Working on a web page with three dropdowns: Agency, Date and City. When a user clicks on one of the choices from the dropdown, a record(s) with all 12 fields are supposed to be returned in a pop up. I only have Agency City and Date being returned. The twelve fields I want returned are listed in the results.php file for output below. Both files are below--can someone please take a look and show me how to remedy this?
Thank you in advance..Bufhal
index.php
Code:
<?php
// Connection to the db server and select active db
$SQLlink = @mysql_connect("t.com", "wny", "947"); //creates a connection
if (!$SQLlink)
Die("Couldn't connect to the db server."); // display error message on error
if (!mysql_select_db("org", $SQLlink))
Die("Couldn't access database."); // display error message on error
// perform query
$data = mysql_query("SELECT date, agency, city FROM agencies");
if (!$data)
Die(mysql_error()); // display MySQL error message on error
$agencies = Array();
while($row = mysql_fetch_array($data)) { // assign results into arrays
$dates[] = $row["date"];
$agencies[] = $row["agency"];
$cities[] = $row["city"];
}
$dates = Array_Unique($dates); // remove duplicate values
$agencies = Array_Unique($agencies);
$cities = Array_Unique($cities);
Sort($dates); // sort arrays
Sort($agencies);
Sort($cities);
$date_out = "<select name='date' onchange=\"window.open('results.php?action=date&value='+this.value, 'agencyWin', 'location=yes,left=20,top=20');\">";
$agency_out = "<select name='agency' onchange=\"window.open('results.php?action=agency&value='+this.value, 'agencyWin', 'location=yes,left=20,top=20');\">";
$city_out = "<select name='city' onchange=\"window.open('results.php?action=city&value='+this.value, 'agencyWin', 'location=yes,left=20,top=20');\">";
$date_out .= "<option>-- select date ---</option>";
$agency_out .= "<option>-- select agency ---</option>";
$city_out .= "<option>-- select city ---</option>";
forEach ($dates as $value)
$date_out .= "<option value='$value'>$value</option>";
forEach ($agencies as $value)
$agency_out .= "<option value='$value'>$value</option>";
forEach ($cities as $value)
$city_out .= "<option value='$value'>$value</option>";
$date_out .= "</select>\n";
$agency_out .= "</select>\n";
$city_out .= "</select>\n";
echo $date_out."";
echo $agency_out."";
echo $city_out."
";
?>
and results.php
Code:
<?PHP
$data = mysql_query("select * FROM agencies");
if (!$data)
Die(mysql_error()); // display MySQL error message on error
$output = "<table border=1>\n"; // I can format table here
if (mysql_num_rows($data) != 0) { // in the case of results
while($agency = mysql_fetch_array($data)) {
$output .= "<tr>";
$output .= "<td>".$agency["date"]."</td>";
$output .= "<td>".$agency["agency"]."</td>";
$output .= "<td>".$agency["city"]."</td>";
$output .= "<td>".$agency["day"]."</td>";
$output .= "<td>".$agency["time"]."</td>";
$output .= "<td>".$agency["location"]."</td>";
$output .= "<td>".$agency["building_room"]."</td>";
$output .= "<td>".$agency["street"]."</td>";
$output .= "<td>".$agency["zip"]."</td>";
$output .= "<td>".$agency["phone"]."</td>";
$output .= "<td>".$agency["contact"]."</td></tr>\n";
echo($output);
$output = ""; // clear buffer
}
}else{ // in case of NO results
echo("<tr><td colspan=11>No results</td></tr>");
}
echo "</table>"; // end of table
?>
//
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