PHP - Keep A Variable From A Mysql Generated Drop Down

hi

i have generated a product list from a mysql table called product_list, once i enter a new product in to the table the product will be shown in the generated list and the list will grow and the table grows.  and i want to allow user the edit/delete/save the products from the generated table, i have no idea how to do it and what is the algorithmic idea to do it so. here is the php and the html code.

<!--Body container  for creating a new product in to the list-->
<div class="body_orderviewform">
  <form name="form1" method="post" action="upload_file.php" enctype="multipart/form-data">
    <p>
      <label for="user_id">User ID:</label>
      <input type="text" name="user_id" id="user_id">
      <label for="customer_name">Customer Name:</label>
      <input type="text" name="customer_name" id="customer_name">
      <label for="customer_family">Customer_family</label>
      <input type="text" name="customer_family" id="customer_family">
      <label for="freelancer_name">Purchaser:</label>
      <input name="freelancer_name" type="text" id="freelancer_name">
     
      <? if (isset($_COOKIE['picAdd'])) echo $_COOKIE['picAdd'];?>
     
    </p>
 <!------------------------------------------------------------------------------------------------------- for generating the list -->
   
<div  class="div.neworder_list" >   

   <div  class="div.neworder_listheader" align="center">
   
    <table width="637" border="1" >
      <tr>
        <td width="193"><label for="link">link:</label>
          <label for="link_new"></label>
          <input type="text" name="link_new" id="link_new"></td>
        <td width="202"><label for="unitprice">Unit Price:</label>
          <label for="Unit_price_new"></label>
          <input type="text" name="Unit_price_new" id="Unit_price_new"></td>
        <td width="220"><label for="qty">Quantity:</label>
          <label for="quantity_new"></label>
          <input type="text" name="quantity_new" id="quantity_new"></td>
      </tr>
      <tr>
        <td><label for="express">Express Fee:</label>
          <label for="express_new"></label>
          <input type="text" name="express_new" id="express_new"></td>
        <td><label for="commission_new">Commission:</label>
          <input type="text" name="commission_new" id="commission_new"></td>
        <td><label for="customer_description">Description</label>
          <label for="description_new"></label>
          <textarea name="description_new" id="description_new" cols="45" rows="5"></textarea></td>
      </tr>
      <tr>
        <td>&nbsp;</td>
        <td colspan="2">Picture Upload:
              <input type="hidden" name="<?php echo ini_get("session.upload_progress.name");?>" value="123" />
              <input name="file" type="file" autofocus="autofocus"/>
              <br />


      </tr>
      <tr>
        <td colspan="3"><input type="reset" name="reset" id="reset" value="Reset">
        <input type="submit" name="submit" id="submit" value="Submit The Product"></td>
      </tr>
    </table>

 
</form>   
</div>
   
<div class="neworder_listview">
  <p>
  <form action="" method="post" name="list">
    <input type="submit" name="del" id="del" value="Save">
    <input type="submit" name="save" id="save" value="Del">
  </p>
  <table width="1022" border="1" align="center">
    <tr>
      <th width="24" scope="col">&nbsp;</th>
    <th width="24" scope="col">Row#</th>
    <th width="137" scope="col">Manager</th>
    <th width="137" scope="col">Purchaser Desc</th>
    <th width="40" scope="col"><p>Link</p>
      <p>/Ссылки</p></th>
    <th width="53" scope="col">ФОТО</th>
    <th width="50" scope="col">Unit Price/Цена за еденицу товара</th>
    <th width="46" scope="col">Quantity/ Кол-во</th>
    <th width="138" scope="col">Total Unit Price/ Общая цена</th>
    <th width="89" scope="col">Express/Доставка по Китаю</th>
    <th width="119" scope="col">Description/Описание</th>
    <th width="89" scope="col">ADDITIONAL LINKS/ЗAMЕНЫ</th>
    </tr>
   


<?php

   $username = "my username";
   $password = "my pass";
   $database = "userinfo";
   $link = mysql_connect("localhost", "$username", "$password");
   
   if(!$link) {echo("Failed to establish connection to mysql server");
               exit();}

   $status = mysql_select_db($database);
   $query = "SELECT * FROM order_list";
   $result = mysql_query($query);
   $num = mysql_num_rows($result);
   $i=0;
 
 
 
  while ($i < $num) {
   $field1_name=mysql_result($result,$i,"admin_st");
   $field2_name=mysql_result($result,$i,"freelancer_st");
   $field3_name=mysql_result($result,$i,"link");
   $field4_name=mysql_result($result,$i,"picture");
   $field5_name=mysql_result($result,$i,"unitprice");
   $field6_name=mysql_result($result,$i,"qty");
   $field7_name=mysql_result($result,$i,"express");
   $field8_name=mysql_result($result,$i,"customer_st");
   $i++;
    }
 
?>







<?php
$i=0;
$row=1;
while ($i < $num) {

$f1=mysql_result($result,$i,"admin_st");
$f2=mysql_result($result,$i,"freelancer_st");
$f3=mysql_result($result,$i,"link");
$f4=mysql_result($result,$i,"pic_address");
$f5=mysql_result($result,$i,"unitprice");
$f6=mysql_result($result,$i,"qty");
$f7=mysql_result($result,$i,"express");
$f8=mysql_result($result,$i,"customer_st");
$totao_unit_price = $f5*$f6;
?>
  <tr>
    <td><input type="checkbox" name="del_chbox" id="del_chbox">
   <td><p><font face="Arial"><input name="row_txtbox" type="text" id="row_txtbox" size="2" value="<?php echo $row; ?>"></font></td>
   <td><p><font face="Arial"> <textarea name="manager_txtbox" cols="10" id="manager_txtbox"><?php echo $f1; ?></textarea></font></td>
   <td><p><font face="Arial">        <textarea name="purchase_txtbox" cols="10" id="purchase_txtbox"><?php echo $f2; ?></textarea>
</font></td>


   <td><font face="Arial"><a href="<?php $f3 ?>" target="_blank"><?php echo $f3; ?></a></font></td>
   <td><font face="Arial"><img src="<?php echo $f4;?>" width="100" align="middle"100></font></td>
   <td><font face="Arial"><input name="unitprice_txtbox" type="text" id="unitprice_txtbox" size="2" value="<?php echo $f5; ?>"></font></td>
   <td><font face="Arial"> <input name="qty_txtbox" type="text" id="qty_txtbox" size="2" value="<?php echo $f6; ?>"></font></td>
   <td><p><font face="Arial"><?php echo $totao_unit_price; ?></font></td>
   <td><p><font face="Arial"><input name="express2" type="text" id="express3" size="2" value="<?php echo $f7; ?>"></font></td>
   <td><p><font face="Arial"> <textarea name="custdesc_txtbox" cols="20" id="custdesc_txtbox"><?php echo $f8; ?></textarea></font></td>
    <td><input name="express2" type="text" id="express3" size="2" value="<?php echo "new link" ?>"></td>
   </tr>

  <p>
    <?php
$i++;
$row++;
}
?>
</table>
   
  </p>
  <p>&nbsp;</p>
  </form>

I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here.  It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere.

<?PHP require('connect.php'); ?>

<form action ='' method='post'> <select name="id">

<?php
$extract = mysql_query("SELECT * FROM cars");
   
   while($row=mysql_fetch_assoc($extract)){
   $id = $row['id'];
   $make= $row['make'];
   $model= $row['model'];
   $year= $row['year'];
   $color= $row['color'];

echo "<option value=$id>$color $year $make $model</option>
";}?>

</select>

Which attribute would you like to change?<br />
<input type="radio" name="getchanged" value="make"/>Make<br />
<input type="radio" name="getchanged" value="model"/>Model<br />
<input type="radio" name="getchanged" value="year" />Year<br />
<input type="radio" name="getchanged" value="color"   />Color<br /><br />

   <br /><input type='text' value='' name='tochange'>
   <input type='submit' value='Change' name='submit'>

</form>

//This is where I need help...

<?PHP
if(isset($_POST['submit'])&&($_POST['tochange'])){
mysql_query("
   UPDATE cars
   SET '$_POST[getchanged]'='$_POST[tochange]'
   where id = '$_POST[id]'
         ");}?>

Hi,

I'm trying to drop a table form a database using the table name as a variable. This variable is set within a session. I am using a form where you input the table name, then the DROP TABLE procedure is executed in the next page.

I know that the variable gets passed because it prints the table name variable out when I "echo".

The execution is simple (code found in deletetable.php)

$temp = $_POST['temp'];
$query = "DROP TABLE $temp";
if(mysql_query($query)){
echo "<h1 style='color:green;'>Table $temp deleted</h1>";}
else{
echo "<h1 style='color:red;'>Deletion of table '$temp' failed!</h1>";}


Here's my code for the two pages:
delete_table.php

<?php
include("includes/header.html");
session_start();
$hostname = $_SESSION['hostname'];
$user = $_SESSION['user'];
$passwd = $_SESSION['passwd'];
$database = $_SESSION['database'];
$tblname = $_SESSION['table'];

$conn = mysql_connect($hostname, $user, $passwd, $database);
/*if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}*/
?>
<div id="container">
<div id="header">Create DB</div>
<div id="breadcrumbs">
<?php
function check_port($port) {
    $conn = @fsockopen("127.0.0.1", $port, $errno, $errstr, 0.2);
    if ($conn) {
        fclose($conn);
        return true;
    }
}

function server_report() {
    $report = array();
    $svcs = array('3306'=>'MySQL');
    foreach ($svcs as $port=>$service) {
        $report[$service] = check_port($port);
    }
    return $report;
}

$report = server_report();
?>
<div id="server"><p>Server is <?php echo $report['MySQL'] ? "running" : "offline"; ?></p></div>
<?php
if(!$_SESSION['hostname']){

echo "<font color='red'>Not connected to server!</font>";
} else {

echo "<font color='green'>Server: ".$_SESSION['hostname']."</font>";
} 
if(!$_SESSION['database']){

echo "";
}else{

echo "<br><font color='green'>Database: ".$_SESSION['database']."</font>";
}
?>
</div>
<div id="nav">
<p><a href="#" onclick="javascript:window.open('connect_db.php','_self');"><img src="server_conn_ico.png" align="left" />Server Connection</a></p>
<p><a href="databases.php"><img src="databases_ico.png" align="left" />View Database</a></p>
<p><a href="create_db.php"><img src="server_folder_ico.png" align="left" />Create Database</a></p>
<p><a href="drop_db.php"><img src="drop_ico.png" align="left" />Drop Database</a></p>
<p><a href="create_table.php"><img src="status_ico.png" align="left" />Add Table</a></p>
<p><a href="add_fields.php"><img src="server_ico.png" align="left" />Add Fields</a></p>
<p class="active"><a href="delete_table.php"><img src="status_delete_ico.png" align="left" />Delete Table</a></p>
<p><a href="view_tables.php"><img src="user_admin_ico.png" align="left" />View Tables</a></p>
<p><a href="view_data.php"><img src="user_admin_ico.png" align="left" />View Table Data</a></p>
<p><a href="http://localhost/phpmyadmin/" target="_blank"><img src="phpmyadmin_ico.png" align="left" />phpMyAdmin</a></p>
<p><a href="connect_info.php"><img src="health_ico.png" align="left" />System Info</a></p>
<p><a href="error_log.php"><img src="logs_ico.png" align="left" />Server Log</a></p>
<p><a href="bug_report.php"><img src="bugs-icon.png" align="left" />Bug Reports</a></p>
</div>
<div id="controls"><a href="javascript:self.close()"><img src="controls.png" /></a></div>
<div id="main">
<div id="content">
<!--<h1>Delete Database Table</h1>-->
<?php
if(empty($database)) {

echo "<p>You must be connected to a database in order to delete a table.</p>";
}else{
//echo 'Connected successfully to \''.$database.'\'';

// Check tables

$sql = "SHOW TABLES FROM $database";

$result = mysql_query($sql);

if (mysql_num_rows($result) > 0) {

echo "<p style='text-align:left;padding-left:24px;'>Available tables:</p>\n";

echo "<pre style='text-align:left;padding-left:24px;font-size:1.2em;'>\n";

while ($row = mysql_fetch_row($result)) {

echo "{$row[0]}\n";

}

echo "</pre>\n";

}
?>
<form action="deletetable.php" method="post">
Enter table you'd like to delete from '<?php echo $database; ?>' database:
<input name="temp" type="text" /><br/>
<input name="Submit" type="submit" value="  Delete Table  " />&nbsp;&nbsp;&nbsp;<input type='button' value='  Cancel  ' onclick="javascript:window.location='connect_db.php';"></p>
</form>
<?php
}
include("includes/footer.html");
?>


Executing file deletetable.php:

<?php
include("includes/header.html");
session_start();
$hostname = $_SESSION['hostname'];
$user = $_SESSION['user'];
$passwd = $_SESSION['passwd'];
$database = $_SESSION['database'];
//$tblname = $_SESSION['table'];

$conn = mysql_connect($hostname, $user, $passwd, $database);
?>
<div id="container">
<div id="header">Create DB</div>
<div id="breadcrumbs">
<?php
function check_port($port) {
    $conn = @fsockopen("127.0.0.1", $port, $errno, $errstr, 0.2);
    if ($conn) {
        fclose($conn);
        return true;
    }
}

function server_report() {
    $report = array();
    $svcs = array('3306'=>'MySQL');
    foreach ($svcs as $port=>$service) {
        $report[$service] = check_port($port);
    }
    return $report;
}

$report = server_report();
?>
<div id="server"><p>Server is <?php echo $report['MySQL'] ? "running" : "offline"; ?></p></div>
<?php
if(!$_SESSION['hostname']){

echo "<font color='red'>Not connected to server!</font>";
} else {

echo "<font color='green'>Server: ".$_SESSION['hostname']."</font>";
} 
if(!$_SESSION['database']){

echo "";
}else{

echo "<br><font color='green'>Database: ".$_SESSION['database']."</font>";
}
?>
</div>
<div id="nav">
<p><a href="#" onclick="javascript:window.open('connect_db.php','_self');"><img src="server_conn_ico.png" align="left" />Server Connection</a></p>
<p><a href="databases.php"><img src="databases_ico.png" align="left" />View Database</a></p>
<p><a href="create_db.php"><img src="server_folder_ico.png" align="left" />Create Database</a></p>
<p><a href="drop_db.php"><img src="drop_ico.png" align="left" />Drop Database</a></p>
<p><a href="create_table.php"><img src="status_ico.png" align="left" />Add Table</a></p>
<p><a href="add_fields.php"><img src="server_ico.png" align="left" />Add Fields</a></p>
<p class="active"><a href="delete_table.php"><img src="status_delete_ico.png" align="left" />Delete Table</a></p>
<p><a href="view_tables.php"><img src="user_admin_ico.png" align="left" />View Tables</a></p>
<p><a href="view_data.php"><img src="user_admin_ico.png" align="left" />View Table Data</a></p>
<p><a href="http://localhost/phpmyadmin/" target="_blank"><img src="phpmyadmin_ico.png" align="left" />phpMyAdmin</a></p>
<p><a href="connect_info.php"><img src="health_ico.png" align="left" />System Info</a></p>
<p><a href="error_log.php"><img src="logs_ico.png" align="left" />Server Log</a></p>
<p><a href="bug_report.php"><img src="bugs-icon.png" align="left" />Bug Reports</a></p>
</div>
<div id="controls"><a href="javascript:self.close()"><img src="controls.png" /></a></div>
<div id="main">
<div id="content">
<?php
$temp = $_POST['temp'];
$query = "DROP TABLE $temp";
if(mysql_query($query)){
echo "<h1 style='color:green;'>Table $temp deleted</h1>";}
else{
echo "<h1 style='color:red;'>Deletion of table '$temp' failed!</h1>";}

include("includes/footer.html");
?>


Your help is greatly appreciated!!

guys i have this  dropdown menu

echo "<form method=\"post\" action=\"\">";

echo "<center><select name=\"mydropdown\" size=\"0\" style=\"height:4em; width:15em;\">";

echo "<option value=\"Milk1\">{$_SESSION['tem1']}</option>";
echo "<option value=\"Milk2\">{$_SESSION['tem2']}</option>";
echo "<option value=\"Milk3\">{$_SESSION['tem3']}</option>";

echo "</select></center>";
echo "</form>";

and i wand to update a variable every time i select one of the contents...


how can i done this???

Ok here is what im trying to do:
1. drop down list pulled from mysql database.( working )
2. after you select the name in the drop down list
3. Use selected name to another page.
here is code


Code: [Select]
<?php
include('include/db_connection.inc');

$table = "truck_master";

$result = mysql_query( "SELECT driver_name FROM $table" );

$options=""; 
while ($row=mysql_fetch_array($result)) {

$driver_name=$row["driver_name"];

$options.="<OPTION VALUE=\"driver_name\">".$driver_name.'</option>';




} ?>
<html>
<head>
</head>
<body>
<table align="center">

<form name="form1" method="GET" align="center" action="issue.php">

<tr>
<td>Driver Name :</td>
<td><SELECT name="driver_name"><OPTION VALUE=''>Choose<?php echo $options?></SELECT> </td>
</tr>
<tr>
<td><input type="submit" name="Submit" value="Submit"> </td>
</tr>
</form>
</table>
</body>
<html>
Second Code IS
Code: [Select]
$driver = $_GET['driver_name'];

Hi all

I am trying to get my drop down menu to use a session variable as the selected menu option. Here's my code:

 echo "<option value=\"\">Delivery Area</option>";
      
       $selected = "";
         
                        if(isset($_SESSION['postal_area']) && $_SESSION['postal_area'] == "England") {

                     $selected = "selected";

                  } elseif(isset($_SESSION['postal_area']) && $_SESSION['postal_area'] == "Wales") {

                     $selected = "selected";

                  } elseif(isset($_SESSION['postal_area']) && $_SESSION['postal_area'] == "Scotland") {

                     $selected = "selected";

                  } else {

                     $selected = "";

                  }
      
         echo " <option value=\"England\" selected=\"".$selected."\">England</option>";
         echo "<option value=\"Scotland\" selected=\"".$selected."\">Scotland</option>";
         echo "<option value=\"Wales\" selected=\"".$selected."\">Wales</option>";
         echo "</select>";

However, this shows the 'Wales' one all the? Please help!

Many thanks
Pete

the same page?

User clicks on a url, ie: example.com/AEQ438J

When I perform this in the code below:

Code: [Select]
    $referrer = $_GET['_url'];
    // echo $referrer displays the referrer ID contents correctly as "AEQ438J"
   
    if ( ! empty($referrer))
    {
    $mysqli->query("UPDATE coming_soon_emails SET clicks = clicks + 1 WHERE code='" . $referrer ."'");
    }
    // this also updates the database correctly as it should

    if (!empty($_POST['email'])){

   // echo $referrer displays the referrer ID contents as BLANK. It should display "AEQ438J"!
    .....
$referrer displays correctly BEFORE if($_POST['form']), however during the if($_POST['form']) $referrer is empty.
How can I fix my code so that $referrer is not empty during the time the user posts their email address in the form?

Thank you!

Complete PHP and HTML

Code: [Select]
    <?php
    
    require "includes/connect.php";
    //var_dump($_GET);die;
    
    function gen_code($codeLen = 7) {     
    $code = '';     
    for ($i=0; $i<$codeLen; $i++) {         
     $d=rand(1,30)%2;       
     $code .= $d ? chr(rand(65,90)) : chr(rand(48,57));      }  
     return $code; 
     }  
    
    
     function add_code($email_id) {
     global $mysqli;
     $code = gen_code(7); 
     $mysqli->query("UPDATE coming_soon_emails SET code='" . $code ."' WHERE email_id='" . $email_id . "'");  
     if($mysqli->affected_rows != 1) {   
     add_code($email_id);  
     } else return $code; } 
    
    $msg = '';
    $referrer = $_GET['_url'];
    // echo $referrer displays the referrer ID contents correctly
    
    if ( ! empty($referrer))
    {
    $mysqli->query("UPDATE coming_soon_emails SET clicks = clicks + 1 WHERE code='" . $referrer ."'");
    }
    
    if (!empty($_POST['email'])){
    
    // echo $referrer displays the referrer ID contents as BLANK
    
        // Requested with AJAX:
        $ajax = ($_SERVER['HTTP_X_REQUESTED_WITH']  == 'XMLHttpRequest');
    
        try{
            if(!filter_input(INPUT_POST,'email',FILTER_VALIDATE_EMAIL)){
                throw new Exception('Invalid Email!');
            }
    
            $mysqli->query("INSERT INTO coming_soon_emails
                            SET email='".$mysqli->real_escape_string($_POST['email'])."'");
    
            if($mysqli->affected_rows != 1){
                throw new Exception('This email already exists in the database.');
            } else {   
              $email_code = add_code($mysqli->insert_id); 
            } 
    
     $msg = "http://www.example.com/" . $email_code;
    
     //the following doesn't work as referrer is now empty :(
     if ( ! empty($referrer))
     {
     $mysqli->query("UPDATE coming_soon_emails SET signup = signup + 1 WHERE code='" . $referrer ."'");
            }
    
     if($ajax){
     die(json_encode(array('msg' => $msg)));
     }
    
        }
        catch (Exception $e){
    
            if($ajax){
                die(json_encode(array('error'=>$e->getMessage())));
            }
    
            $msg = $e->getMessage();        
        }
    }
    
    ?>
   
   
    <!DOCTYPE html>
    <html>
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title></title>
   
    <link rel="stylesheet" type="text/css" href="css/styles.css" />
   
    </head>
   
    <body>
   
    <div id="launch">
   
       
        <form id="form" method="post" action="">
        <input type="text" id="email" name="email" value="<?php echo $msg;?>" />
            <input type="submit" value="Submit" id="submitButton" />
        </form>
       
        <div id="invite">
        <p style="margin-top:20px;">The ID of who referred you: <?php echo $referrer; //this displays correctly?>)</p>
        <p style="margin-top:20px;"><span id="code" style="font-weight:bold;">&nbsp;</span></p>
    </div>
   
       
    </div>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.3/jquery.min.js"></script>
    <script src="js/script.js"></script>
    </body>
    </html>
script.js

Code: [Select]
    $(document).ready(function(){
   
    // Binding event listeners for the form on document ready
   
    $('#email').defaultText('Your Email Address');
   
    // 'working' prevents multiple submissions
    var working = false;
   
    $('#form').submit(function(){
   
    if(working){
    return false;
    }
    working = true;
   
    $.post("./index.php",{email:$('#email').val()},function(r){
   
    if(r.error){
    $('#email').val(r.error);
    } else {
        $('#email').val(r.msg); // not needed but gets hidden anyways...
    $('#launch form').hide();
    $("#code").html(r.msg);
    $("#invite").fadeIn('slow');
    }
   
   
    working = false;
    },'json');
   
    return false;
    });
    });
   
    // A custom jQuery method for placeholder text:
   
    $.fn.defaultText = function(value){
   
    var element = this.eq(0);
    element.data('defaultText',value);
   
    element.focus(function(){
    if(element.val() == value){
    element.val('').removeClass('defaultText');
    }
    }).blur(function(){
    if(element.val() == '' || element.val() == value){
    element.addClass('defaultText').val(value);
    }
    });
   
    return element.blur();
    }
htaccess

Code: [Select]
    RewriteEngine on
   
    RewriteCond %{HTTP_HOST} ^my-url.com
    RewriteRule (.*) http://www.my-url.com/$1 [R=301,L]
   
    RewriteCond %{REQUEST_FILENAME} !-f
    RewriteCond %{REQUEST_FILENAME} !-d
    RewriteRule ^([a-z0-9]+)$ /index.php?_url=$1 [NC,L,QSA]
table.sql

Code: [Select]
    CREATE TABLE IF NOT EXISTS `coming_soon_emails` (   
      `email_id` int(11) NOT NULL auto_increment,   
      `email` varchar(64) collate utf8_unicode_ci NOT NULL,   
      `code` char(7) collate utf8_unicode_ci DEFAULT NULL,
      `clicks` int(64) collate utf8_unicode_ci DEFAULT 0,
      `signup` int(64) collate utf8_unicode_ci DEFAULT 0,
      `ts` timestamp NOT NULL default CURRENT_TIMESTAMP,   
      PRIMARY KEY  (`email_id`),   
      UNIQUE KEY `email` (`email`),
      UNIQUE KEY `code` (`code`)
    ) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

hey guys just a quick one

i want a drop down box which is populated from a mysql query which is eay enough but also at the same time i also want the dropdown box to look at the current record and have the value in the drop down box

e.g.

the drop down has values 1, 2, 3, 4, 5 in it which are pulled from a table called values

the current record is from a table called staff and it has an id of 45 and a user level of 3

i want the drop down to have all the values selectable to update the record but on load it will have 3 as the current, i want to be able to chage this 3 to a 5 and update the record

how can i do this.? just a brief code example would be awsome and ill adopt it to fit

I got some help here about 6 months ago with drop down menus & I need a little more.
I'm working on a form that has about 20 drop down menus,  each populated from a mysql database table with about 50 entries each.
I need to keep the selected menu option on the form after the form is submitted but each time the form is submitted the selected  menu option reverts back to the first item in the database table. Is there any practical way to fix this problem other than using javascript to finish  up? 

Hi,
At present the PHP page reads the MySQL data and prints on screen and if the result if 'FREE' it shows a clickable image for the user to book. I am looking for something that allows the user to drag the result they fund to another location with in the table. I.e. user has box 1 filled with their name, they can drag their name to box 9 and it saves it, so they can go back on the it will be their.

If anyone knows where I can firn maybe a tutorial, or guide to get started with this, as I have looked and cannot see anything.

Cheers.
Stu

I'm a newbie on php. I'm really a system administrator and I was just task to do this simple task. For me its hard but I'm sure for a programmer this is very simple. My agenda is to pull out data on one of my column in mysql, select it and dump it on mysql. Here is the php for retrieving mysql data

Code: [Select]
<?php
function database_connect($users)
{
    $resource_link = mysql_connect("localhost", "root", "root");
    if (mysql_select_db($users, $resource_link)) {
        return $resource_link;
    } else {
        echo "Cannot connect to DB";
        return false;
    }
}

function print_dropdown($query, $link){
    $queried = mysql_query($query, $link);
    $menu = '<select username="username">';
    while ($result = mysql_fetch_array($queried)) {
        $menu .= '
      <option value="' . $result['id'] . '">' . $result['username'] . '</option>';
    }
    $menu .= '</select>';
    return $menu;
}

//Some other form elements, or just start a form.
echo '<form method="post" action="create2.php">';

//The important bit
echo print_dropdown("SELECT username FROM mailbox;", database_connect("users"));

//Some other form elements, or just end the form.
echo '<input type="submit" name="submit" value="submit"/></form>';

Here is the content of my create2.php. This is the php page who do the insert on my mysql.

Code: [Select]
<?php
// open the connection
$conn = mysql_connect("localhost", "root", "root");
// pick the database to use
mysql_select_db("users",$conn);
// create the SQL statement
$sql2 = "INSERT INTO mailbox values ('','locked','','$_POST[username]','',NOW(),'','locked','')";
// for troubleshooting
$result = mysql_query($sql2, $conn) or die(mysql_error());
// execute the SQL statement
//if (mysql_query($sql2, $conn)) {
//    echo "Success";
//} else {
 //   echo "Fail";
//}
}
?>

When I click the submit button, I don't see any record being inserted on my table. I'm using the create2.php on my other page though it is only an insert/fill up form not like this one that I need to pull up the date, select and insert to mysql.

Hey all -

I (think) that I'm following the code exactly, but no luck...  Essentially I want to populate a drop down menu from text stored in mysql (varchar). 

Here is my code (and where I'm pulling the code found below).  Any ideas what's going wrong.  (My error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in....).



$sql = "SELECT * FROM table";
$results = mysql_query($sql); //or die("Query failed with error: ".mysql_error());       // execute the query
$options="";

while ($row=mysql_fetch_array($results)){
$id=$row["id"];
$thing=$row["thing"];
$options.="<OPTION VALUE=\"$id\">".$thing;
}



<select name="id" style="font-size:20px;font-family:Arial;width:275px">
            <option value=0>Choose<? echo $options?></option>
</select>

Here is where I'm getting my code sample: http://forums.devarticles.com/mysql-development-50/drop-down-menu-populated-from-a-mysql-database-1811.html

Thank for any advice!!

Hello Friends!.... here is the big idea. I am trying to make a form in which there will be 2 drop down lists which will be populating directly from MySQL DB ....
Actually i am developing a student management system as my first PHP project... here i want to have 2 drop down lists first is roll number and second is student name. i want that when some one select the roll number in the first drop down the student name against it is automatically populated in the next drop down.


please help me friends i am new to php and dont know soo much.! any help PLZZZZZZZZZZ

I've got a HTML drop down box as similar to this:

<select name="dropdown">
<option>Option 1</option>
<option>Option 2</option>
<option>Option 3</option>
</select>

If I run a mysql query and get a result of "Option 3", is there anyway using PHP to give Option 3 the selected value?