04-03-2008
Getting error in command line arguments
Hi,
When i am running the following script 1.sh (without giving the command line arguments) then i am getting the following error.
if [ $1 = dm5admin ]
then
echo "UID and PWD are correct"
elif [ $1 != dm5admin ]
then
echo "Either UID or PWD is wrong. Please check your UID and PWD"
else
echo "UID and PWD can't be blank"
exit;
fi
by running this 1.sh script i am getting the following error.
>1.sh
./1.sh: [: =: unary operator expected
./1.sh: [: !=: unary operator expected
UID and PWD can't be blank
Can anybody rectify this?????
Thanks.
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LEARN ABOUT PHP
sqlsrv_connect
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)