I am worse than a newbie at C, but...
On 4/7/08, dark0s dark0s <[EMAIL PROTECTED]> wrote:
> Hi all, after writing my program, I typed:
>
> bash-3.1# gcc -lsqlite3 CreaDB.c -o creadb
> CreaDB.c: In function 'main':
> CreaDB.c:35: error: too few arguments to function 'sqlite3_bind_text'
> CreaDB.c:36: error: too few arguments to function 'sqlite3_bind_text'
> CreaDB.c:37: error: too few arguments to function 'sqlite3_bind_text'
>
> But I don't see the problem, my program is:
>
> int main(int argc, char** argv[]) {
>
> int rc, i;
> sqlite3* db;
> sqlite3_stmt* stmt;
> char* sql;
> const char* tail;
>
> rc = sqlite3_open("prova.db", &db);
> if (rc) {
> fprintf(stderr, "E' impossibile aprire il file %s\n", sqlite3_errmsg(db));
> sqlite3_close(db);
> exit(1);
> }
>
> sql = "create table modulo(id);";
>
> rc = sqlite3_prepare(db, sql, strlen(sql), &stmt, &tail);
> if (rc != SQLITE_OK) {
> fprintf(stderr, "Errore SQL: %s\n", sqlite3_errmsg(db));
> }
>
> rc = sqlite3_step(stmt);
> sqlite3_reset(stmt);
>
> sql = "insert into modulo(id, nome, classe, istanza) values(?,?,?,?)";
> sqlite3_prepare(db, sql, strlen(sql), &stmt, &tail);
>
> sqlite3_bind_int(stmt, 1, 1);
> sqlite3_bind_text(stmt, 2, "nome1");
> sqlite3_bind_text(stmt, 3, "classe1");
> sqlite3_bind_text(stmt, 4, "istanza1");
aren't you binding 4 values to the SQL statement, but providing only 3?
> sqlite3_step(stmt);
>
> while (rc == SQLITE_ROW) {
> for (i = 0; i < sqlite3_column_count(stmt); i++)
> fprintf(stderr, "'%s' ", sqlite3_column_text(stmt, i));
> fprintf(stderr, "\n");
> rc = sqlite3_step(stmt);
> }
>
> sqlite3_finalize(stmt);
> sqlite3_close(db);
>
> return 0;
>
> }
>
> Thank you very much in advance for answers,
> savio
>
>
>
> ---------------------------------
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