_type_double:
      case json_type_int:
      case json_type_string: print_json_value(val);
                            db_json(&val);
                           break;
      case json_type_object: printf("json_type_object\n");
                           jobj = json_object_object_get(jobj, key);
                           json_parse(jobj);
                           break;
      case json_type_array: printf("type: json_type_array, ");
                          json_parse_array(jobj, key);
                          break;
    }
  }
} 

int db_json(char *val) {

typedef struct {

char data1[500];

} pearson_record;

pearson_record s;


int i =0;
 DB *dbp;
    DBT key, data;
    int ret, t_ret;
    int recno;



    if ((ret = db_create(&dbp, NULL, 0)) != 0) {
        fprintf(stderr, "db_create: %s\n", db_strerror(ret));
        exit (1);
    }

    // if ((ret = dbp->set_flags(dbp, DB_RECNUM)) != 0) {
        // fprintf(stderr, "db_create: %s\n", db_strerror(ret));
        // exit (1);
    // }

    if ((ret = dbp->open(dbp,
        NULL, DATABASE, NULL, DB_BTREE, DB_CREATE, 0664)) != 0) {
        dbp->err(dbp, ret, "%s", DATABASE);
        goto err;
    }

    strncpy(s.data1, val, strlen(val)+1);

    //printf("chk %\n",jvalue);

    recno = 10;

    memset(&key, 0, sizeof(key));
    memset(&data, 0, sizeof(data));
    //memset(&s, 0, sizeof(struct pearson_record));
    key.data = &recno;
    key.size = sizeof(recno);
    data.data = &s;
    data.size = sizeof(s);


     if ((ret = dbp->put(dbp, NULL, &key,&data,0)) == 0)
    printf("db: %d: key stored.\n", *(int *)key.data);
    else
    {
        dbp->err(dbp, ret, "DB->put");
        goto err;
    }

    pearson_record *ppr;
         if ((ret = dbp->get(dbp, NULL, &key, &data, 0)) == 0) {

    ppr = (pearson_record *) data.data;

    printf("db: %d: key retrieved: data was %s %d\n",
            *(int *)key.data, ppr->data1, data.size);
    }
    else {
        dbp->err(dbp, ret, "DB->get");
        goto err;
    }

    err:    if ((t_ret = dbp->close(dbp, 0)) != 0 && ret == 0)
        ret = t_ret; 

        exit(ret);
    }

int main() {

  char * string = "{\"Extracurricular\" : \"5,6,7,41,40,22,9,12,91,97\"}";
  printf("JSON string: %s\n", string);
  json_object * jobj = json_tokener_parse(string);
  json_parse(jobj);
}

当我编译代码时，我看到以下警告:

try.c:81:8:警告:函数“db_json”的隐式声明

我的输出是:JSON字符串:{“课外”:“5,6,7,41,40,22,9,12,91,97”} 类型: 类型:json_type_string 值:5、6、7、41、40、22、9、12、91、97 db: 10: 存储 key 。 db: 10: 检索到 key :数据为 p4È@4ȸ4È 500

基本上我想将 JSON 字符串的值传递到 Berkeley DB 数据库中。我无法确定将值传递给我的函数是否存在问题，或者是否还有其他问题。我确信我在指针和函数中犯了一些小问题。任何形式的帮助将不胜感激。提前致谢。

Answer 1

一个问题是 db_json 函数接受 char* 作为参数，但对它的调用似乎是将指针的地址传递给 >json_object。我不熟悉这些函数，但您可能应该将 json_object_get_string(jobj) 作为参数传递给 db_json，因为这似乎会产生一个字符串数据的 JSON 表示形式。请注意，这只是通过简要查看代码得出的猜测；我对涉及的类型不太熟悉。