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PHP - Need Help With Dropdown Menu Displaying Results From Mysql Please
Hello everyone,
So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help. A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right. Some of this probably doesnt even make sense but I'll show you the kind of things I've tried... <select name="gender" size="1" id="gender"> <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option> <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option> </select> OR <select name="gender" id="gender"> <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option> <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option> <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option> </select> OR <select name="gender" size="1" id="gender"> <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option> <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option> <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option> </select> OR <select name="gender" id="gender"> <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option> <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option> </select> Honestly man, I've got no idea. The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen. And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol. Any help would be tremendous and greatly appreciated. Thanks very much, Learner P.S Man! Similar TutorialsI am querying a database and trying to display the names of categories in a drop-down menu. My problem is that the names are not visible. I have the following code: $result = mysql_query('SELECT name FROM category') or die(mysql_error()); echo('Choose a category'); echo('<select name="category">'); echo('<option value="general" >general</option>'); while($row = mysql_fetch_array($result)){ echo('<option style="color:black;" value = "'.$row['name'].'">'.$row['name'].'</option>'); } echo('</select><br /><br />'); The query is good and executing the query retrieves the expected number of results. My dropdown box is the proper length and width (showing that it is trying to print the names); however, nothing is displayed. Any ideas? Hi I have got results being displayed after clicking the search button in a form on my home page but it brings up all the results which is ok but how do I get onlt the results a user searches for for example a location or property type etc as its for a property website The coding is below for the results page Also sorry how do I add a background image to the php page, I tried using css but wouldn't work Code: [Select] <style type="text/css"> body {background-image:url('images/greybgone.png');} </style> <?php mysql_connect ("2up2downhomes.com.mysql", "2up2downhomes_c","mD8GsJKQ") or die (mysql_error()); mysql_select_db ("2up2downhomes_c"); echo $_POST['term']; $sql = mysql_query("select * from properties where typeProperty like '%$term%' or location like '%$term%'"); while ($row = mysql_fetch_array($sql)){ echo 'Type of Property: '.$row['typeProperty']; echo '<br/> Number of Bedrooms: '.$row['bedrooms']; echo '<br/> Number of Bathrooms: '.$row['bathrooms']; echo '<br/> Garden: '.$row['garden']; echo '<br/> Description: '.$row['description']; echo '<br/> Price: '.$row['price']; echo '<br/> Location: '.$row['location']; echo '<br/> Image: '.$row['image']; echo '<br/><br/>'; } ?> Hello, I have a quick question about methods for retrieving records from a mysql table and displaying them as a links For example, imagine I have three tables called countries, cities and city_info. I'd like to be able to select a country and have a list of that country's city names returned as links. I'd then like to be able to click on the link for London, say, and that would trigger a mysql query to retrieve the entry in city_info about London. Are there any functions that allow this? If anyone could point me in the right direction for further research I'd be grateful. Thanks. Hey guys, Having a slight problem with part of the code in my index.php file Code: [Select] mysql_select_db('db_name', $con); $result = mysql_query("SELECT * FROM spy ORDER BY id desc limit 25"); $resulto = mysql_query("SELECT * FROM spy ORDER BY id desc"); $count = mysql_num_rows($resulto); while($row = mysql_fetch_array($result)) { ?> <div class="contentDiv">Someone is looking at <?=$row[title];?> Stats for "<a href="/<?=$row[type];?>/<?=$row[code];?>/<?=$row[city];?>"><?=$row[code];?> <?=$row[city];?></a>"</div> <?}?> </div> <div id="login"></div> <? include("footer.php"); ?> </div> </body> </html> I'm getting the following error when viewing the file Code: [Select] Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 70 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 71 where lines 70 and 71 are $count = mysql_num_rows($resulto); while($row = mysql_fetch_array($result)) Any ideas on how to fix this? Hi, This has been baffling me for a couple hours now and i cant seem to figure it out. I have some code which creates an array and gets info from a mysql database and then displays in a list. This works great but after adding more and more rows to my database the list is now becoming quite large and doesnt look great on my site. Is it possible to split the list into multiple columns of about 25 and if possible once 3 or 4 columns have been created start another column underneath. To help explain i would be looking at a layout as follows: Code: [Select] line 1 line 1 line 1 line 2 line 2 line 2 ... ... ... line 25 line 25 line 25 line 1 line 1 line 1 line 2 line 2 line 2 ... ... ... line 25 line 25 line 25Im guessing there should be some sort of if statement to check how many items are being displayed and to create a new column if necessary. Is this correct? Thanks, Alex Hi all, I'm trying to display a form based on a dropd own selection. The drop down is propagated from a database query. I have the drop down list displaying, but when I click submit, I can't get the result to display. Code: [Select] $sql="SELECT id, company_name FROM webprojects ORDER BY company_name ASC"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["id"]; $company_name=$row["company_name"]; $options.="<OPTION VALUE=\"$id\">".$company_name; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>PHP Dynamic Drop Down Menu</title> </head> <body> <form id="form1" name="form1" method="post" action="selected.php"> <SELECT NAME=id> <OPTION VALUE=0>Choose <?=$options.="<OPTION VALUE=\"$id\">".$company_name.'</option>';?> </SELECT> <label> <input type="submit" name="submit" id="submit" value="Submit" /> </label> </form> </body> </html> Any help is greatly appreciated i am storing my menu in the database, i want to be able to output it by priority, heres so far what i have. I have no idea were to start. Database dump: -- -- Table structure for table `menu` -- CREATE TABLE IF NOT EXISTS `menu` ( `menu_access_lvl` int(2) NOT NULL, `priority` int(11) NOT NULL, `name` varchar(200) NOT NULL, `comment` text NOT NULL, `location` text NOT NULL, `creator_id` varchar(255) NOT NULL ) ENGINE=InnoDB DEFAULT CHARSET=latin1; -- -- Dumping data for table `menu` -- INSERT INTO `menu` (`menu_access_lvl`, `priority`, `name`, `comment`, `location`, `creator_id`) VALUES (0, 1, 'Home Page', 'Home page', 'index.php', 'admin'), (0, 3, 'Contact', 'Contact', 'index.php?PG=contact', 'admin'), (0, 2, 'Events & Meetings', 'Events & Meetings', 'index.php?PG=events', 'admin'), (0, 4, 'About', 'About', 'index.php?PG=about', 'admin'), (2, 5, 'Admin', 'Admin', 'index.php?PG=admin', 'admin'); And here is the php code displaying it //gets the role of the user if set, otherwise role = 0 if(isset($_SESSION['SESS_MEMBER_ID']))$lvl = $_SESSION['SESS_ROLE']; else $lvl = 0; // this loads the menu buttons that correspond to the users role $menuqry="SELECT * FROM menu WHERE menu_access_lvl<='$lvl'"; $menuresult=mysql_query($menuqry); while($row = mysql_fetch_array($menuresult)){ echo "<li class=\"menuitem\"><a href=\"".$row['location']."\">".$row['name']."</a></li>"; } What this currently displays: Home Contact Events & meetings About I want it to be according to priority in the menu like: Home Events & meetings Contact About need a little help guys! I use the script below to display profile images, trouble is it shows 1 on top of the other, and i need it to double up 2 profile images on top of 2 profile images ect any ideas how i can do this. require("./include/mysqldb.php"); $con = mysql_connect("$dbhost","$dbuser","$dbpass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("$dbame", $con); $result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40"); print "<table width=\"293\" height=\"111\" border=\"0\"> <tr>\n"; while($row = mysql_fetch_array($result)) { print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /></td>\n"; print " </tr>\n"; print " <tr> \n"; print "<td>" . $row['star'] . "</td>\n"; print " </tr>\n"; print " <tr>\n"; print "<td>" . $row['username_search'] . "</td>\n"; print " </tr>\n"; print " <tr> \n"; print "<td>" . $row['phone_search'] . "</td>\n"; print " </tr> \n"; } print "</table>"; mysql_close($con); ?> Hello, Im trying to display some results from mysql database, however none display. Can anyone tell me where im going wrong please? Code: [Select] </head><body> <div id="listhold"> <div class="list"> <a href="Restaurants.html">Restaurants</a><br /> <?php mysql_connect("","",""); mysql_select_db("") or die("Unable to select database"); $result = mysql_query("SELECT name FROM business WHERE type ='restaurant' ORDER BY name"); $number_of_results = mysql_num_rows($result); $results_counter = 0; if ($number_of_results != 0) {while ($array = mysql_fetch_array($result)) $results_counter++; if ($results_counter >= $number_of_results);} ?> </div> I have a PHP while loop that pulls from an SQL database and displays the contents in a table with two columns.
// Check Connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Select Data Table
$result = mysqli_query($con,"SELECT * FROM Recommendations") or die(mysqli_error);
// Split Data
$mid = ceil(mysqli_num_rows($result)/2);
// Display Content
while ($rows = mysqli_fetch_array($result)) {
$Name = $rows['Name'];
$Author = $rows['Author'];
$Cover = $rows['Link to Cover'];
$Link = $rows['Link to Profile'];
echo "<table><tr><td>
<a href='" . $Link . "' >$Name</a>
<br />
$Author
<br />
<a href='" . $Link . "' ><img src='" . $Cover . "' /></a>
</td>
<td>
<a href='" . $Link . "' >$Name</a>
<br />
$Author
<br />
<a href='" . $Link . "' ><img src='" . $Cover . "' /></a>
</td></tr></table>";
}
?>
I want to be able to display the looped results side by side in columns of two.
Example:
1 2
3 45 6 I've tried using pseudo classes to display only the even and odd results in the different table columns, but honestly have no idea how to do this. I'm new to PHP, so my apologies if the results are really obvious. Thanks in advance! Dear all, I am new in this forum. This is my code $query = " SELECT webdb.id, webdb.writer, writer.picLoc, webdb.title FROM webdb, writer WHERE webdb.writer=writer.name and category = 'Researchworks' and language = 'Farsi' ORDER BY writer DESC"; $resultaat = mysql_query($query, $LinkID); $column_count = mysql_num_fields($resultaat) or die (mysql_error()."<br>Couldn't execute query: $SQLquery"); $counter=1; echo "<table border=\"0\" width=\"700\" border color=white><tr>"; while ($row = mysql_fetch_row($resultaat)) { if ($author !== $row[1]) { $author = $row[1]; echo "<td align=right width=220 valign=top style=\"margin: 5px; float: right border-bottom-color:#000; border-left-color:#000;\">"; echo "<img width=\"50\" height=\"80\" src=\"admin/writers/$row[2]\" border =\"0\"><br>".$row[1]."<br>"; echo "<a href=\"poems.php?writer=$row[1]\">".$row[3]."</a><br>"; echo "</td>"; if($counter%3==0) { echo"</tr><tr>"; } $counter++; } } echo"</tr></table>"; i have authors with different articles on a certain topic. What i want is, displaying the name of the author only once and all his titles under his name. I also want a dynamic table where i display three authors in each row and soon as there a fourth author a new row must start. My problem now is is the title is also being filtered and i can only display one title. Thanks in advance I'm realitivly new to PHP and was hoping somebody could help. I have a mysql database that stores information about books. I am currently using the code below to query the database and extract the 3 most recent entries and showing them in a dynamic list: Code: [Select] $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 3"); $productCount = mysql_num_rows($sql); if ($productCount > 0) { // ensure a book exists while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $title = $row["title"]; $author = $row["author"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $dynamiclist .= //My table showing the products } } else { $dynamicList = "There are currently no Books listed in this store"; } This works well when showing the most recent books 1 below the other. However, I would like to show these products side by side, horizontally across the page. Can somebody please point me in the right direction? Many thanks I'm trying to create a list that groups information by username. Only part of it is working. The first query ($get_item_sql) is grouping the information perfectly but the second query ($get_sold) is lumping the $item_price and $item_amount_due as one total for each one and outputting the same amounts into every username. I'm stuck on this and would appreciate your help. For example: Username item fees image fees item sales item price total due Jim 2 $0.40 $100.00 $3.00 $3.40 Kelly 5 $1.00 $100.00 $3.00 $4.00 This example shows the columns in red as being the problem where Kelly didn't sell anything so her "item sales" and "item price" should be $0.00 but is carrying Jim's totals into hers. Hope this helps! Thank you! $get_item_sql = mysql_query("SELECT id, username, date, ROUND(price,2) AS price, SUM(item_fee) AS fee, item_fee, SUM(sold) AS sales, SUM(ROUND(price,2)) AS total FROM product WHERE MONTH(date) = MONTH(DATE_ADD(CURDATE(),INTERVAL -1 MONTH)) GROUP BY username" ) or die(mysql_error()); if (mysql_num_rows($get_item_sql) < 1) { //invalid item $display_block .= "<p><em>Invalid item selection.</em></p>"; } else { //valid item, get info while ($item_info = mysql_fetch_array($get_item_sql)) { $item_username = $item_info['username']; $item_date = $item_info['date']; $item_price = $item_info['price']; $item_fee = $item_info['fee']; $image_fees = $item_fee * .20; $item_sold = $item_info['sales']; $get_sold = mysql_query("SELECT SUM(ROUND(price,2)) AS total, SUM(ROUND(sold,2)) AS sales, date, username FROM product WHERE sold = '1' AND MONTH(date) = MONTH(DATE_ADD(CURDATE(),INTERVAL -1 MONTH)) GROUP BY username") or die(mysql_error()); if (mysql_num_rows($get_sold) < 1) { //invalid item $display_block .= "<p><em>Invalid item selection.</em></p>"; } else { //valid item, get info while ($item_sold2 = mysql_fetch_array($get_sold)) { $item_sales = $item_sold2['total']; $item_price = ($item_sold2['total']) * .03; $item_amount_due = $image_fees + $item_price; $content .= "<form action=\"add_artist.php\" method=\"post\"><table class=\"anotherfont\" width=\"670\" border=\"0\"> <tr><td width=\"201\">{$item_username}</td> <td width=\"109\">{$item_fee}</td> <td width=\"109\">{$image_fees}</td> <td width=\"109\"> {$item_sales}</td> <td width=\"109\"> {$item_price}</td> <td width=\"109\"><input name=\"balance_due\" type=\"text\" value=\"{$item_amount_due}\" /></td> </tr><br /></table></form>"; } } } } Hi everyone. I'm stuck on the following query. I need to display all the fields listed below on a page, but linked via communications.CommID. I'd appreciate any assistance you can provide. thank you. Code: [Select] <?php $result = mysql_query("SELECT records.NameFirst_1, records.NameLast_1, records.CompanyName, records.CompanyBranch, records.CompanyReferenceNumber, records.CaseOwnerSelect, communications.ConversionType, communications.Contact, communications.ContactFrom, communications.CommID, communications.ContactPosition, communications.ContactTelephone, communications.ContactEmail, communications.ContactFax, communications.CallDate, communications.CallTime, communications.ActionTextField FROM records INNER JOIN communications ON records.IDNumber = '$IDNumber'") or die(mysql_error()); $row = mysql_fetch_array($result); ?> I'm trying to pull results from a database based on where the user is located based upon the variables $usr_lat & $usr_lng, and search for by a radius of x amount of miles/km (need to make it optional). I can't seem to find exactly what I'm looking for on google so I thought I'd asked here. Any help would be appreciated.
Code: [Select] <?php $result = mysql_query("SELECT * FROM FamilyTbl INNER JOIN PeopleTbl ON (FamilyTbl.Name_ID = PeopleTbl.NameID) WHERE FamilyTbl.House_ID = '$address' ORDER BY NameLast, NameFirst ") OR die(mysql_error()); WHILE ($row = mysql_fetch_array($result) ) { echo $row[NameLast]. ", ". $row[NamePrefix]. " ". $row[NameFirst]. " ". $row[NameMiddle]. $row[NameSuffix]. " "; } ?> OK, some of these queries return A LOT of names. I'd like to be able to display them in columns in a table like so: Code: [Select] Charne, Mr. Michael Glanger, Mrs. Karin Kling, Mr. Wayne Charne, Mrs. Suzette Glanger, Mr. Trevor Lazarow, Mrs. Fiona Charney, Mrs. Linda Jochelson, Mrs. Barbara Lazarow, Mr. Mark Charney, Mr. Norman Jochelson, Mr. Neil Norton, Mr. Charles Cohen, Mr. Brendan Karlan, Mr. Dennis Norton, Mrs. Jodi Cohen, Mrs. Joanna Karlan, Mrs. Helen Roy, Mr. Michael Flekser, Mrs. Jean Kling, Mrs. Danielle Roy, Mrs. Nicki Frysh, Dr. Howard Kling, Mrs. Melanie Tsafrir, Mrs. Lauren Frysh, Mrs. Sandra Kling, Mr. Nevil Tsafrir, Mr. Thomer That way it reads top to bottom THEN left to right. math-wise, it's simple to set up: Code: [Select] $num_results = number_of_$results; $numcols = 3; $numrows = trunc($numresults/numcols); <table> for row_loop=1 to numrows <tr> for col_loop = 1 to numcols <td> echo result(col_loop-1)*(numrows)+row_loop </td> next col_loop </tr> next row_loop </table> Anyone want to give this a shot? Revraz already directed me to http://www.phpfreaks.com/forums/index.php/topic,95426.0.html but that displayed the data from left to right then up to down like so: Code: [Select] Charne, Mr. Michael Charne, Mrs. Suzette Charney, Mrs. Linda Charney, Mr. Norman Cohen, Mr. Brendan Cohen, Mrs. Joanna Flekser, Mrs. Jean Frysh, Dr. Howard Frysh, Mrs. Sandra Glanger, Mrs. Karin Glanger, Mr. Trevor Jochelson, Mrs. Barbara Jochelson, Mr. Neil Karlan, Mr. Dennis Karlan, Mrs. Helen Kling, Mrs. Danielle Kling, Mrs. Melanie Kling, Mr. Nevil Kling, Mr. Wayne Lazarow, Mrs. Fiona Lazarow, Mr. Mark Norton, Mr. Charles Norton, Mrs. Jodi Roy, Mr. Michael Roy, Mrs. Nicki Tsafrir, Mrs. Lauren Tsafrir, Mr. Thomer It's a good temp solution, but if anyone is good with for loops, I'd appreciate the help, thanks! -Dave Hi, fairly new to PHP over the last couple weeks. Been having a problem with certain queries. I have a database with football results, games, teams etc. I can filter these using drop down and that's all well and good. The problem I'm having is displaying the data via gameweek. I've been asked to display the table like so - Gameweek1 will display week1 teams, results etc. Gameweek2 will display week2... and so on.
I can manage to do this in a drop down. But I've been asked to display this using links like "Previous, 1, 2, 3 Next". I've tried pagination but I couldn't figure it out. Can anyone point me in the right direction? If I need a GET() method, how would I go about coding that so it will be used in a link(s)? Been searching and searching to find an answer but to no avail...
//Database connection etc...
$gameweek = "SELECT * FROM games WHERE gameweek= 1";
//if(isset($_GET['gameweek']))
//{
// $gameweek = $_GET['gameweek'];
//
//}
//....
$result=mysqli_query($connection, "select * from games WHERE gameweek= 1");
//Print table and table headings...
mysqli_close($connection);
?>
<a href="http://weeks.php?gameweek=2">Week 2</a>
<a href="http://weeks.phpgameweek=3">Week 3</a>
</body>
</html>
Hello Everyone was wondering if I could get some help with the following code? I am querying a database for results of listings that are in a database these listings are displayed on the page in a form. I am wanting each listing to be on a different page. Below is my code. Code: [Select] $lim=1; if (!isset($s) || $s < 1 || !is_numeric($s)) { $s = 1; } $start = ($s - 1) * $lim; $sql = "select id,bussimg,imagewidth,imageheight,email,usridm,company,businesscategory,address1,address2,state,city,zip,website,email,repname,description,phonenumber,country,status from $approvecheckbusinesses where usridm='$user_id'"; $result=db_query($sql); $countpages = $sql; $sql = $sql . " order by id asc limit $start, $lim"; $result=db_query($sql); $pages = ceil(mysql_num_rows(mysql_query($countpages)) / $lim); $result=db_query($sql); for ($i = 0; $i < mysql_num_rows($result); $i++) { $Listid= mysql_result($result, $i, "id"); $usridm= mysql_result($result, $i, "usridm"); $CompanyName= mysql_result($result, $i, "company"); $realname= mysql_result($result, $i, "repname"); $email= mysql_result($result, $i, "email"); $BusinessCategory= mysql_result($result, $i, "businesscategory"); $status= mysql_result($result, $i, "status"); echo ("FORM IS TO BE DISPLAYED HERE"); } if ($pages > 1) { echo("<p align=left style='font-size: 85% color=white'>"); for ($i = 1; $i <= $pages; $i++) { echo("["); if ($i == $s) {echo("<b>");} else {echo("<a id=home_offerLink href='index.html?EditMemberListing&user_id=$user_id&s=$i'>");} echo("Page $i"); if ($i == $s) {echo("</b>");} else {echo("</a>");} echo("] "); } echo("</p>") Page Numbers here using the above code.. The problem I seem to be running into is that it only displats the first record. The page numbers show up page 1 page 3 page 2 and three are blank there is no mysql error or anything for some reason I only get that first result out of three Hello all, I'm relatively new to all of this, but making progress.... I figure this has come up before, but couldn't fin anything by search or browsing. I have an sql table that is cron updated every two minutes from externally generated data in table X columns a,b,c,... . As part of more complicated site, I have one body element page (selected by a tab in the header) that does various sql queries on the data in table X and displays the data. I need this to rerun the queries and update the displayed results every 2 minutes or so as well. However, the standard solutions do not seem to be working: <META HTTP-EQUIV="refresh" CONTENT="15"> A refresh button refreshes the entire site, and not just this element. <FORM> <INPUT TYPE="button" onClick="history.go(0)" VALUE="Force Flight Data Refresh"> </FORM> I hope it is clear. Any ideas of how I could get this to work? Thanks, Kalle Here is a snippet of the code on that page for what it's worth... </head> <body> <div id="header"> <center><h2><font color="red">TRAFFIC INFORMATION</font></h2> </div> <div id="navigation"> </div> <div id="content"> <!-- refresh button --> <FORM> <INPUT TYPE="button" onClick="history.go(0)" VALUE="Force Flight Data Refresh"> </FORM> </center> <?php $IDS->db_build($db); $IDS->db_query($db,$res,"SELECT * FROM `Pilots` WHERE `dest`='KORD'"); echo "<h5>Arrivals to O'Ha </h5>"; while ($arr_row = mysql_fetch_assoc($res)) { echo "<font color=yellow>".$arr_row['callsign']."</font> using route: ". $arr_row['route']."<br>"; } //end queries //close connect to sql database mysql_close($con) </div> <div id="footer"> </div> </body> </html> |
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