JSON Parser using c

Question

I have to parse a JSON using c code(not lib because want to make things look as simple as possible) for some real-time handling. Below is data need to be parsed which I will be getting from some calculation generated by the code itself. Please help me out.

[
{
    "Letter": 0 ,
    "Freq": 2858    
},
.
.
.
.
.
{
    "Letter" : 31,
    "Freq" : 0
}
]

Are you sure, implementing a JSON parser will be simpler than just using an existing lib? — Gerhardh
– Gerhardh, Commented Apr 29, 2019 at 10:43
Is the JSON always in the same format with the same number of labels? — David Ranieri
– David Ranieri, Commented Apr 29, 2019 at 10:59
@KeineLust Yes, in this particular case only frequency changes. — ashu nath
– ashu nath, Commented Apr 29, 2019 at 17:30

Shubham · Accepted Answer · 2019-04-29 10:46:09Z

4

These are two C libraries I have used.

https://github.com/DaveGamble/cJSON : this can parse your string and can prepare json strings.
https://github.com/zserge/jsmn : this is only for parsing json strings.

Both libraries are well documented and has test code available.

answered Apr 29, 2019 at 10:46

Shubham

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4 Comments

Moritz Schmidt Over a year ago

He says he doesnt want to use libraries

Shubham Over a year ago

oh, sorry I missed it. Then in this case using strstr may be helpful in parsing string on our own

ashu nath Over a year ago

@shubham Can you please elaborate ? I am calculating data in the code and want to update it in json file.

Shubham Over a year ago

Can you clarify whether you want to update json to a file or just parse it? cJSON_Test, here is the test code for parsing as well as updating the json read/update from/to file. And usage of strstr and strtok is explained in belowed answer by Keine Lust

David Ranieri · Accepted Answer · 2019-04-29 18:01:59Z

3

It seems that you only want to extract the "Freq" value, in this case this code is enough:

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

char *str = "[{\"Letter\": 0 ,\"Freq\": 2858},{\"Letter\" : 31,\"Freq\" : 0}]";

int main(void)
{
    char *ptr = str;
    long value;

    while (ptr) {
        ptr = strstr(ptr, "\"Freq\"");
        if (ptr == NULL) {
            break;
        }
        ptr = strchr(ptr, ':');
        if (ptr == NULL) {
            break;
        }
        ptr++;
        value = strtol(ptr, &ptr, 10);
        if (*ptr != '}') {
            break;
        }
        ptr++;
        printf("%lu\n", value);
    }
    return 0;
}

answered Apr 29, 2019 at 18:01

David Ranieri

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