Website form not submitting data to mssql database - php

Question

I have spent days trying to figure how to get form data on my webpage to insert into my employee table on the forklift database mssql. when i click submit on the form it refreshes the page with Connection established but no data in the database.
<?php
/* Connect using Windows Authentication. */
$serverName = "EXAMPLE";
$connectionOptions = array("Database"=>"FORKLIFT");
/* Connect using Windows Authentication. */
$conn = sqlsrv_connect( $serverName, $connectionOptions);
if($conn)
{
echo "Connection established.<br />";
}
else
{
echo "Connection could not be established.<br />";
die(print_r(sqlsrv_errors(), true));
}
if(empty($_POST) === false && empty($errors)=== true)
{
//assign form input to variables
$FIRSTNAME = $_POST["FIRSTNAME"];
$LASTNAME = $_POST["LASTNAME"];
$DATEOFBIRTH = $_POST["DATEOFBIRTH"];
$PHONENUMBER = $_POST["PHONENUMBER"];
$ADDRESS = $_POST["ADDRESS"];
/*Insert data.*/
$INSERT_ROW = $query = "INSERT INTO
EMPLOYEE(FIRSTNAME,LASTNAME,DATEOFBIRTH,PHONENUMBER,ADDRESS)
VALUES ('$FIRSTNAME','$LASTNAME','$DATEOFBIRTH','$PHONENUMBER','$ADDRESS')";
$result = sqlsrv_prepare($conn,$query)or die('Error querying MSSQL
database');
sqlsrv_execute($result);
}
?>
HTML
<form name="submit" action="employee.php" method="POST" >
<h2>Register New Member</h2>
<table border="0">
<tr>
<td>FIRSTNAME</td>
<td>
<input type="text" name="FIRSTNAME" id="FIRSTNAME"/>
</td>
</tr>
<tr>
<td>LASTNAME</td>
<td>
<input type="text" name="LASTNAME" id="LASTNAME"/>
</td>
</tr>
<tr>
<td>DATE_OF_BIRTH</td>
<td>
<input type="date" name="DATE_OF_BIRTH" id="DATE_OF_BIRTH"/>
</td>
</tr>
<tr>
<td>PHONENUMBER</td>
<td>
<input type="text" name="PHONENUMBER" id="PHONENUMBER"/>
</td>
</tr>
<tr>
<td>ADDRESS</td>
<td>
<input type="text" name="ADDRESS" id="ADDRESS"/>
</td>
</tr>
<tr>
<td></td>
<td align="right"><input type="submit" name="submit" value="REGISTER"/></td>
</tr>
</table>

Answer 1

I noticed that you are attempting to post data to a database, that is not actually named correctly on the PHP side, this would just result in the information related to your DATE OF BIRTH not being posted in the first place, as it has to match the PHP side exactly when you declare to retrieve posted data. I am pretty sure from memory that you only need to change the name in the form on the client side to DATEOFBIRTH, or you will have to change the PHP side to DATE_OF_BIRTH instead of this, whilst keeping the client side as DATE_OF_BIRTH
<input type="date" name="DATE_OF_BIRTH" id="DATE_OF_BIRTH"/>
NEEDS TO BE:
<input type="date" name="DATEOFBIRTH" id="DATEOFBIRTH"/>

Answer 2

I made a couple of minor alterations but cannot test to see - run it and see what the resultant sql statement looks like. I hope it helps.
<?php
if( !empty( $_POST ) && isset( $_POST["FIRSTNAME"], $_POST["LASTNAME"], $_POST["DATE_OF_BIRTH"], $_POST["PHONENUMBER"], $_POST["ADDRESS"] ) ) {
/*
There is no point creating a db connection if a later condition fails
so create the db conn after testing key variables are set
*/
$server = "EXAMPLE";
$options = array( "Database"=>"FORKLIFT" );
$conn = sqlsrv_connect( $server, $options );
/* In production environment do not display actual errors! */
if( !$conn ) die( print_r( sqlsrv_errors(), true ) );
/* as pointed out by #Daniel Rutter, the field is actually called `DATE_OF_BIRTH` */
$FIRSTNAME = $_POST["FIRSTNAME"];
$LASTNAME = $_POST["LASTNAME"];
$DATEOFBIRTH = $_POST["DATE_OF_BIRTH"];
$PHONENUMBER = $_POST["PHONENUMBER"];
$ADDRESS = $_POST["ADDRESS"];
$sql = "INSERT INTO EMPLOYEE
( FIRSTNAME,LASTNAME,DATEOFBIRTH,PHONENUMBER,ADDRESS )
VALUES
( '$FIRSTNAME','$LASTNAME','$DATEOFBIRTH','$PHONENUMBER','$ADDRESS' )";
/* Comment out the following line after running the script and observing the output here */
exit( $sql );
/* I assume that `sqlsrv_prepare` returns either true or false */
$stmt = sqlsrv_prepare( $conn, $sql ) or die( 'Error querying MSSQL database' );
if( $stmt ) sqlsrv_execute( $stmt );
/* debug post vars */
} else { echo 'Error::' . print_r($_POST,true ); }
?>
Having briefly had a look at the online PHP manual for the various sqlsrv_* commands perhaps you ought to try along these lines - it follows closely with one of the examples given - though it is untested due to not having a mssql server instance to play with.
<?php
if( !empty( $_POST ) && isset( $_POST["FIRSTNAME"], $_POST["LASTNAME"], $_POST["DATE_OF_BIRTH"], $_POST["PHONENUMBER"], $_POST["ADDRESS"] ) ) {
/*
There is no point creating a db connection if a later condition fails
so create the db conn after testing key variables are set
*/
$server = "EXAMPLE";
$options = array( "Database"=>"FORKLIFT" );
$conn = sqlsrv_connect( $server, $options );
/* In production environment do not display actual errors! */
if( !$conn ) die( print_r( sqlsrv_errors(), true ) );
$FIRSTNAME = $_POST["FIRSTNAME"];
$LASTNAME = $_POST["LASTNAME"];
$DATEOFBIRTH = $_POST["DATE_OF_BIRTH"];
$PHONENUMBER = $_POST["PHONENUMBER"];
$ADDRESS = $_POST["ADDRESS"];
$sql = "INSERT INTO EMPLOYEE
( FIRSTNAME, LASTNAME, DATEOFBIRTH, PHONENUMBER, ADDRESS )
VALUES
( ?,?,?,?,? )";
$params=array(
&$FIRSTNAME,
&$LASTNAME,
&$DATEOFBIRTH,
&$PHONENUMBER,
&$ADDRESS
);
/* add the posted variables as an array here as the third arg */
$stmt = sqlsrv_prepare( $conn, $sql, $params ) or die( 'Error querying MSSQL database' );
if( $stmt ) sqlsrv_execute( $stmt );
/* debug post vars */
} else {
echo 'Error::' . print_r( $_POST, true );
}
?>
From the PHP online manual regarding sqlsrc_prepare:
Prepares a query for execution. This function is ideal for preparing a
query that will be executed multiple times with different parameter
values.
and
When you prepare a statement that uses variables as parameters, the
variables are bound to the statement. This means that if you update
the values of the variables, the next time you execute the statement
it will run with updated parameter values. For statements that you
plan to execute only once, use sqlsrv_query().