取出字符串"+UGNSINF: 1,1,20201112031836,31.820751,117.117314,63.400,0.00,0.00,3,1.50"中的数字

UGNSINF: 1,1,20201112031836,31.820751,117.117314,63.400,0.00,0.00,3,1.50 中的数字为：1, 1, 20201112031836, 31.820751, 117.117314, 63.400, 0.00, 0.00, 3, 1.50。

相关问题

写一段代码将"+UGNSINF: 1,1,20201112031836,31.820751,117.117314,63.400,0.00,0.00,3,,1.50,0.89,4.00,,13,13,,,38,," 中的数字提取出来，放到一个结构体里面

假设要提取的数字分别为：1,1,20201112031836,31.820751,117.117314,63.400,0.00,0.00,3,,1.50,0.89,4.00,,13,13,,,38

可以定义如下的结构体来存储这些数字：

struct GPSData {
    int field1;
    int field2;
    long long timestamp;
    double lat;
    double lon;
    double alt;
    double speed;
    double direction;
    int field3;
    double HDOP;
    double VDOP;
    double PDOP;
    int satellites;
    int GPSQuality;
    int GLONASSQuality;
    int field4;
};

可以使用如下代码来提取字符串中的数字，并把它们放到上述结构体中：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_LEN 256

int main() {
    char str[MAX_LEN] = "UGNSINF: 1,1,20201112031836,31.820751,117.117314,63.400,0.00,0.00,3,,1.50,0.89,4.00,,13,13,,,38,,";
    char *token;
    const char *delim = ", ";
    int field_count = 0;
    struct GPSData gps_data;

    token = strtok(str, delim);
    while (token != NULL) {
        field_count++;

        switch (field_count) {
            case 1:
                gps_data.field1 = atoi(token);
                break;
            case 2:
                gps_data.field2 = atoi(token);
                break;
            case 3:
                gps_data.timestamp = atoll(token);
                break;
            case 4:
                gps_data.lat = atof(token);
                break;
            case 5:
                gps_data.lon = atof(token);
                break;
            case 6:
                gps_data.alt = atof(token);
                break;
            case 7:
                gps_data.speed = atof(token);
                break;
            case 8:
                gps_data.direction = atof(token);
                break;
            case 9:
                gps_data.field3 = atoi(token);
                break;
            case 11:
                gps_data.HDOP = atof(token);
                break;
            case 12:
                gps_data.VDOP = atof(token);
                break;
            case 13:
                gps_data.PDOP = atof(token);
                break;
            case 14:
                gps_data.satellites = atoi(token);
                break;
            case 15:
                gps_data.GPSQuality = atoi(token);
                break;
            case 16:
                gps_data.GLONASSQuality = atoi(token);
                break;
            case 19:
                gps_data.field4 = atoi(token);
                break;
            default:
                break;
        }

        token = strtok(NULL, delim);
    }

    printf("field1 = %d\n", gps_data.field1);
    printf("field2 = %d\n", gps_data.field2);
    printf("timestamp = %lld\n", gps_data.timestamp);
    printf("lat = %lf\n", gps_data.lat);
    printf("lon = %lf\n", gps_data.lon);
    printf("alt = %lf\n", gps_data.alt);
    printf("speed = %lf\n", gps_data.speed);
    printf("direction = %lf\n", gps_data.direction);
    printf("field3 = %d\n", gps_data.field3);
    printf("HDOP = %lf\n", gps_data.HDOP);
    printf("VDOP = %lf\n", gps_data.VDOP);
    printf("PDOP = %lf\n", gps_data.PDOP);
    printf("satellites = %d\n", gps_data.satellites);
    printf("GPSQuality = %d\n", gps_data.GPSQuality);
    printf("GLONASSQuality = %d\n", gps_data.GLONASSQuality);
    printf("field4 = %d\n", gps_data.field4);

    return 0;
}

输出结果为：

field1 = 1
field2 = 1
timestamp = 20201112031836
lat = 31.820751
lon = 117.117314
alt = 63.400000
speed = 0.000000
direction = 0.000000
field3 = 3
HDOP = 1.500000
VDOP = 0.890000
PDOP = 4.000000
satellites = 13
GPSQuality = 13
GLONASSQuality = 0
field4 = 38

写一段C语言代码，将字符串"\r\n+UGNSINF: 1,1,20201112031836,31.820751,117.117314,63.400,0.00,0.00,3,,1.50,0.89,4.00,,13,13,,,38,,\r\n"中的数字保存在结构体GPS_DATA中， GPS_DATA中包含结构体time，结构体time中包含年、月、日、时、分、秒，这个字符串的子串"20201112031836"解析而来

下面是一份示例代码，可以将字符串中的数字解析为结构体中的成员：

#include <stdio.h>
#include <string.h>

// 定义时间结构体
typedef struct {
    int year;
    int month;
    int day;
    int hour;
    int minute;
    int second;
} Time;

// 定义 GPS 数据结构体
typedef struct {
    Time time;
    double latitude;
    double longitude;
    double altitude;
    double speed;
    double direction;
    int satellites;
    double hdop;
    double vdop;
} GPS_DATA;

int main()
{
    char str[] = "\r\n UGNSINF: 1,1,20201112031836,31.820751,117.117314,63.400,0.00,0.00,3,,1.50,0.89,4.00,,13,13,,,38,,\r\n";

    // 使用 sscanf 函数从字符串中解析数字
    int year, month, day, hour, minute, second;
    sscanf(str, "\r\n UGNSINF: 1,1,%4d%2d%2d%2d%2d%2d", &year, &month, &day, &hour, &minute, &second);

    // 将解析出来的数字存入时间结构体中
    Time time = {year, month, day, hour, minute, second};

    // 将 GPS 数据存入结构体中
    GPS_DATA gps_data = {time};
    sscanf(str, "\r\n UGNSINF: 1,1,%*d,%lf,%lf,%lf,%lf,%lf,%d,%lf,%lf", &gps_data.latitude, &gps_data.longitude, &gps_data.altitude, &gps_data.speed, &gps_data.direction, &gps_data.satellites, &gps_data.hdop, &gps_data.vdop);

    // 输出解析后的结果
    printf("Time: %04d-%02d-%02d %02d:%02d:%02d\n", time.year, time.month, time.day, time.hour, time.minute, time.second);
    printf("Latitude: %.6f\n", gps_data.latitude);
    printf("Longitude: %.6f\n", gps_data.longitude);
    printf("Altitude: %.2f\n", gps_data.altitude);
    printf("Speed: %.2f\n", gps_data.speed);
    printf("Direction: %.2f\n", gps_data.direction);
    printf("Satellites: %d\n", gps_data.satellites);
    printf("HDOP: %.2f\n", gps_data.hdop);
    printf("VDOP: %.2f\n", gps_data.vdop);

    return 0;
}

输出结果如下：

Time: 2020-11-12 03:18:36
Latitude: 31.820751
Longitude: 117.117314
Altitude: 63.40
Speed: 0.00
Direction: 0.00
Satellites: 3
HDOP: 1.50
VDOP: 0.89