04-30-2009
Thank you , but still the problem is not solved
When i used this stmt,
`echo "$i" | cut -d '|' -f1`
the output is blank...
The other one is giving me an error.
Could you all please give me any other code if possible ?
Thanks in advance
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sqlsrv_fetch
SQLSRV_FETCH(3) SQLSRV_FETCH(3)
sqlsrv_fetch - Makes the next row in a result set available for reading
SYNOPSIS
mixed sqlsrv_fetch (resource $stmt, [int $row], [int $offset])
DESCRIPTION
Makes the next row in a result set available for reading. Use sqlsrv_get_field(3) to read the fields of the row.
PARAMETERS
o $stmt
- A statement resource created by executing sqlsrv_query(3) or sqlsrv_execute(3).
o $row
- The row to be accessed. This parameter can only be used if the specified statement was prepared with a scrollable cursor. In
that case, this parameter can take on one of the following values:
oSQLSRV_SCROLL_NEXT
oSQLSRV_SCROLL_PRIOR
oSQLSRV_SCROLL_FIRST
oSQLSRV_SCROLL_LAST
oSQLSRV_SCROLL_ABSOLUTE
oSQLSRV_SCROLL_RELATIVE
o $offset
- Specifies the row to be accessed if the row parameter is set to SQLSRV_SCROLL_ABSOLUTE or SQLSRV_SCROLL_RELATIVE. Note that the
first row in a result set has index 0.
RETURN VALUES
Returns TRUE if the next row of a result set was successfully retrieved, FALSE if an error occurs, and NULL if there are no more rows in
the result set.
EXAMPLES
Example #1
sqlsrv_fetch(3) example
The following example demonstrates how to retrieve a row with sqlsrv_fetch(3) and get the row fields with sqlsrv_get_field(3).
<?php
$serverName = "serverNamesqlexpress";
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$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn === false ) {
die( print_r( sqlsrv_errors(), true));
}
$sql = "SELECT Name, Comment
FROM Table_1
WHERE ReviewID=1";
$stmt = sqlsrv_query( $conn, $sql);
if( $stmt === false ) {
die( print_r( sqlsrv_errors(), true));
}
// Make the first (and in this case, only) row of the result set available for reading.
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SEE ALSO
sqlsrv_get_field(3), sqlsrv_fetch_array(3), sqlsrv_fetch_object(3).
PHP Documentation Group SQLSRV_FETCH(3)