PHP - Php Data Matching Dropdown With Qr Code Scan

Hi all.

I'm trying to build a dynamic multiple tiered navigation menu that operates on the existence of directories.  So far, I have some code that will read through the document root, and one sub level of directories but I'm not sure how to get into the third level of directories.  Here's what I have so far...

The code:

      $docroot=scandir($_SERVER['DOCUMENT_ROOT']);

      unset($docroot[array_search('.',$docroot)]);

      unset($docroot[array_search('..',$docroot)]);

      unset($docroot[array_search('php',$docroot)]);

      unset($docroot[array_search('css',$docroot)]);

      unset($docroot[array_search('input_forms',$docroot)]);

      $docroot=array_values($docroot);

      

      //Build tier 1

      echo "<div class=\"tier\">";

      echo "<a href=\"/index.php\">Home</a>";

      foreach($docroot as $value1){

         if(is_dir($value1)){

            $t2=scandir($_SERVER['DOCUMENT_ROOT'] . "/" . $value1);

            unset($t2[array_search('.',$t2)]);

            unset($t2[array_search('..',$t2)]);

            $t2=array_values($t2);

            $t1[]=$t2;

            

            echo "<a href=\"/$value1/\">";

            $value1=str_replace("_"," ",$value1);

            $value1=ucwords($value1);

            echo $value1;

            echo "</a>";

         }

      }

      echo "</div>";


The output:
Code: [Select]
[it automatically builds the first tier of the menu with some stylesheet definitions and places it here.  I did this just for testing purposes.]
Array
(
    [0] => Array
        (
            [0] => bus_mgmt_context.pdf
            [1] => financial_management
            [2] => index.php
            [3] => nav.php
            [4] => service_level_management
            [5] => service_portfolio
            [6] => workforce_management
        )

    [1] => Array
        (
            [0] => element_inventory
            [1] => index.php
            [2] => mobile_administration
            [3] => monthly_bill_review
            [4] => nav.php
        )

    [2] => Array
        (
            [0] => business_it_requests
            [1] => change_management
            [2] => demand_management_context.pdf
            [3] => index.php
            [4] => nav.php
            [5] => project_management
            [6] => release_management
        )

    [3] => Array
        (
            [0] => availability_management
            [1] => capacity_management
            [2] => configuration_management
            [3] => index.php
            [4] => nav.php
            [5] => planning
            [6] => problem_management
            [7] => system_planning
        )

    [4] => Array
        (
            [0] => context
            [1] => input
            [2] => reports
            [3] => sop
            [4] => t3.php
            [5] => video
        )

    [5] => Array
        (
            [0] => executive_reports
            [1] => index.php
            [2] => kpi_index
            [3] => nav.php
        )

    [6] => Array
        (
            [0] => compliance
            [1] => hipaa_definitions.pdf
            [2] => index.php
            [3] => information_security
            [4] => nav.php
            [5] => risk_context.pdf
            [6] => service_continuity
            [7] => threat_management
        )

    [7] => Array
        (
            [0] => incident_management
            [1] => index.php
            [2] => nav.php
            [3] => release_management
            [4] => service_desk
            [5] => service_requests
            [6] => supplier_management
        )

)

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> 

Hi, 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?

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!

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 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? 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");   
   }
}
?>

 

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,

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.


                                       

hello everybody.. i need help.. i want to make a dropdown menu with php coding that will show the result from my database.. can anyone help me to generate the coding? this is an example form:   <form action="" method="get"><select name="">
  <option value="select">select</option>
  <option value="1">1 </option>
  <option value="2">2</option>
  <option value="3">3</option>
 
</select>
  <label>
  <input type="submit" name="button" id="button" value="Submit" />
  </label>
</form>   if user select 1, the php code will show a list of name by their identification card, if user select 2 the php code will show a list of name by country.

I have a sql query similar to below in  a .php file.       $sql = "select id,organisation,price from table where category = '$category';     These are the four unique values of the category column for your reference.   "A-t1" "B-t1" "C-t1" "D-t1"       Now I want to create a dropdown list or listbox in .php file with option to select multiple values. If customer selects multiple values, it should fetch query for the selected categories.   E.g. dropdown list should be similar to below : "A-t1" "B-t1" "C-t1" "D-t1"   If an user selects "A-t1"  and "C-t1", it should give output/query for the selected categories.    

Hi,

I am in the procress of creating discussion system however I am a bit puzzled about the best way to go about it.

I am starting the discussion by creating an ID number and then match the answer to the initial ID number.  However, I dont know whether if is best to put the responses into a different database.

I'm a bit puzzled how ID matching systems works.

Lets say:

Question 1 =    ID1
Question 2 =    ID2
Question 3 =    ID3
Question 1 Answer 1    =   ID4   (How is this matched to ID1)
Question 2 Answer 1    =   ID5   (How is this matched to ID2)

is this based on preg_match?