Re: PHP 4.0 Bug #7032: Multiple mysql connections over-write eachother
| From: | chrisv at b0rked dot dhs dot org | Date: | Wed, 04 Oct 2000 20:35:47 +0000 |
| Subject: | Re: PHP 4.0 Bug #7032: Multiple mysql connections over-write eachother | ||
| References: | 1 | Groups: | php.dev |
| Request: | Send a blank email to php-dev+get-34260@lists.php.net to get a copy of this message | ||
This is clearly documented within the mysql_connect() documentation.
If a second call is made to mysql_connect() with the same arguments, no
new link will be established, but instead, the link identifier of the
already opened link will be returned.
Try using mysql_db_query() instead --
-------------------------
$conn_id = mysql_connect("localhost","root","");
mysql_db_query("Database1", "select * from table1", $conn_id);
mysql_db_query("Database2", "select * from table2", $conn_id);
-------------------------
Chris
On 5 Oct 2000 bob@ark.org wrote:
> From: bob@ark.org
> Operating system: Unix
> PHP version: 4.0.2
> PHP Bug Type: MySQL related
> Bug description: Multiple mysql connections over-write eachother
>
> Assume you have 2 database in one mysql server:
>
> Database1 (contains table: table1)
> Database2 (contains table: table2)
>
> Try this:
> -----------------------
> $conn_id1 = mysql_connect("localhost", "root", "");
> $conn_id2 = mysql_connect("localhost", "root", "");
>
> mysql_select_db("Database1", $conn_id1);
> mysql_select_db("Database2", $conn_id2);
>
> mysql_query("select * from table1", $conn_id1);
> mysql_query("select * from table2", $conn_id2);
>
> -----------------
> the first query will fail, b/c the 'select_db' for conn_id1 was overwritten in the
> 'select_db' for conn_id2: it is looking for table1 in Database2.
>
> This hold true even inside objects, so it is difficult to create an object that accesses the
> database and use it in a script that accesses a different database.
>
>
>
>
>
> --
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