Re: new json funcs

Josh Berkus wrote:
> On 01/27/2014 01:06 PM, Alvaro Herrera wrote:
> > Andrew Dunstan escribió:
> >
> >> I'm not sure I understand the need. This is the difference between
> >> the _text variants and their parents. Why would you call
> >> json_object_field when you want the dequoted text?
> >
> > Because I first need to know its type. Sometimes it's an array, or an
> > object, or a boolean, and for those I won't call the _text version
> > afterwards but just use the original.
>
> It would make more sense to extract them as JSON, check the type, and
> convert.

That's what I'm already doing. My question is how do I convert it?
I have this, but would like to get rid of it:

/*
* dequote_jsonval
* Take a string value extracted from a JSON object, and return a copy of it
* with the quoting removed.
*
* Another alternative to this would be to run the extraction routine again,
* using the "_text" variant which returns the value without quotes; but this
* complicates the logic a lot because not all values are extracted in
* the same way (some are extracted using json_object_field, others
* using json_array_element). Dequoting the object already at hand is a
* lot easier.
*/
static char *
dequote_jsonval(char *jsonval)
{
char *result;
int inputlen = strlen(jsonval);
int i;
int j = 0;

result = palloc(strlen(jsonval) + 1);

/* skip the start and end quotes right away */
for (i = 1; i < inputlen - 1; i++)
{
/*
* XXX this skips the \ in a \" sequence but leaves other escaped
* sequences in place. Are there other cases we need to handle
* specially?
*/
if (jsonval[i] == '\\' &&
jsonval[i + 1] == '"')
{
i++;
continue;
}

result[j++] = jsonval[i];
}
result[j] = '\0';

return result;
}

--
Álvaro Herrera http://www.2ndQuadrant.com/
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