How to: Specify SQL Server Data Types When Using the SQLSRV Driver

Microsoft Drivers for PHP for SQL Server

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This topic demonstrates how to use the SQLSRV driver to specify the SQL Server data type for data that is sent to the server. This topic does not apply when using the PDO_SQLSRV driver.

To specify the SQL Server data type, you must use the optional $params array when you prepare or execute a query that inserts or updates data. For details about the structure and syntax of the $params array, see sqlsrv_query or sqlsrv_prepare.

The following steps summarize how to specify the SQL Server data type when sending data to the server:

Note Note

If no SQL Server data type is specified, default types will be used. For information about default SQL Server data types, see Default SQL Server Data Types.

  1. Define a Transact-SQL query that inserts or updates data. Use question marks (?) as placeholders for parameter values in the query.

  2. Initialize or update PHP variables that correspond to the placeholders in the Transact-SQL query.

  3. Construct the $params array to be used when preparing or executing the query. Note that each element of the $params array must also be an array when you specify the SQL Server data type.

  4. Specify the desired SQL Server data type by using the appropriate SQLSRV_SQLTYPE_* constant as the fourth parameter in each sub-array of the $params array. For a complete list of the SQLSRV_SQLTYPE_* constants, see the SQLTYPEs section of Constants (Microsoft Drivers for PHP for SQL Server). For example, in the code below, $changeDate, $rate, and $payFrequency are specified respectively as the SQL Server types datetime, money, and tinyint in the $params array. Because no SQL Server type is specified for $employeeId and it is initialized to an integer, the default SQL Server type integer is used.

      Copy imageCopy Code 
    $employeeId = 5;
$changeDate = "2005-06-07";
$rate = 30;
$payFrequency = 2;
$params = array(
            array($employeeId, null),
            array($changeDate, null, null, SQLSRV_SQLTYPE_DATETIME),
            array($rate, null, null, SQLSRV_SQLTYPE_MONEY),
            array($payFrequency, null, null, SQLSRV_SQLTYPE_TINYINT)
          );

Example

The following example inserts data into the HumanResources.EmployeePayHistory table of the Adventureworks database. SQL Server types are specified for the $changeDate, $rate, and $payFrequency parameters. The default SQL Server type is used for the $employeeId parameter. To verify that the data was inserted successfully, the same data is retrieved and displayed.

This example assumes that SQL Server and the AdventureWorks database are installed on the local computer. All output is written to the console when the example is run from the command line.

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<?php
/* Connect to the local server using Windows Authentication and 
specify the AdventureWorks database as the database in use. */
$serverName = "(local)";
$connectionInfo = array( "Database"=>"AdventureWorks");
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn === false )
{
     echo "Could not connect.\n";
     die( print_r( sqlsrv_errors(), true));
}

/* Define the query. */
$tsql1 = "INSERT INTO HumanResources.EmployeePayHistory (EmployeeID,
                                                        RateChangeDate,
                                                        Rate,
                                                        PayFrequency)
           VALUES (?, ?, ?, ?)";

/* Construct the parameter array. */
$employeeId = 5;
$changeDate = "2005-06-07";
$rate = 30;
$payFrequency = 2;
$params1 = array(
               array($employeeId, null),
               array($changeDate, null, null, SQLSRV_SQLTYPE_DATETIME),
               array($rate, null, null, SQLSRV_SQLTYPE_MONEY),
               array($payFrequency, null, null, SQLSRV_SQLTYPE_TINYINT)
           );

/* Execute the INSERT query. */
$stmt1 = sqlsrv_query($conn, $tsql1, $params1);
if( $stmt1 === false )
{
     echo "Error in execution of INSERT.\n";
     die( print_r( sqlsrv_errors(), true));
}

/* Retrieve the newly inserted data. */
/* Define the query. */
$tsql2 = "SELECT EmployeeID, RateChangeDate, Rate, PayFrequency
          FROM HumanResources.EmployeePayHistory
          WHERE EmployeeID = ? AND RateChangeDate = ?";

/* Construct the parameter array. */
$params2 = array($employeeId, $changeDate);

/*Execute the SELECT query. */
$stmt2 = sqlsrv_query($conn, $tsql2, $params2);
if( $stmt2 === false )
{
     echo "Error in execution of SELECT.\n";
     die( print_r( sqlsrv_errors(), true));
}

/* Retrieve and display the results. */
$row = sqlsrv_fetch_array( $stmt2 );
if( $row === false )
{
     echo "Error in fetching data.\n";
     die( print_r( sqlsrv_errors(), true));
}
echo "EmployeeID: ".$row['EmployeeID']."\n";
echo "Change Date: ".date_format($row['RateChangeDate'], "Y-m-d")."\n";
echo "Rate: ".$row['Rate']."\n";
echo "PayFrequency: ".$row['PayFrequency']."\n";

/* Free statement and connection resources. */
sqlsrv_free_stmt($stmt1);
sqlsrv_free_stmt($stmt2);
sqlsrv_close($conn);
?>

See Also

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