Question

任何人都可以帮助编写C解析器（使用Codewarrior）来解析以下文件吗？我尝试了许多C语言json解析器，但遗憾的是没有得到这个想法，或者他们让它变得困难。

[
[0,"Door1","Door1Drv","Calculator","Alarm"],
[1,"Door3","Door3Drv","Calculator","Alarm"],
[2,"Door2","Door2Drv","Calculator","Alarm"]]

Answer 1

对于高级应用程序，建议使用Frozen。如果您的应用程序具有如上所示的修复数据格式，请参阅下面的参考。

相当基本的版本如下所示：

#include <stdio.h>

#define MAX_PRAM    4

struct param_t {
    int id;

    // FIXME: use pointer & malloc to allocate if possible
    char data[MAX_PRAM][30];
};

static int parse(const char *buff, int start, const int end,
    struct param_t *out)
{
    char *ptr = out->data[0];
    int state = 0;
    int index = 0;
    int param_index = 0;

    printf("\t> parse next record<%d:%d>\n", start, end);
    while((start < end) && (param_index < MAX_PRAM)) {
        char ch = buff[start];

        // printf("\t> state: %d, char '%c'\n", state, ch);
        switch (state) {
            case 0: // searching for start point
                if ((ch >= '0') && (ch <= '9')) {
                    out->id = ch - '0';
                    state++;
                }

                break;

            case 1: // parse number
                if ((ch < '0') || (ch > '9')) {
                    printf ("\t> number %d\n", out->id);

                    state++;
                    if (ch == '"')
                        state++;
                } else
                    out->id = (out->id * 10) + (ch - '0');
                break;

            case 2: // search string
                if (ch == '"') {
                    index = 0;
                    state++;
                }

                break;

            case 3: // get string
                if (ch == '"') { // finish one string, parse next one if need
                    ptr[index] = '\0';

                     printf ("\t> string '%s', length %d\n",
                        out->data[param_index], index);

                    param_index++;
                    ptr = out->data[param_index];
                    state--;
                } else
                    ptr[index++] = ch;

                break;

            default: // EOS - exit
                break;
        }

        start++;
    }

    return (param_index >= 4) ? start : 0;
}

//
// main entry
//
int main(int argc, char *argv[]) {
    int file = open("test.txt", 0 /* 0 ~ O_RDONLY in fcntl.h */ );

    if (file) {
        char buff[1024] = { 0 };
        struct param_t param = { 0 };
        int len = read(file, buff, 1024);
        int pos = 0;

        while((pos = parse(buff, pos, len, &param)) > 0) {
            printf ("\t%d\n", param.id);

            printf("[%4d] %d - %-10s | %-10s | %-15s | %-10s \n",
                pos, param.id, param.data[0], param.data[1], param.data[2],
                param.data[3]);
        }

        close(file);
    }
}

输入：

[[10,"Door0",

"Door0Drv","Calculator0","Alarm0"],

[15,"Door1","Door1Drv","Calculator1","Alarm1"],

[230,"Door2","Door2Drv","Calculator2",
    "Alarm2"]]

输出如下：

        > number 10
        > string 'Door0', length 5
        > string 'Door0Drv', length 8
        > string 'Calculator0', length 11
        > string 'Alarm0', length 6
        10
[  49] 10 - Door0      | Door0Drv   | Calculator0     | Alarm0
        > number 15
        > string 'Door1', length 5
        > string 'Door1Drv', length 8
        > string 'Calculator1', length 11
        > string 'Alarm1', length 6
        15
[ 100] 15 - Door1      | Door1Drv   | Calculator1     | Alarm1
        > number 230
        > string 'Door2', length 5
        > string 'Door2Drv', length 8
        > string 'Calculator2', length 11
        > string 'Alarm2', length 6
        230
[ 158] 230 - Door2      | Door2Drv   | Calculator2     | Alarm2

随意定制。

使用μC/ OS-II直接解析json

1 个答案: