PHP connect multiple mysql database sample code at the same time_PHP tutorial

Example:

Copy code The code is as follows:

$conn1 = mysql_connect("127.0.0.1 ", "root","root","db1");
mysql_select_db("db1", $conn1);
$conn2 = mysql_connect("127.0.0.1", "root","root", "db2");
mysql_select_db("db2", $conn2);

$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. When the program is executed An error will be reported: PHP Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ....

Cause analysis:

The program starts to establish two database links, the function mysql_query( ) Prototype:

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 last opened connection is used. If there is no open connection, this function will try to call the mysql_connect() function without parameters to establish a connection and use it. Query results will be cached.

In this example, since link_identifier is not specified, when executing the first sql, the previous open link, $conn2, is used by default, but in fact the first sql statement should use It is $conn1, so an error is reported, so in order to be able to link multiple mysql databases, you can use the following method:

Method 1: Specify the connection used in the mysql_query function, that is:

Copy code The code is as follows:

$conn1 = mysql_connect("127.0.0.1", "root","root"," db1");
mysql_select_db("Muma", $conn1);
$conn2 = mysql_connect("127.0.0.1", "root","root","db2");
mysql_select_db(" product", $conn2);

$sql = "select * from ip";
$query = mysql_query($sql,$conn1); //Add connection $conn1
if($ row = mysql_fetch_array($query))
echo $row[0]."n";

$sql = "select * from web ";
$query = mysql_query($sql, $ conn2);
if($row = mysql_fetch_array($query))
echo $row[0];
?>

Method 2: Association in sql statement For the database used, the second parameter of mysql_query can be omitted at this time, that is:

Copy code The code is as follows:

$conn1 = mysql_connect("127.0.0.1", "root","root","db1");
mysql_select_db("db1", $conn1);
$conn2 = mysql_connect( "127.0.0.1", "root","root","db2");
mysql_select_db("db2", $conn2);

$sql = "select * from db1.ip"; //Associated 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.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/742452.htmlTechArticleExample: Copy the code as follows: ?php $conn1 = mysql_connect("127.0.0.1", "root", "root","db1"); mysql_select_db("db1", $conn1); $conn2 = mysql_connect("127.0.0.1", "root","root"...