PHP connects multiple MySQL Database sample code at the same time

Source: Internet
Author: User
Tags db2 mysql database

  This article mainly introduces PHP to connect multiple MySQL database implementation, the need for friends can refer to the following

Instance:    code is 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", "DB2");  mysql_select_db ("DB2", $conn 2);     $sql = "SELECT * from IP";  $query = mysql_query ($sql);  if ($row = mysql_fetch_array ($query)) &nbsp ; 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, and the error occurs when the program executes: PHP Warning:mysql_fetch_array () expects parameter 1 to B E resource, Boolean given in ....    cause analysis:    program started to establish two database links, function mysql_query () prototype:    Resource m Ysql_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 the link_identifier is not specified, the first SQL is executed by using theThe previous open link, that is, $conn2, and in fact the first SQL statement should be using $CONN1, which causes an error, so in order to be able to link multiple MySQL databases, you can use the following methods:    Method 1: In Mysql_ The query function specifies the connection used, that is, the:    code as follows: <?php  $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: To associate the database used in the SQL statement, you can omit the second parameter of mysql_query,:    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", "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 ($row = mysql_fetch_array ($que RY)   echo $row [0]; ?> 
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