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我有一个 bash 脚本，它以格式输出字符串，Hostname IP MacAddr并由我用 C 编写的脚本读取。我试图将这 3 个拆分为一个数组，并使其能够将它们存储到Json-c 对象生成看起来像{Clients: [{Hostname: Value, IP: Value, MacAddr: Value}]}.

目前我的程序能够逐行读取每个字符串并将其存储到一个数组中（该数组被初始化错误只是为了测试目的，我将改变它）：

int get_list_of_connected_clients(json_object *input, json_object *output) {
    FILE *fp;
    char path[1035];
    int i = 0;
    char a[2][100];
    fp = popen("./Sample_Bash_Script_Test.sh", "r");
    if (fp == NULL) {
        printf("Failed To Run Script \n");
        exit(1);
    }
    while (fgets(path, sizeof(path) - 1, fp) != NULL) {
        stpcpy(a[i], path);
        i++;
    }
    pclose(fp);
}

有没有人能够帮助我并引导我朝着正确的方向前进？C 中的字符串操作对我来说相对较新，我仍在努力解决它！

编辑：

我的函数现在看起来像这样：

int get_list_of_connected_clients(json_object* input, json_object* output){
    FILE *filepath;
    char output_line[1035];
    int index=0;
    char arr_clients[30][100];

    filepath = popen("./Sample_Bash_Script_Test.sh", "r");
    if (filepath == NULL){
        printf("Failed To Run Script \n");
        exit(1);
    }
    while (fgets(output_line, sizeof(output_line)-1, filepath) != NULL){
        stpcpy(arr_clients[index], output_line);
        index++;
    }
    pclose(filepath);

    /*Creating a json object*/
    json_object * jobj = json_object_new_object();

    /*Creating a json array*/
    json_object *jarray = json_object_new_array();

    json_object *jstring1[2][2];

    for (int y=0; y < 2; y++) {
        int x = 0;
        char *p = strtok(arr_clients[y], " ");
        char *array[2][3];

        while (p != NULL) {
            array[y][x++] = p;
            p = strtok(NULL, " ");
        }

        for (x = 0; x < 3; ++x) {
            jstring1[y][x] = json_object_new_string(array[y][x]);
            /*Adding the above created json strings to the array*/
            json_object_array_add(jarray,jstring1[y][x]);

        }
    }
    /*Form the json object*/
    json_object_object_add(jobj,"Clients", jarray);

    /*Now printing the json object*/
    printf ("%s",json_object_to_json_string(jobj));

    return 0;
    }

当我运行它时，输出如下所示：{ "Clients": [ "Hostname", "192.168.1.18", "XX:XX:XX:XX", "Hostname", "192.168.1.13", "XX:XX:XX:XX" ] }

有没有人知道我做错了什么来阻止它在每个客户之后打破列表？IE

{
  "Clients" : [
    {
      "Hostname" : "example.com",
      "IP" : "127.0.0.1",
      "MacAddr" : "mactonight"
    },
    {
      "Hostname" : "foo.biz",
      "IP" : "0.0.0.0",
      "MacAddr" : "12:34:56:78"
    }
  ]
}
4

1 回答 1

2

与其尝试将 JSON 构建为字符串，不如使用json-glib等库为您构建 JSON。这更灵活，可以处理各种边缘情况。它提供了JsonBuilder来构建 JSON 结构。

我们首先获取一个文件指针，其他东西应该打开文件。然后我们启动一个 JsonBuilder 并开始构建声明{ "Clients"对象的 JSON 结构并启动数组。

JsonNode *bash_connected_clients_to_json(FILE *fp) {
    JsonBuilder *builder = json_builder_new();

    // { "Clients": [ ...
    json_builder_begin_object(builder);
    json_builder_set_member_name(builder, "Clients");
    json_builder_begin_array(builder);

现在我们读取每一行并将其和构建器发送到一个函数中以处理该行并将其添加到开放数组中。

    char line[1024];
    while (fgets(line, sizeof(line), fp) != NULL) {
        bash_connected_clients_line_to_json(line, builder);
    }

最后关闭数组和对象，返回我们刚刚构建的JsonNode 。

    // ... ] }
    json_builder_end_array(builder);
    json_builder_end_object(builder);

    return json_builder_get_root(builder);
}

然后可以打印JsonNode。

int main() {
    JsonNode *json = bash_connected_clients_to_json(stdin);
    printf("%s", json_to_string(json, TRUE));
}

处理每一行从解析它开始。这可以在各种情况下完成。sscanf工作正常。

void bash_connected_clients_line_to_json( const char *line, JsonBuilder *builder ) {    
    char hostname[1024], ip[1024], macaddr[1024];
    if( sscanf(line, "%1023s %1023s %1023s", hostname, ip, macaddr) != 3 ) {
        fprintf(stderr, "Could not parse line: '%s'\n", line);
        return;
    }

然后我们将一个 JSON 对象添加到我们已经打开的数组中，将我们的每个元素添加到该对象中，然后关闭该对象。

    // { "Hostname": "foo", "IP", "bar", "MacAddr", "baz" }
    json_builder_begin_object(builder);

    json_builder_set_member_name(builder, "Hostname");
    json_builder_add_string_value(builder, hostname);

    json_builder_set_member_name(builder, "IP");
    json_builder_add_string_value(builder, ip);

    json_builder_set_member_name(builder, "MacAddr");
    json_builder_add_string_value(builder, macaddr);

    json_builder_end_object(builder);
}

$ cat > test.txt
example.com 127.0.0.1 mactonight
foo.biz 0.0.0.0 12:34:56:78

$ ./test < test.txt
{
  "Clients" : [
    {
      "Hostname" : "example.com",
      "IP" : "127.0.0.1",
      "MacAddr" : "mactonight"
    },
    {
      "Hostname" : "foo.biz",
      "IP" : "0.0.0.0",
      "MacAddr" : "12:34:56:78"
    }
  ]
}

或者你可以用几行 Ruby 来完成。

require 'json'

clients = []

STDIN.each do |line|
  fields = line.split(/\s+/)
  clients << {
    Hostname: fields[0],
    IP: fields[1],
    MacAddr: fields[2]
  }
end

connections = {}
connections[:Clients] = clients
puts connections.to_json

对于 json-c 基本上是一样的。主要区别在于它不是bash_connected_clients_line_to_json将 JSON 对象添加到构建器，而是返回一个 JSON 对象。

json_object* bash_connected_clients_line_to_json( const char *line ) {    
    char hostname[1024], ip[1024], macaddr[1024];
    if( sscanf(line, "%1023s %1023s %1023s", hostname, ip, macaddr) != 3 ) {
        fprintf(stderr, "Could not parse line: '%s'\n", line);
        return NULL;
    }

    json_object *json = json_object_new_object();

    json_object_object_add(json, "Hostname", json_object_new_string(hostname));
    json_object_object_add(json, "IP", json_object_new_string(ip));
    json_object_object_add(json, "MacAddr", json_object_new_string(macaddr));

    return json;
}

然后将其添加到客户端的 JSON 数组中。

json_object *bash_connected_clients_to_json(FILE *fp) {
    json_object *clients = json_object_new_array();

    char line[1024];
    while (fgets(line, sizeof(line), fp) != NULL) {
        json_object_array_add(
            clients,
            bash_connected_clients_line_to_json(line)
        );
    }

    json_object *json = json_object_new_object();
    json_object_object_add(json, "Clients", clients);

    return json;
}
于 2019-04-13T01:15:25.290 回答