I want to replace a string which contains "/" in vi but what is the escape character for forward slash?
e.g. I have a text file with the contents below and I want to replace "/Top/Sub/Sub1" with "ABC".
/Top/Sub/Sub1
The replace command I am using is ... (4 Replies)
Hi ,
I want to change space to ' in my script.
I tried doing this,
sed 's/ /\'/g' filename
but i could not get it.
can some one help me please.
Thanks,
Deepak (4 Replies)
Hi All,
How do i write in sed for the 6th and 7th field of etc/passwd file as it involves "/" character?
Does mine below is correct? It's incomplete script as i need help with syntax as i always getting may errors :(
Example of etc/passwd file:
blah:x:1055:600:blah... (6 Replies)
Hello experts
I am trying to write a shell script which will add ' ' to a unix variable and then pass it to oracle for inserting to a table.
I am running the script as root and I have to do a su -c .
The problem is the character ' is not recognised inside sed even after adding escape... (1 Reply)
my @array;
my $sepa = "|";
print $sepa;
open FH, "<100_20091023_2.txt";
while(<FH>){
push @array, split(/\$sepa/, $_);
print "@array\n\n";
}
I am not able split the line which have | separated (1 Reply)
All ,
i have input line as below .
abc\ , ewioweioi \,
and want the output as below removing the "\"
abc , ewioweioi ,
could anyone help me out (2 Replies)
I need to change a pattern with single quotes
# echo "serversignature: 'On'"
serversignature: 'On'
I did
# echo "serversignature: 'On'" | sed 's/.*serversignature.*/serversignature: 'Off'/'
serversignature: Off
The output I need is with single quotes. But its swallowing it.
... (2 Replies)
friends,
I have a situation where i am using a $RANDOM function along with the filename, I want this to be escaped by the OS in the first assignment (works as expected) and executed in the second assignment (does not execute $RANDOM)
filename1=filename1_\$RANDOM
echo $filename1... (3 Replies)
Hi ,
I am looking for a function which will do the following.
1. I have a variable which will hold few special chracter like
SPECIAL_CHARS="& ;"2. I have an escape character.
ESCAPE_CHAR="\"3. Now when I passed some string in the function it will return the same string but now it will... (8 Replies)
Discussion started by: Anupam_Halder
8 Replies
LEARN ABOUT PHP
sqlsrv_connect
SQLSRV_CONNECT(3)SQLSRV_CONNECT(3)sqlsrv_connect - Opens a connection to a Microsoft SQL Server databaseSYNOPSIS
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)