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Top Forums Shell Programming and Scripting Command substitution inside of a variable expression (AIX, KORN) Post 302761159 by Subbeh on Friday 25th of January 2013 08:39:43 AM
Old 01-25-2013
Your code works fine for me under Solaris.

Maybe try cd ${1:-$PWD} instead.
 

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SQLSRV_CONNECT(3)														 SQLSRV_CONNECT(3)

sqlsrv_connect - Opens a connection to a Microsoft SQL Server database

SYNOPSIS
resource sqlsrv_connect (string $serverName, [array $connectionInfo]) DESCRIPTION
Opens a connection to a Microsoft SQL Server database. By default, the connection is attempted using Windows Authentication. To connect using SQL Server Authentication, include "UID" and "PWD" in the connection options array. PARAMETERS
o $serverName - The name of the server to which a connection is established. To connect to a specific instance, follow the server name with a forward slash and the instance name (e.g. serverNamesqlexpress). o $connectionInfo - An associative array that specifies options for connecting to the server. If values for the UID and PWD keys are not specified, the connection will be attempted using Windows Authentication. For a complete list of supported keys, see SQLSRV Connection Options. RETURN VALUES
A connection resource. If a connection cannot be successfully opened, FALSE is returned. EXAMPLES
Example #1 Connect using Windows Authentication. <?php $serverName = "serverNamesqlexpress"; //serverNameinstanceName // Since UID and PWD are not specified in the $connectionInfo array, // The connection will be attempted using Windows Authentication. $connectionInfo = array( "Database"=>"dbName"); $conn = sqlsrv_connect( $serverName, $connectionInfo); if( $conn ) { echo "Connection established.<br />"; }else{ echo "Connection could not be established.<br />"; die( print_r( sqlsrv_errors(), true)); } ?> Example #2 Connect by specifying a user name and password. <?php $serverName = "serverNamesqlexpress"; //serverNameinstanceName $connectionInfo = array( "Database"=>"dbName", "UID"=>"userName", "PWD"=>"password"); $conn = sqlsrv_connect( $serverName, $connectionInfo); if( $conn ) { echo "Connection established.<br />"; }else{ echo "Connection could not be established.<br />"; die( print_r( sqlsrv_errors(), true)); } ?> Example #3 Connect on a specifed port. <?php $serverName = "serverNamesqlexpress, 1542"; //serverNameinstanceName, portNumber (default is 1433) $connectionInfo = array( "Database"=>"dbName", "UID"=>"userName", "PWD"=>"password"); $conn = sqlsrv_connect( $serverName, $connectionInfo); if( $conn ) { echo "Connection established.<br />"; }else{ echo "Connection could not be established.<br />"; die( print_r( sqlsrv_errors(), true)); } ?> NOTES
By default, the sqlsrv_connect(3) uses connection pooling to improve connection performance. To turn off connection pooling (i.e. force a new connection on each call), set the "ConnectionPooling" option in the $connectionOptions array to 0 (or FALSE). For more information, see SQLSRV Connection Pooling. The SQLSRV extension does not have a dedicated function for changing which database is connected to. The target database is specified in the $connectionOptions array that is passed to sqlsrv_connect. To change the database on an open connection, execute the following query "USE dbName" (e.g. sqlsrv_query($conn, "USE dbName")). SEE ALSO
sqlsrv_close(3), sqlsrv_errors(3), sqlsrv_query(3). PHP Documentation Group SQLSRV_CONNECT(3)
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