If the client_photo blob is initialized it won't seg. fault.
It looks like there is no check when a sql_blob type is not
initialized. (when used in this manner)
It should be able to handle this without seg. faulting.
Changing the populate_client_info() to the following works without
seg. faulting:
void populate_client_info(clients& client_info)
{
client_info.billing_id = 1;
client_info.client_first_name = "Graham";
client_info.client_middle_initial = "T";
client_info.client_last_name = "Reitz";
char data = 'A'; // new
ColData cd(&data, 1); // new
client_info.client_photo = cd; // new
}
Thanks,
Graham
On Aug 10, 2007, at 7:41 PM, Graham Reitz wrote:
> Thanks for the reply Warren,
>
> If the sql_blob is there it seg faults, if not it's ok.
>
> The following code will get it to crash: (Can you see anything?)
>
> #include <string>
> #include <mysql++.h>
> #include <custom.h>
>
> using namespace std;
> using namespace mysqlpp;
>
> sql_create_6(clients, 1, 6,
> unsigned int, client_id,
> unsigned int, billing_id,
> string, client_first_name,
> string, client_middle_initial,
> string, client_last_name,
> sql_blob, client_photo)
>
> void populate_client_info(clients& client_info)
> {
> client_info.billing_id = 1;
> client_info.client_first_name = "Joe";
> client_info.client_middle_initial = "T";
> client_info.client_last_name = "Shmo";
> }
>
> int main (int argc, char* const argv[])
> {
> Connection connection("db_name","ip_addres","user",
> "pw",3306, false);
>
> Query client_query = connection.query();
>
> clients a_client;
> populate_client_info(a_client);
>
> client_query.insert(a_client);
>
> return 0;
> }
>
>
>
>
> On Aug 9, 2007, at 7:04 AM, Warren Young wrote:
>
>> Graham Reitz wrote:
>>> I was able to narrow it down to this:
>>> In the sql_create_N macro I have a blob entry:
>>> mysqlpp::sql_blob, my_blob
>>> If the sql_blob type is in the sql_create call it seg. faults,
>>> otherwise it succeeds.
>>
>> Can you modify examples/load_jpeg.cpp to exhibit the same problem?
>> Providing an example of the problem in a way anyone can run will
>> increase the chances that someone can help you.
>>
>> --
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>>
>
