MYSQL API in C - linux

Question

I can't get the following code to work. It's working with normal select or insert, but when I try to insert a variable to it it didn't work. Can some let me know please what's wrong here?

Please note that the output is below.

int main(int argc, char **argv)
{

  MYSQL *conn;
  char str[100] = "test";
  conn = mysql_init(NULL);

  char stmt_buf[100];
  sprintf (stmt_buf, "insert into test values ('%s')", str);

  printf("\n%s\n",stmt_buf);

  mysql_query (conn, stmt_buf);
  mysql_close(conn);
  return 0;
}

~$./version

insert into test values ('test')
Segmentation fault

You should run your program through GDB to find out where it's crashing. gdb ./version -> run — spencercw
– spencercw, Commented Feb 25, 2012 at 18:19
Did you try to compile your code with all warnings and debugging information (e.g. gcc -Wall -g), to improve your code till no warnings are given by the compiler, and to use a debugger (e.g. gdb)?? — Basile Starynkevitch
– Basile Starynkevitch, Commented Feb 25, 2012 at 18:21
I did but the output wasn't helpful as below. Program received signal SIGSEGV, Segmentation fault. 0x00199634 in mysql_send_query () from /usr/lib/libmysqlclient.so.16 — Bag Dev
– Bag Dev, Commented Feb 25, 2012 at 18:21
-1: If you'd done any debugging whatsoever, you'd see that conn is 0x0. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Feb 25, 2012 at 18:48

Dietrich Epp · Accepted Answer · 2012-02-25 18:19:56Z

2

You don't do any error checking. My guess is conn == NULL.

For the record, don't use sprintf ever. Use snprintf instead, that is, if you're okay with SQL injections.

answered Feb 25, 2012 at 18:19

Dietrich Epp

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Comments

Shiplu Mokaddim · Accepted Answer · 2012-07-09 15:40:44Z

1

I didn't see you are connected to database by mysql_real_connect also there is no mysql_select_db and mysql_exec_sql call.

Here is an example for you.

int main(int argc, char **argv)
{
    MYSQL mysql;
    char stmt_buf[100];
    if (mysql_init(&mysql) == NULL) {
        printf("Can not initialize");
        exit(1);
    }
    if (!mysql_real_connect
        (&mysql, "localhost", "USERNAME", "PASSWORD", NULL, 0, NULL, 0)) {
        mysql_error(&mysql);
        exit(1);
    }

    if (mysql_select_db(&mysql, "DATABASENAME")) {
        mysql_error(&mysql);
        exit(1);
    }

    sprintf(stmt_buf, "insert into test values ('%s')", "test");

    if (mysql_exec_sql(&mysql, stmt_buf))
        mysql_error(&mysql));

    mysql_close(&mysql);
}

edited Jul 9, 2012 at 15:40

answered Feb 25, 2012 at 18:44

Shiplu Mokaddim

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1 Comment

BraveNewCurrency Over a year ago

Please note: this is a SQL injection waiting to happen if someone replaces "test" with untrusted user input. xkcd.com/327

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