PHP simultaneously connect multiple MySQL Database sample code _php tutorial

Instance:
Copy CodeThe code is as follows:
$conn 1 = mysql_connect ("127.0.0.1", "root", "root", "db1");
mysql_select_db ("DB1", $conn 1);
$conn 2 = mysql_connect ("127.0.0.1", "root", "root", "DB2");
mysql_select_db ("DB2", $conn 2);

$sql = "SELECT * from IP";
$query = mysql_query ($sql);
if ($row = mysql_fetch_array ($query))
echo $row [0]. " \ n ";

$sql = "SELECT * from Web";
$query = mysql_query ($sql);
if ($row = mysql_fetch_array ($query))
echo $row [0];
?>

There is a problem with this code, which will be error when the program executes: PHP Warning:mysql_fetch_array () expects parameter 1 to be resource, boolean given in ....

Cause Analysis:

The program starts building two database links, function mysql_query () Prototypes:

Resource mysql_query (String $query [, Resource $link _identifier])

Sends a query to the currently active database in the server associated with the specified connection identifier. If Link_identifier is not specified, the previous open connection is used. If there is no open connection, this function attempts to call the mysql_connect () function without arguments to establish a connection and use it. The query results are cached.

In this case, because Link_identifier is not specified, the first SQL is executed by default with the previous open link, which is $conn2, and the first SQL statement should actually use $CONN1, which results in an error, So in order to be able to link multiple MySQL databases, you can use the following methods:

Method 1: Specify the connection used in the mysql_query function, i.e.:
Copy CodeThe code is as follows:
$conn 1 = mysql_connect ("127.0.0.1", "root", "root", "db1");
mysql_select_db ("Muma", $conn 1);
$conn 2 = mysql_connect ("127.0.0.1", "root", "root", "DB2");
mysql_select_db ("Product", $conn 2);

$sql = "SELECT * from IP";
$query = mysql_query ($sql, $conn 1); Add Connection $conn1
if ($row = mysql_fetch_array ($query))
echo $row [0]. " \ n ";

$sql = "SELECT * from Web";
$query = mysql_query ($sql, $conn 2);
if ($row = mysql_fetch_array ($query))
echo $row [0];
?>

Method 2: Associate the database used in the SQL statement, at which point the second parameter of mysql_query can be omitted, namely:
Copy CodeThe code is as follows:
$conn 1 = mysql_connect ("127.0.0.1", "root", "root", "db1");
mysql_select_db ("DB1", $conn 1);
$conn 2 = mysql_connect ("127.0.0.1", "root", "root", "DB2");
mysql_select_db ("DB2", $conn 2);

$sql = "SELECT * from Db1.ip"; Association database
$query = mysql_query ($sql);
if ($row = mysql_fetch_array ($query))
echo $row [0]. " \ n ";

$sql = "SELECT * from Db2.web";
$query = mysql_query ($sql);
if ($row = mysql_fetch_array ($query))
echo $row [0];
?>

http://www.bkjia.com/PHPjc/742452.html www.bkjia.com true http://www.bkjia.com/PHPjc/742452.html techarticle Example: Copy code as follows: PHP $conn 1 = mysql_connect ("127.0.0.1", "root", "root", "db1"), mysql_select_db ("DB1", $conn 1); $conn 2 = mysql_connect ("127.0.0.1", "root", "root " ...)

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