Subj : Re: server-side compilation using spidermonkey
To   : wyl_lyf
From : Brendan Eich
Date : Thu Jan 06 2005 11:40 am

wyl_lyf@yahoo.com wrote:
> Hello,
> 
> thanks for the information. It would be nice to execute the script in
> runtime. But I'm not sure how to do this without running apache or
> opening the browser.
> 
> Example:
> Given that I only have a javascript with the response.write ("..").
> So I will have to create a program that links with spidermonkey
> libraries pass it my javascript server response and it will be able to
> check syntax and compile?
> 
> is this correct?
> 
> many thanks!


Sure, it's this easy:

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

JSClass global_class = {
     "global",
     JS_PropertyStub, JS_PropertyStub, JS_PropertyStub, JS_PropertyStub,
     JS_EnumerateStub, JS_ResolveStub, JS_ConvertStub, JS_FinalizeStub
     JSCLASS_NO_OPTIONAL_MEMBERS
};

int
main(int argc, char **argv)
{
     JSRuntime *rt;
     JSContext *cx;
     JSObject *global;
     JSBool ok;

     if (argc == 1) {
         fprintf(stderr, "usage: %s source\n", argv[0]);
         return -1;
     }
     rt = JS_NewRuntime(4 * 1024 * 1024);
     if (!rt)
         return 2;
     cx = JS_NewContext(rt, 8192);
     if (!cx)
         return 2;
     global = JS_NewObject(cx, &global_class, NULL, NULL);
     if (!global)
         return 2;
     if (!JS_InitStandardClasses(cx, global))
         return 2;
     ok = JS_BufferIsCompilableUnit(cx, global, argv[1],
                                    strlen(argv[1]));
     return ok ? 0 : 1;
}

This program exits -1 on usage error, 2 on out of memory or similar 
problems, 1 on syntax error in argv[1], and 0 if argv[1] is a 
well-formed JS program.

/be

.