Bug #16903 Updated: CREATE DATABASE - No Database Selected
| From: | zak@php.net | Date: | Thu, 27 Jun 2002 11:59:54 +0000 |
| Subject: | Bug #16903 Updated: CREATE DATABASE - No Database Selected | ||
| References: | 1 | Groups: | php.bugs |
| Request: | Send a blank email to php-bugs+get-12234@lists.php.net to get a copy of this message | ||
ID: 16903
Updated by: zak@php.net
Reported By: hz11@nyu.edu
-Status: Open
+Status: Feedback
Bug Type: MySQL related
Operating System: FreeBSD 4.5
PHP Version: 4.1.2
-Assigned To:
+Assigned To: zak
New Comment:
Hi Hans,
I cannot reproduce the problem under PHP-4.3.0 Dev
Please try upgrading to the latest version of
PHP. If the problem still exists, please also
provide details on the version of MySQL that you
are trying to access.
Regarding your question, please submit a bug
report with a category of Feature Request!
Thanks!
Previous Comments:
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[2002-04-29 09:44:06] hz11@nyu.edu
I've found what looks to be a wierd little problem. When executing
something like:
mysql_query('CREATE DATABASE mynewdb', $mysql_link);
I get the error 'No database Selected'. In the mysql client (local,
CLI) this of course works as it should. I've looked at
mysql_create_db() in the manual, but it says it's deprecated and that a
'CREATE DATABASE' query should be executed using mysql_query().
There's something wrong here.
On a sidenote, I'm curious if there is any way (or plans to allow)
multiple queries per mysql_query() run? For instance:
mysql_query('CREATE DATABASE mynewdb; CREATE TABLE newtable (id TINYINT
NOT NULL);', $mysql_link);
Currently, I would get a SQL syntax error at the first semicolon. Is
there a way to do this? Or is there plans to add it in the future?
Thank you,
Hans
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Edit this bug report at http://bugs.php.net/?id=16903&edit=1