I am trying to create a drop down menu that will display Project Names off all Projects present in the table Project in the database Testing... The drop down is created but it is not accessing the database and retrieving the required data... The code is as follows:
<html>
<head>
<title>Dynamic Drop Down List</title>
</head>
<form id="form1" name="form1" method="post" action="<?php echo $PHP_SELF; ?>">
Project List :
<select Project Name='NEW'>
<option value="">--- Select ---</option>
<?
$serverName = "Swagatha-PC";
$connectionInfo = array( "Database"=>"Testing");
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn )
{
echo "Connection established.<br />";
}
else
{
echo "Connection could not be established.<br />";
die( print_r( sqlsrv_errors(), true));
}
if (isset ($select)&&$select!="")
{
$select=$_POST ['NEW'];
}
?>
<?
$query=mysql_query("select ProjectName from dbo.Project");
$menu=" ";
while($row = sqlsrv_fetch_array($query))
{
$menu ="<option>" . $row["ProjectName"] . "</option>";
}
?>
</select>
<input type="submit" name="Next" value="Select" />
</form>
</body>
</html>
What Do i do to fix this??
The solution is
while($row = sqlsrv_fetch_array($query))
{
echo "<option>" . $row["ProjectName"] . "</option>";
}
See? You output data with echo instead assinging data to variable.
And as @FirstOne noticed using mysql_query with sqlsrv_connect is an error too. I hope it's just your typo:
$query=sqlsrv_query($conn, "select ProjectName from dbo.Project");
As another sidenote
if (isset ($select)&&$select!="")
{
$select=$_POST ['NEW'];
}
this if will always be false, because you first check $select and then define it. It definitely should be:
if (isset($_POST['NEW']) && $_POST['NEW'] != "")
{
$select = $_POST['NEW'];
}
Here is you fixed code. You can try it
<html>
<head>
<title>Dynamic Drop Down List</title>
</head>
<form id="form1" name="form1" method="post" action="<?php echo $PHP_SELF; ?>">
Project List :
<select Project Name='NEW'>
<option value="">--- Select ---</option>
<?php
$serverName = "Swagatha-PC";
$connectionInfo = array( "Database"=>"Testing");
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn )
{
echo "Connection established.<br />";
}
else
{
echo "Connection could not be established.<br />";
die( print_r( sqlsrv_errors(), true));
}
if (isset ($select)&&$select!="")
{
$select=$_POST ['NEW'];
}
$query=sqlsrv_query("select ProjectName from dbo.Project");
$menu=" ";
while($row = sqlsrv_fetch_array($query))
{
echo "<option>" . $row["ProjectName"] . "</option>";
}
?>
</select>
<input type="submit" name="Next" value="Select" />
</form>
</body>
</html>