Unknown column 'agency' in 'field list' "

[bufhal]

Hello;
I am a newbie trying to create three dropdowns on a webpage for Date, Agency and City. However, when I open the webpage on the server, I get a blank page with "Unknown column 'agency' in 'field list'

I included the following code with the MySQL commands with different log on info (I don't think that has anything to do with the problem but I included it just in case).

<?PHP
$hostName = "localhost.com";
$userName = "bufhal";
$password = "pass";
$dbName = "inyorg";

mysql_connect($hostName, $userName, $password) or die("Unable to connect to host $hostName");

mysql_select_db($dbName) or die("Unable to select database $dbName");

$date = '<select name="date">';
$agency = '<select name="agency">';
$city = '<select name="city">';

$result = mysql_query("SELECT date,agency,city FROM agencies") or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$date .= '<option value="'.$row["date"].'>'.$row["date"].'</option>';
$agency .= '<option value="'.$row["agency"].'>'.$row["agency"].'</option>';
$city .= '<option value="'.$row["city"].'>'.$row["city"].'</option>';
}
$date .= '</select>';
$agency .= '</select>';
$city .= '</select>';
?>

This is in the body section of Web page for dropdowns:
date: <?=$date?><br/>
city: <?=$city?><br/>
agency: <?=$agency?><br/>

Can someone tell me why I cannot get my page to open up with the dropdowns?
I also want to "print" to a pop up form the contents when a user clicks on an entry in dropdown.
I greatly appreciate any help..
bufhal

[HalfaBee]

Look at the HTML source your script creates and you will see the problem.

[bufhal]

can you please elaborate a bit?

[bufhal]

thanks, got it cooking..

[bufhal]

Hello HalfaBee;
I am trying to get the results to print but keep getting parse error: Parse error: parse error, unexpected T_STRING in E:\www\immunizewny_org\index4.php on line 126
Can you please take a look when you find time.

<?PHP

//error_reporting(E_ALL^E_NOTICE);
//insert mysql_connect(); and mysql_select_db(); calls here
// make connection to database

$hostName = "db.com";
$userName = "wny";
$password = "immunize";
$dbName = "wnyorg";

// make connection to database
mysql_connect($hostName, $userName, $password) or die("Unable to connect to host $hostName");

mysql_select_db($dbName) or die("Unable to select database $dbName");

$i = 0;

$data = mysql_query("SELECT date, agency, city FROM agencies");
while($rs = mysql_fetch_array($data))
{
$date[$i] = $rs['date'];
$agency[$i] = $rs['agency'];

$city[$i] = $rs['city'];

$i++;
}

$date_out = '<select name="date" onchange="window.open('results.php?date='+this.value, 'agencyWin', 'location=yes,left=20,top=20');">
';
foreach($date as $key => $val)
{
$date_out .= '<option>'.$val.'</option>';
}
$date_out .= '</select>';

$agency_out .= '<select name="agency" onchange="window.open('results.php?agency='+this.value, 'agencyWin', 'location=yes,left=20,top=20');">';
foreach($agency as $key => $val)
{
$agency_out .= '<option>'.$val.'</option>';
}
$agency_out .='</select>';

$city_out = '<select name="city" onchange="window.open('results.php?city='+this.value, 'agencyWin', 'location=yes,left=20,top=20');">
';
foreach($city as $key => $val)
{
$city_out .= '<option>'.$val.'</option>';
}
$city_out .='</select>';

echo $date_out;
echo $agency_out;
echo $city_out;
?>

Thank you very much...🙂

[HalfaBee]

$city_out = '<select name="city" onchange="window.open('results.php?city='+this.value, 'agencyWin', 'location=yes,left=20,top=20');">