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Top Forums UNIX for Dummies Questions & Answers counter / increment problem within echo stmt Post 1091 by blaze on Wednesday 7th of February 2001 04:57:06 PM
Old 02-07-2001
Question

Simple script trying to increment a counter within an echo statement never gets past 1 - PLEASE HELP!

Thanks.
~~~~~~~~~~~

#!/bin/sh
stepup()
{
STEP=`expr $STEP + 1`
echo $STEP
}

#
# Initialize variables
#
STEP=0

echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
echo "Counter Value: `stepup`"
 

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SQLSRV_NEXT_RESULT(3)													     SQLSRV_NEXT_RESULT(3)

sqlsrv_next_result - Makes the next result of the specified statement active

SYNOPSIS
mixed sqlsrv_next_result (resource $stmt) DESCRIPTION
Makes the next result of the specified statement active. Results include result sets, row counts, and output parameters. PARAMETERS
o $stmt - The statment on which the next result is being called. RETURN VALUES
Returns TRUE if the next result was successfully retrieved, FALSE if an error occurred, and NULL if there are no more results to retrieve. EXAMPLES
Example #1 sqlsrv_next_result(3) example The following example executes a batch query that inserts into a table and then selects from the table. This produces two results on the statement: one for the rows affected by the INSERT and one for the rows returned by the SELECT. To get to the rows returned by the SELECT, sqlsrv_next_result(3) must be called to move past the first result. <?php $serverName = "serverNamesqlexpress"; $connectionInfo = array("Database"=>"dbName", "UID"=>"userName", "PWD"=>"password"); $conn = sqlsrv_connect( $serverName, $connectionInfo); $query = "INSERT INTO Table_1 (id, data) VALUES (?,?); SELECT * FROM TABLE_1;"; $params = array(1, "some data"); $stmt = sqlsrv_query($conn, $query, $params); // Consume the first result (rows affected by INSERT) without calling sqlsrv_next_result. echo "Rows affected: ".sqlsrv_rows_affected($stmt)."<br />"; // Move to the next result and display results. $next_result = sqlsrv_next_result($stmt); if( $next_result ) { while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC)){ echo $row['id'].": ".$row['data']."<br />"; } } elseif( is_null($next_result)) { echo "No more results.<br />"; } else { die(print_r(sqlsrv_errors(), true)); } ?> SEE ALSO
sqlsrv_query(3), sqlsrv_fetch_array(3), sqlsrv_rows_affected(3). PHP Documentation Group SQLSRV_NEXT_RESULT(3)
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