| Bug #6047 | permission problem when executing mysql_stmt_execute with derived tables | ||
|---|---|---|---|
| Submitted: | 12 Oct 2004 17:12 | Modified: | 22 Oct 2004 23:30 |
| Reporter: | Georg Richter | Email Updates: | |
| Status: | Can't repeat | Impact on me: | |
| Category: | MySQL Server | Severity: | S2 (Serious) |
| Version: | 4.1 | OS: | Linux (Linux) |
| Assigned to: | Michael Widenius | CPU Architecture: | Any |
[22 Oct 2004 23:30]
Michael Widenius
Not enough information was provided for us to be able to handle this bug. Please re-read the instructions at http://bugs.mysql.com/how-to-report.php If you can provide more information, feel free to add it to this bug and change the status back to 'Open'. Thank you for your interest in MySQL. Additional info: I tried to repeat this in the current 4.1 tree but was not able to do that. I did however create a test case for this in ps.test
Description: When using "UPDATE `t1` AS P1 INNER JOIN (SELECT N FROM `t1` GROUP BY N HAVING Count(M) > 1) AS P2 ON P1.N = P2.N SET P1.M = 2" from derived.test with prepared statements mysql_stmt_execute fails with permission problem: Error in mysql_stmt_execute: Access denied for user ''@'localhost' to database '' The same statement works fine when using mysql_query. How to repeat: #include <stdio.h> #include <mysql.h> int main () { MYSQL *mysql = mysql_init(NULL); MYSQL_STMT *stmt; MYSQL_BIND bind[1]; char query[500]; int b; if (!mysql_real_connect(mysql, "localhost", "", "", "test", 0, NULL, 0)) { printf("Can't connect! Error: %s\n", mysql_error(mysql)); return 1; } mysql_query(mysql, "DROP TABLE IF EXISTS t1"); mysql_query(mysql, "CREATE TABLE t1 (N int, M tinyint)"); mysql_query(mysql, "INSERT INTO a1 VALUES (1,0),(1,0),(2,0),(2,0),(3,0)"); stmt = mysql_stmt_init(mysql); strcpy(query, "UPDATE t1 AS P1 INNER JOIN (SELECT N FROM t1 GROUP BY N HAVING Count(M) > 1)" \ "AS P2 ON P1.N = P2.N SET P1.M = 2"); if (mysql_stmt_prepare(stmt, query, strlen(query))) { printf("Error in mysql_stmt_prepare: %s\n", mysql_error(mysql)); return 1; } if (mysql_stmt_execute(stmt)) { printf("Error in mysql_stmt_execute: %s\n", mysql_stmt_error(stmt)); return 1; } mysql_stmt_close(stmt); mysql_close(mysql); }