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PHP - Data-driven Dropdown
I currently have an HTML form where the options for a certain drop-down menu are hard-coded. Instead, I want to use PHP to...
1. Look up the values in a column (cities) in a MySQL table (locations) 2. Make those values the only options in the dropdown menu. Any ideas how I would do this? This is what I have so far. <label for="city">What is your destination city?</label> <select class="form-control" id="city" name="city"> <?php //connect to the database $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); //grab the city names from the MySQL table $query = "SELECT cities FROM locations"; $res = mysqli_query($dbc, $query); while ($data = mysqli_fetch_assoc($res)) { echo '<option value="'.$data['cities'].'">'.$data['cities'].'</option>'; } //close the db connection mysqli_close($dbc); ?> </select> Similar TutorialsHi, I'm trying to get data from one field in a table (database). But I get undesirable result: Here is my code -> <?php $result2 = mysql_query("SELECT DISTINCT theme FROM mytable ") or die(mysql_error()); while($row2 = mysql_fetch_array( $result2 )) { ?> <form method="post" action='<?php echo $_SERVER["PHP_SELF"]; ?>'> <select name='themes'"> <?php $arr= array($row2['theme']); foreach($row2 as $value) { echo "<option value='$value'><b>". $value."</b> </option><br> "; } } ?> The attached image file show the result that I don't wont. (It's not a dropdown). Is there anyone who may help me, I spent a lot of time to find out but I can't. Thanks a lot for your help I'm not sure what to search for or if this can be done... Let's say I have a mysql table named "genre". Now I have "Male" and "Female" in a dropdown menu like this in a form: <select name="genre"> <option>Male</option> <option>Female</option> </select> How would you display, for example, the option Female in a update.php file if that's the genre stored in the mysql database when you fetch the results? Hello there! I've been banging my head on this for a while and I just can't seem to get it to work properly. I have a dropdown menu which selects information from table1 using a select statement (this table is called 'lid'). It selects the firstname, lastname and member id from this table and shows it in the dropdown menu. I'm glad I got that part working but the hard thing is inserting the data that the user selects into another table. So when you select the id member from this dropdown menu it only inserts a blank row into table2 (which is called 'teamlid'). Can you guys help me? How can I insert the id member into my table2? What am I doing wrong here? Thanks a million! This is my first post so if I'm doing anything wrong, let me know and I'll fix it asap! My code:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Boast & Drive</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.css">
<style type="text/css">
.wrapper{
width: 650px;
margin: 0 auto;
}
.page-header h2{
margin-top: 0;
}
table tr td:last-child a{
margin-right: 15px;
}
</style>
</head>
<body>
<div class="container-fluid">
<div class="row">
<div class="col-md-12">
<div class="page-header clearfix">
<h2 class="pull-left">Teamleden</h2>
<div class="btn-toolbar">
<a href="read.php" class="btn btn-primary btn-lg pull-right">Terug</a>
</div>
</div>
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
error_reporting(E_ALL);
//Verbinding maken met de database
require_once "login.php";
$sql = "SELECT tl.teamnaam,
tl.tl_ID,
tl.lidnummer,
l.voornaam,
l.achternaam
FROM teamlid tl
JOIN lid l ON tl.lidnummer = l.lidnummer
ORDER BY tl.teamnaam;";
if($result = mysqli_query($conn, $sql)) {
if(mysqli_num_rows($result) > 0) {
echo "<table class='table table-bordered table-striped'>";
echo "<thead>";
echo "<tr>";
echo "<th>Teamnaam</th>";
echo "<th>Tl_ID</th>";
echo "<th>Lidnummer</th>";
echo "<th>Voornaam</th>";
echo "<th>Achternaam</th>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['teamnaam'] . "</td>";
echo "<td>" . $row['tl_ID'] . "</td>";
echo "<td>" . $row['lidnummer'] . "</td>";
echo "<td>" . $row['voornaam'] . "</td>";
echo "<td>" . $row['achternaam'] . "</td>";
echo "<td>";
echo "<a href='update.php?id=". $row['lidnummer'] ."' title='Gegevens wijzigen' data- toggle='tooltip'><span class='glyphicon glyphicon-pencil'></span></a>";
echo "<a href='delete.php?id=". $row['lidnummer'] ."' title='Lid verwijderen' data- toggle='tooltip'><span class='glyphicon glyphicon-trash'></span></a>";
echo "</td>";
echo "</tr>";
}
echo "</tbody>";
echo "</table>";
mysqli_free_result($result);
} else{
echo "<p class='lead'><em>Er zijn geen gegevens om weer te geven.</em></p>";
}
} else{
echo "De volgende fout is gevonden: " . mysqli_error($conn);
}
?>
<form name="dropdown" method="post">
<div class="page-header clearfix">
<h2 class="pull-left">Teamlid toevoegen</h2>
</div>
<p>Selecteer hieronder met behulp van het dropdown menu een lid welke je aan bovenstaand team wilt toevoegen</p>
<div class="container-fluid">
<div class="row">
<?php
// Variabelen aanmaken en tonen met lege waardes
$teamnaam = $lidnummer = '';
// Code voor dropdown. Selecteert voornaam, achternaam en lidnummer van tabel lid)
$sql = "SELECT voornaam, achternaam, lidnummer FROM lid ORDER BY achternaam";
$result = mysqli_query($conn, $sql);
echo "<select id='teamLid' name='teamLid'>";
echo "<option>--Selecteer Lid--</option>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['lid'] . "'>" . $row['voornaam'] . " " . $row['achternaam'] . " " . $row['lidnummer'] . "</option>";
}
echo "</select>";
if
(isset($_POST["id"]) && !empty($_POST["id"])) {
$id = $_POST["teamLid"];
$stmt = $conn->prepare("INSERT INTO teamlid (teamnaam, lidnummer) VALUES (?,?)");
$stmt->bind_param('si', $param_teamnaam, $param_lidnummer);
$param_teamnaam = $teamnaam;
$param_lidnummer = $lidnummer;
$stmt->execute();
}
// Verbinding sluiten
mysqli_close($conn);
?>
<div>
<input type="hidden" name="id" value="<?php echo $id; ?>" />
<input type="submit" name="submit" class="btn btn-primary" value="Toevoegen">
</div>
</div>
</div>
</form>
</div>
</div>
</body>
</html>
Hi? im just a beginner in php i just want to ask how to insert a data into a table from a dropdown list. I have concatenate the itemid and description to form the dropdown list. But when i viewed my item_table the itemid and description columns are null. can you help me with this.. this is my php code for the dropdown list... <?php $query = "SELECT CONCAT(itemid,' ', '-',' ', description) AS Item FROM item_table"; $result = mysql_query($query) or die(mysql_error()); $dropdown = "<SELECT CONCAT(itemid,' ' '-',' ', description) AS Item FROM item_table>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['Item']}'>{$row['Item']}</option>"; } $dropdown .= "\r\n</select>"; echo $dropdown; ?> this my code for inserting data into the item_table... <?php if(isset($_POST ['submit'])) { $itemid = $_POST['itemid']; $description = $_POST['description']; $datein = $_POST['datein']; $qtyin = $_POST['qtyin']; $unitprice = $_POST['unitprice']; $unit = $_POST['unit']; $category = $_POST['category']; $empid = $_POST['empid']; $message =''; if(($itemid && $description == "")||($itemid && $description == null)) { header("location:IncomingEntry.php?msg=Incorrect"); exit(); } else { $link = mysql_connect('localhost', 'root', '') or die(mysql_error()); $db_selected = mysql_select_db('inventory', $link); $message=''; $query = "INSERT INTO incoming_table (itemid , description, datein, qtyin, unitprice, unit, category, empid) VALUES ('".$itemid."', '".$description."', '".$datein."', '".$qtyin."', '".$unitprice."', '".$unit."', '".$category."', '".$empid."')"; if (!mysql_query($query,$link)) { die('Error: ' . mysql_error()); } header("location: IncomingEntry.php?msg=1 record added"); } } ?>
Good morning. Thanks in advance to everyone.
<!doctype html>
<html dir=ltr style="overflow-x: hidden;padding: 0px;width: 100%;">
<head>
<meta charset=utf-8>
<meta http-equiv=X-UA-Compatible content="IE=edge">
<meta name=viewport content="width=device-width, initial-scale=1">
<title></title>
<meta name="description" content="">
<meta name="author" content="SoftMat">
<link media="all" href="css/style.css" rel="stylesheet" />
<!-- qrcode-reader core CSS file -->
<link rel="stylesheet" href="css/qrcode-reader.min.css">
<!-- jQuery -->
<script src="js/jquery.min.js"></script>
<!-- qrcode-reader core JS file -->
<script src="js/qrcode-reader.min.js"></script>
<script>
$(function(){
// overriding path of JS script and audio
$.qrCodeReader.jsQRpath = "js/jsQR.min.js";
$.qrCodeReader.beepPath = "audio/sound.mp3";
// bind all elements of a given class
$(".qrcode-reader").qrCodeReader();
// bind elements by ID with specific options
$("#openreader-multi2").qrCodeReader({multiple: true, target: "#multiple2", skipDuplicates: false});
$("#openreader-multi3").qrCodeReader({multiple: true, target: "#multiple3"});
// read or follow qrcode depending on the content of the target input
$("#openreader-single2").qrCodeReader({callback: function(code) {
if (code) {
window.location.href = code;
}
}}).off("click.qrCodeReader").on("click", function(){
var qrcode = $("#single2").val().trim();
if (qrcode) {
window.location.href = qrcode;
} else {
$.qrCodeReader.instance.open.call(this);
}
});
});
</script>
</head>
<body>
<div align="center" class="container">
<div align="center" class="col-xs-12">
<div align="center" class="container" style="background-color:white; box-shadow:0px 2px #00000085;">
<img style="margin-top:15px; margin-bottom:15px;" src="img/logo.png">
</div>
<div align="center" class="col-xs-12" style="background-image: url(img/blakcstonemain.png); background-repeat:no-repeat; background-position:center; background-size: cover; margin-top:10px;">
<br>
<!--Lavorazione : Scelta OP-->
<h1 style="margin-bottom: 0px;">Seleziona Ordine di Produzione</h1>
<?php
$conn = odbc_connect('', '', '');
if(! $conn){
print( "..." );
exit;
}
//definisco gli ordini di produzione
$sql="SELECT *
FROM dbo.OP_Ordini
LEFT JOIN dbo.MG_AnaART
ON dbo.OP_Ordini.OPOR_MGAA_Id = dbo.MG_AnaArt.MGAA_Id ";
$rs=odbc_exec($conn,$sql);
if (!$rs)
{exit("Errore nella tabella!");}
echo"<center>";
echo"<br>";
echo"<select>";
echo"<option>--ORDINI--</option>";
while(odbc_fetch_row($rs))
{
$ord_id=odbc_result($rs,"OPOR_Id");
$ord_anno=odbc_result($rs,"OPOR_Anno");
$ord_ordine=odbc_result($rs,"OPOR_Ordine");
$ord_lotto=odbc_result($rs,"OPOR_Lotto");
$ord_desc=odbc_result($rs,"OPOR_Descr");
$mgaa_matr=odbc_result($rs,"MGAA_Matricola");
echo"<option>| $ord_id | $ord_anno | $ord_ordine | $ord_lotto | $ord_desc | $mgaa_matr</option>";
}
echo"</select>";
echo"<br>";
echo"<br>";
?>
<!--Lavorazione : Scansione-->
<div align="center" class="col-xs-12">
<h1 style="margin-bottom: 0px;">Scansione Materiale</h1>
<br>
<label for="single"></label>
<input id="single" type="text" size="50">
<button type="button" class="qrcode-reader" id="openreader-single" data-qrr-target="#single" data-qrr-audio-feedback="false" data-qrr-qrcode-regexp="^https?:\/\/"><img style="width:20%;" src="img/qr1.png"></button>
<!--Lavorazione : Matching-->
<h1 style="margin-bottom: 0px;">Esamina</h1>
<img style="width:20%;" src="img/compara.png">
<input id="submit" type="submit" value="MANDALA!">
<!--<video id="preview" class="p-1 border" style="width:75%;border: solid #d34836;box-shadow: 5px 5px #000000a6;"></video>-->
</div>
</div>
</div>
</div>
</body>
</html>
Edited September 1, 2020 by requinix please use the Code <> button when posting code I have a form created with code already written. I am in need of a push in the right direction or a potential tutorial on this issue i have. I am running a fanatsy golf website where the user will pick one golfer each week and the cannot select them again. Is there a way I can remove that data from the list for the next week when the user makes his selection or can I have that data another color and unselectable. If you want code, please let me know and i can provide it. Thanks in advance for your help. p.s. the list data is stored in a MySQL database. 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 Hi, I am hoping someone can help me out with a slight issue I have with php and mySQL. I have an ajax-powered form with a select (dropdown) field populated through a php function. Based on the user-selected values in this field, data is displayed on the webpage; i.e. selected value 1 returns values x and y on the page. I am now trying to call additional data (value z) from a different table in the same database, and as before, use the selected values from the dropdown to display the data. For some reason, value z is not changing according to the user-selected value. This is my code: [The function to populate the select field] Code: [Select] function kfl_get_funds_names() { $result = array(); $result['CDF'] = 'Crosby Dragon Fund'; $result['CPF'] = 'Crosby Phoenix Fund'; $result['AMZPIF'] = 'AMZ Plus Income Fund'; $result['KASBIIF'] = 'KASB Islamic Income Opportunity'; $result['KASBCPGF'] = 'KASB Capital Protected Gold Fund'; $result['KASBLF'] = 'KASB Income Opportunity Fund'; $result['KASBCF'] = 'KASB Cash Fund'; $result['KASBBF'] = 'KASB Asset Allocation Fund'; $result['KASBSMF'] = 'KASB Stock Market Fund'; return $result; } [the code calling and using the function to interact with the database] Code: [Select] $funds_to_display = kfl_get_funds_names(); $current_symbol = key( $funds_to_display ); $current_nav_rates = kfl_get_latest_rates( $current_symbol ); [the code calling additional data, value z, from the database, and using the info in the select field to filter it] Code: [Select] $cutoff = kfl_cutoff( $current_symbol ); The display of each of these items is as follows: Code: [Select] <?php echo $current_nav_rates['nav_date']; ?> <?php echo $funds_to_display[$current_symbol]; ?> <?php echo $cutoff['cutoff']; ?> I can't get the $cutoff code to display the correct values. It picks up the first symbol to display and doesn't change with user selection. The code for the selection box, by the way: Code: [Select] <select id="dailynav-funds" autocomplete="off" name="dnf"> <?php foreach ($funds_to_display as $fund_symbol => $fund_name) { echo '<option'; if( $fund_symbol == $current_symbol ) { echo ' selected="selected"'; } echo ' value="' . $fund_symbol . '">'; echo $fund_name; echo '</option>'; } ?> </select> I've tried to get data using $_GET['dnf'] into the cutoff code, but that throws up parse errors. What am I doing wrong, and how can I resolve this issue? Thanks in advance! Hi, I have a MySQL database called "2011_database" that has a table called "2011_list." In that table I have fields, amongst others, called "name" and "district." I need to find way to get the data from the table and put them into a drop down list on other PHP page. But they need to be listed as "name - district" on one line. I am PHP beginner and if I understand it correctly there need to be two references to get all the data in all records, a third reference to merge them together with " - " in between; and what eludes me the most, putting them in a drop down menu. Any help is greatly appreciated Thanks This is actually a page to edit the drama details. Database- 3 tables 1. drama dramaID drama_title 1 friends 2. drama_genre drama_genreID dramaID genreID 1 1 2 2 1 1 3 1 3 3. genre genreID genre 1 comedy 2 romance 3 family 4 suspense 5 war 6 horror <?php $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = ''; $dbname = 'drama'; $link = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $id = $_GET['dramaID']; $link2 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $sql2 = "SELECT * from drama , drama_genre , genre WHERE drama.dramaID='".$id."' AND drama_genre.dramaID='".$id."' AND drama.dramaID = drama_genre.dramaID AND drama_genre.genreID = genre.genreID"; $status2 = mysqli_query($link2,$sql2) or die (mysqli_error($link2)); $link3 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $sql3 = "SELECT * from genre "; $status3 = mysqli_query($link3,$sql3) or die (mysqli_error($link3)); <form method='Post' action='doUpdate.php' enctype="multipart/form-data"> <?php while ($row2 = mysqli_fetch_assoc($status2)) { ?> <td height="1"></td> <td><select name="gdrop"> <option value="<?php echo $row2['genreID'];?>"><?php echo $row2['genre'];?></option> <?php while ($row3 = mysqli_fetch_assoc($status3)) { ?> <option value="<?php echo $row3['genreID'];?>"><?php echo $row3['genre'];?></option> <?php } mysqli_close($link3); ?> <input type='hidden' id='drama_genreID' name='drama_genreID' value = "<?php echo $row2['drama_genreID']; ?>"/> </select> <input type="hidden" id="dramaID" name="dramaID" value = "<?php echo $row['dramaID'];?>"/> <input type="submit" Value="Update"/> The result was there were 3 dropdown menus but only the first dropdown menu has all 6 genres from my database and also the genre that belongs to the drama. I'm also wondering how I can bring all the drama_genreIDs to my save(doupdate.php) page and update all 3 of them because it seems like only the last dropdown menu's data is saved. And also how can I display only 6 genres instead of 7 with the genre that belongs to the drama , being set as the default selection. Hey folks! a two-part problem he PART 1: here's a weird problem i bumped into today. when i leave OUT the 'varTable' section (see the class designations), it generates a table (allbeit messed up for reasons that i'll get into in part 2). but when i put 'varTable' back in, it generates the first array output just fine, but nothing shows for the table. (i need the first to be able to display the array) why does that happen? PART 2: when the table DOES generate, it gives a lot of errors and $i counts up to 7! whaaaaat!? thanks in advance for your help. this has me stumped. i've tried resetting the array, different variable names...nothing worked. i also included the array output at the very bottom. seems like a lot of extra data in there to me...and the [ 0 ] is printing oddly here (spaces prevent odd formatting...). WR! DATABASE TABLE: | id | item | cost | use | -------------------------------- | 0 | basket | 35 | 1 | | 1 | bicycle | 165 | 1 | | 2 | glass | 5 | 1 | | 3 | running shoe | 156 | 1 | | 4 | silver | 300 | 1 | PHP CODE: Code: [Select] <?php include("constants.php"); <?php $connexion = mysql_connect(DB_SERVER, DB_USER, DB_PASS); if(!$connexion) { die("Database Connection Failed: ".mysql_error()); } $selectDB = mysql_select_db("DB_CHOICE", $connexion); if(!$selectDB) { die("Database Selection Failed: ".mysql_error()); } $query = "SELECT * "; $query .= "FROM basic_table"; $results = mysql_query($query, $connexion); if(!$results) { die("Database Query Failed: ".mysql_error()); } ?> <div> <pre class="varTable"> <?php while($checkTable = mysql_fetch_array($results)) { print_r($checkTable); } ?> </pre> </div> <br><br> <div id='codeBlock'> <table class='phpTable' width='500' cellpadding="10"> <?php // generate table header row while ($tableRow = mysql_fetch_array($results)) { echo "<tr>"; for($i=0; $i<count($tableRow); $i++) { echo "<td>" . $tableRow[$i] . "</td>"; } echo "</tr>\n"; } ?> </table> ARRAY OUPUT: Code: [Select] Array ( [0] => 1 [id] => 1 [1] => basket [item] => basket [2] => 35 [cost] => 35 [3] => 1 [usable] => 1 ) Array ( [0] => 2 [id] => 2 [1] => bicycle [item] => bicycle [2] => 165 [cost] => 165 [3] => 1 [usable] => 1 ) Array ( [0] => 3 [id] => 3 [1] => glass [item] => glass [2] => 5 [cost] => 5 [3] => 1 [usable] => 1 ) Array ( [0] => 4 [id] => 4 [1] => running shoe [item] => running shoe [2] => 14 [cost] => 14 [3] => 1 [usable] => 1 ) Array ( [0] => 5 [id] => 5 [1] => silver [item] => silver [2] => 300 [cost] => 300 [3] => 1 [usable] => 1 ) I am trying to write out a calendar style html table with data in each cell, and with a column on the left that shows the times. Basically it should look like this when written out to the webpage: Time Mon Tues Wed Thur Fri Sat 8:00 data data2 9:30 data3 data4 10:00 data5 data6 11:00 data7 data8 The times on the left will vary each month, and the 'data' (these are actually fitness classes held each month) will appear on various days of the week. There is no need to toggle from one month to the next and such, it's just a listing of the current month to display the classes held on which days and at which times. Very easy to do statically in html, but a bit more challenging dynamically. Using MySQL, the table that holds the data looks like this: Code: [Select] schedule_id class_time class_title day_order 1 08:00 data 1 2 08:00 data2 3 3 09:30 data3 2 4 09:30 data4 4 5 10:00 data5 3 6 10:00 data6 5 7 11:00 data7 1 8 11:00 data8 3 The day_order field corresponds to the weekday - Mon is 1, Tues is 2, etc. My query looks like this: Code: [Select] SELECT schedule_id, class_time, class_title, day_orderFROM `class_schedule` order by class_time, day_order Here is the html/ php code to write it out: Code: [Select] <table border="0" cellpadding="2" cellspacing="0" width="625"> <tr> <td class="calheader">Time</td> <td class="calheader">Mon</td> <td class="calheader">Tue</td> <td class="calheader">Wed</td> <td class="calheader">Thu</td> <td class="calheader">Fri</td> <td class="calheader">Sat</td> <td class="calheader">Sun</td> </tr> <? $i=0; $first_time = ""; while($i<$num_rows){ $sid = mysql_result($sql,$i,"schedule_id"); $ctime = mysql_result($sql,$i,"class_time"); $cdate = mysql_result($sql,$i,"CURDATE()"); $strtime = date("h:i:s",$cdate.$ctime); $ctitle = stripslashes(mysql_result($sql,$i,"class_title")); $order = mysql_result($sql,$i,"day_order"); if ($ctime != $first_time){ if (isset($first_time)){ echo ""; } if ($first_time == ""){ echo "<tr>\n"; echo "<td class=\"timecell\">". $ctime ."</td>\n"; } $first_time = mysql_result($sql,$i,"class_time"); } $j = 1; while ($j<8 && $first_time != ""){ if ($order == $j){ $thetitle = $ctitle; } else { $thetitle = " "; } echo "<td class=\"calcell\">". $thetitle ."</td>\n"; $j++; } unset($first_time); if (!isset($first_time)) { echo "</tr>\n"; } $i++; } unset($ctitle); unset($thetitle); unset($weekday); unset($sid); unset($clink); unset($order); mysql_close(); The above code sort of works, but the trouble I am having is that it is creating a new table row for each class, even if they occur at the same time slot. It should be putting them on the same row, to the right of the time slot across the days of the week. If anyone could assist, that would be wonderful - I am really stuck with this... Thanks Hello, I am making a small CMS and I having some trouble with making a dynamic menu. I am wondering for a few days how I should make this menu system. I have the follow MySQL table: Code: [Select] CREATE TABLE `pages` ( `id` int(4) unsigned NOT NULL AUTO_INCREMENT, `time` varchar(10) DEFAULT NULL, `lastby` varchar(2) DEFAULT NULL, `order_id` varchar(2) DEFAULT NULL, `text` text, PRIMARY KEY (`id`) ) ENGINE=InnoDB; Now I would like to have a menu system as following that I make menu's in a menu table with a name and a menu ID and that I could place those pages as parent or as subparent. How could I manage to turn my theory into reality? This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=351877.0 Sorry for my bad english but i hope you would understand my query. How can i set 404 not found for database driven pages..? I am using header("HTTP/1.0 404 Not Found"); in my 404 error page and ErrorDocument 404 /404.html, whereas 404.html is my error page. i have an URL http://www.example.com/folder/my-test-for-url-95/ which is opening correctly because it has existence on database. but when i am opening http://www.example.com/folder/my-test-for-url-951/ which has existence on database is also opening without any information but i want here a 404 not found page. How can i do so. Thanks Hello all; I have a client that has a members area. He asked me to password protect it, which I did simply by assigning one static password. Now he wants a full username/login system where the member can set their own password, which I have never done before. I assume I'd just set up a Table with three fields, (one for name, one for password, one for the type of access they have) then check against it for access, but experience has taught me that whenever something seems simple, it's actually very complex. Do any of you know of any good premade templates for this kind of thing? Ideally it'd be session-based (obviously). I found one system he http://frozenade.wordpress.com/2007/11/24/how-to-create-login-page-in-php-and-mysql-with-session/ but it's several years old, and the misspellings in the comments tend to scare me away a bit. Thanks for any help you might be able to provide. Hi all, I am wondering what the best way to do this is... Take GoDaddy for example, if your domain is about to expire you will get an automatic email. I am wanting to do something similar, If someone orders a Soap Dispenser, I want to send an email (if they accept to receiving emails on the order) each month asking/reminding them to top up on soap. I have been reading but cant find anything adequate, I have seen something about Cron Jobs, but never heard of this before, please could you point me in the right direction. Thanks in advance. Peter Whats the best way to make a database driven menu ? Right now I have a db that looks like id | boats | Inshore id | boats | Offshore id | boats | Bay Boats and so on I'm think of going to database 1 id-1 | boats database 2 id | 1 | inshore or would a multideminsional array work best. My goal is to have the short term format be Boats Inshore Offshore Bay Boats and in the long term a format whose style I can change later, I wanted to get ideas from people in the know and have done this before. Thanks in advance |
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