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Fix value returned by mysql_warning_count() after fetching a prepared
statement that generated a warning. (Bug #15510)
This commit is contained in:
@ -14590,6 +14590,40 @@ static void test_bug14845()
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myquery(rc);
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}
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/*
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Bug #15510: mysql_warning_count returns 0 after mysql_stmt_fetch which
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should warn
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*/
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static void test_bug15510()
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{
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MYSQL_STMT *stmt;
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MYSQL_RES *res;
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int rc;
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const char *query= "select 1 from dual where 1/0";
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myheader("test_bug15510");
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rc= mysql_query(mysql, "set @@sql_mode='ERROR_FOR_DIVISION_BY_ZERO'");
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myquery(rc);
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stmt= mysql_stmt_init(mysql);
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rc= mysql_stmt_prepare(stmt, query, strlen(query));
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check_execute(stmt, rc);
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rc= mysql_stmt_execute(stmt);
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check_execute(stmt, rc);
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rc= mysql_stmt_fetch(stmt);
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DIE_UNLESS(mysql_warning_count(mysql));
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/* Cleanup */
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mysql_stmt_close(stmt);
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rc= mysql_query(mysql, "set @@sql_mode=''");
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myquery(rc);
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}
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/*
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Read and parse arguments and MySQL options from my.cnf
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*/
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@ -14849,6 +14883,7 @@ static struct my_tests_st my_tests[]= {
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{ "test_bug13488", test_bug13488 },
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{ "test_bug13524", test_bug13524 },
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{ "test_bug14845", test_bug14845 },
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{ "test_bug15510", test_bug15510},
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{ 0, 0 }
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};
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