PHP - Displaying Data From A Mysql Dropdown Selection

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!

So i pull some records out of a mysql table and i want to display them in 5 even columns.
I'm not entirely sure how to do the math & logic to accomplish this.

The pull is simple

$qry = "SELECT DIST_PART_NUM FROM $tablename";
$sql = mysql_query($qry) or die(mysql_error());
while($res = mysql_fetch_assoc($sql))

{

// CREATE 5 even columns here.

}


so let's say i just retrieved 5,000 part numbers, i'd like to display then in a table of 5 columns with 1000 records per column. This is easy math, but i need the script to automatically figure out the #'s.

Also the tricky part is that i dont want to display the part numbers like so

11111     22222           33333        44444          55555
66666     77777           88888        99999          00000

but rather

11111     44444     77777
22222     55555     88888
33333     66666     99999
                              00000
the remainder if there is one can go in the last column or whatever is easier.

I'd tried googling this, but it's not easy to phrase what i'm looking for.


Thanks for the help.

PS: I'm not looking to copy and paste code, if possible please explain your way so that i can learn the logic.

Hey guys have been trying to get this script to work for a while now, i am new to php and mysql so i am sure i am missing something simple. I have DB setup and need to pull data based on the key item code and get the following

I want to get the fields
item_code
description
allergy_statement
useable_units
region_availability
order_lead_time
ingredients

for item_code 12-100 LITERALLY 12-100, no range, but like i said before i am really new to php and mysql. I have 1187 items that when a user clicks a link in search results it takes them to the product details page for that item code All that data is in my database just can't figure out how to get it out of the database. Is this even the right script to achieve that result.

here is the code to get the data from database
Code: [Select]
<?php require_once('includes/mysql_connect_nfacts_ro.php'); 

$query = "SELECT item_code, description, allergy_statement, useable_units, region_availability, order_lead_time, ingredients " . 
 "FROM products  " .
 "WHERE item_code = '12-100' ";
$resuts = mysql_query($query) 
or die(mysql_error());


?>

And need to display the data like so :

Code: [Select]
<td width="715" align="center" valign="top">&nbsp;
        <h1>Product Details</h1>
        <h3>DISPLAY description HERE</h3>
        <table width="420" border="0">
            <td class="ingreg">&nbsp;</td>
        </table>
        <h5>Item Number</h5>
        <table width="420" border="0">
          <tr>
            <td class="ingreg">DISPLAY ITEM_CODE HERE</td>
          </tr>
        </table>
        <h3>Ingredients:</h3>
        <table width="420" border="0">
          <tr>
            <td class="ingreg">DISPLAY INGREDIENTS HERE</td>
          </tr>
        </table>
        <h4>Allergy Statement:</h4>
        <table width="420" border="0">
          <tr>
            <td class="ingreg">DISPLAY Allergy Statement HERE</td>
          </tr>
        </table>
        <h4>Useable Units Per Package:</h4>
        <table width="420" border="0">
          <tr>
            <td class="ingreg">DISPLAY Useable Units Per Package HERE</td>
          </tr>
        </table>
        <h4>Region Availability: </h4>
        <table width="420" border="0">
          <tr>
            <td class="ingreg">&DISPLAY ITEM_CODE HERE</td>
          </tr>
        </table>
        <h4>Order Lead Time:</h4>
        <table width="420" border="0">
          <tr>
            <td class="ingreg">&DISPLAY order lead time HERE</td>
          </tr>
        </table>
        <p>&nbsp;</p>
      <div align="right"></div></td>
    </tr>
  </table>

 how do i get data in database to display where i need it to? Can any one shine some light on this

Ok... I need some help - I want to show a players balance in a game beside there name (Balance is in mysql database)I can do that but... - I also want to show if there online or offline at the same time( This is stored in a different database)
I have the code which says whether they are online or offline
<?PHP

// Conect to the Mysql Server
$connect = mysql_connect("localhost","scswccla_bukkit","********");

//connect to the database
mysql_select_db("scswccla_bukkit");

//query the database
$query = mysql_query("SELECT * FROM users_online WHERE online = 1");

//title
echo "<font size='100'>NovaCraft Users</font><br>";

// fetch the results / convert into an array

        WHILE($rows = mysql_fetch_array($query)):
 
 
        $users = $rows['name'];



echo "<font color='black'>|Online|<br><font color='green'>$users</font></font><br>";

 endwhile;
 
//query the database
$query = mysql_query("SELECT * FROM users_online WHERE online = 0");

//title
echo "<font color='black'>|Offline|</font><br>";

// fetch the results / convert into an array

        WHILE($rows = mysql_fetch_array($query)):
 
 
        $users = $rows['name'];



echo "<font color='red'>$users</font><br>";

 endwhile;

?>

Here is the page: www.scswc.com/Offline_Users.php displaying that

But I want to Create something like this:

Nocvacraft Players

|Online|
Name:Player Balance:$20
|Offline|
Name:Player Balance:$15

Here is what I have tried:
<?PHP

// Conect to the Mysql Server
$connect = mysql_connect("localhost","scswccla_bukkit","**********");

//connect to the database
mysql_select_db("scswccla_bukkit");

//query the database
$query = mysql_query("SELECT * FROM users_online WHERE online = 1");
$query2 =mysql_query("SELECT * FROM iBalances WHERE player = $users");

//title
echo "<font size='100'>NovaCraft Users</font><br>";

// fetch the results / convert into an array

        WHILE($rows = mysql_fetch_array($query) $rows2 = mysql_fetch_array($query2)):
 
 
        $users = $rows['name'];
$balance = $rows2['balance'];



echo "<font color='black'>|Online|<br><font color='green'>Name:$usersBalance:$balance</font></font><br>";

 endwhile;
 
//query the database
$query = mysql_query("SELECT * FROM users_online WHERE online = 0");

//title
echo "<font color='black'>|Offline|</font><br>";

// fetch the results / convert into an array

        WHILE($rows = mysql_fetch_array($query)):
 
 
        $users = $rows['name'];



echo "<font color='red'>$users</font><br>";

 endwhile;

?>

I know I am trying to use a variable before it is been set - but if I don't how

I have tried this as well...
<?PHP

// Conect to the Mysql Server
$connect = mysql_connect("localhost","scswccla_bukkit","**********");

//connect to the database
mysql_select_db("scswccla_bukkit");

//query the database
$query = mysql_query("SELECT * FROM users_online WHERE online = 1");

//title
echo "<font size='100'>NovaCraft Users</font><br>";

// fetch the results / convert into an array

        WHILE($rows = mysql_fetch_array($query)):
 
 
        $users = $rows['name'];




echo "<font color='black'>|Online|<br><font color='green'>Name:$users</font></font><br>";

 endwhile;
 
//query the database
$query = mysql_query("SELECT * FROM users_online WHERE online = 0");

//title
echo "<font color='black'>|Offline|</font><br>";

// fetch the results / convert into an array

        WHILE($rows = mysql_fetch_array($query)):
 
 
        $users = $rows['name'];



echo "<font color='red'>$users</font><br>";

 endwhile;
$query = mysql_query("SELECT * FROM iBalances WHERE player = $users");

WHILE($rows = mysql_fetch_array($query)):
 
 
        $balance = $rows['balance'];



echo "<font color='red'>$users $balance</font><br>";

 endwhile;
//

?>


Can you use variables in mysql_query()?Is that why it isn't working?

This is my first php script so if I need to give you more information for you to help me just tell me

Thanks


Here is database pictures

iBalances

users_online

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>