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#1 |
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Newbie
Join Date: May 2004
Posts: 1
Rep Power: 0
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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 11 fields:
(city date day time agency location building_room street zip phone contact) are supposed to be returned in a pop up. I only have Agency City and Date being returned. Both files are below--can someone please take a look and show me in simple terms how to remedy this? Thank you in advance..Bufhal <?php
// Connection to the db server and select active db
$SQLlink = @mysql_connect("sql.com", "wnyaic", "immunize"); //creates a connection
if (!$SQLlink)
Die("Couldn't connect to the db server."); // display error message on error
if (!mysql_select_db("immunizewnyorg", $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."<BR>";
echo $agency_out."<BR>";
echo $city_out."<BR>";
?><?php
if (!IsSet($action) || !IsSet($value)) // check if both vars are set
Die("Both vars must be set");
if (Trim($value) == "") // check if value is non-blank
Die("Value can't be left blank");
if ($action != "date" && $action != "agency" && $action != "city")
Die("Unknown action requested");
// Connection to the db server and select active db
$SQLlink = @mysql_connect(".com", "wnyaic", "947"); //creates a connection
if (!$SQLlink)
Die("Couldn't connect to the db server."); // display error message on error
if (!mysql_select_db("immunizewnyorg", $SQLlink))
Die("Couldn't access database."); // display error message on error
// escape data from user
if (ini_get('magic_quotes_gpc')) { // unescaping data if needed
$value = StripSlashes($value);
}
$value = mysql_escape_string($value); // escaping data for MySQL db
$data = mysql_query("SELECT date, agency, city FROM agencies WHERE $action = '$value'"); // perform a query
if (!$data)
Die(mysql_error()); // display MySQL error message on error
$agencies = Array();
while($row = mysql_fetch_array($data)) {
$agencies[] = $row;
}
$output = "<table border=1>\n";
if (mysql_num_rows($data) == 0) { // in the case of no results found - display alert message
$output .= "<tr><td colspan=3>No results found</td></tr>";
} else {
forEach ($agencies as $agency) { // display row for each result (eg. you can have more agencies in one town)
$output .= "<tr>";
$output .= "<td>".$agency["date"]."</td>";
$output .= "<td>".$agency["agency"]."</td>";
$output .= "<td>".$agency["city"]."</td>";
$output .= "</tr>\n";
}
}
$output .= "</table>\n";
echo $output;
?> |
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#2 |
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PFO Founder
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well your not really telling the reply.php to print out anything more then date, agency, and city. if you want the rest of the stuff printed they need to be added to the output. that is all i can see as to why its not printing but i could be wrong
i will keep looking
__________________
BIG K aka Kyle Programming Forums Kyle K Online Please do not PM or email me programming questions. Post them in the forums instead. |
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#3 | |
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Programming Guru
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As fars as i can tell, big_k105 is right.
Quote:
echo "<table border=1>";
if (mysql_num_rows($data) <> 0)
{
forEach ($agencies as $agency)
{
echo "<tr>";
echo "<td>{$agency['date']}</td>";
echo "<td>{$agency['agency']}</td>";
echo "<td>{$agency['city']}</td>";
echo "<td>{$agency['day']}</td>";
echo "<td>{$agency['time']}</td>";
echo "<td>{$agency['location']}</td>";
echo "<td>{$agency['building_room']}</td>";
echo "<td>{$agency['street']}</td>";
echo "<td>{$agency['zip']}</td>";
echo "<td>{$agency['phone']}</td>";
echo "<td>{$agency['contact']}</td>";
echo "</tr>";
}
}
else
{
echo "<tr>";
echo "<td>No Results Found!</td>";
echo "</tr>";
}
echo "</table>";But that's just how i would do it. I find this way a little easyer to read, thus making it easyer to debug. Post any questions that you have and i am sure we can get this problem fixed. Happy Codding! -Pizentios
__________________
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