Dynamic dropdown menu!

Rafi2012 · April 12, 2012

I want make the following,

(I have already a database with three tables (Countries, Timeline and Category)).

1: list of countries (drop down menu 1), Timeline of the countries history (drop down 2) and Category (drop down 3).

2: The selected values of the drop down menus must show take the information from the database.

Can any one help me with the coding?

Muddy_Funster · April 12, 2012

let's see what you've got and where it's going wrong

Rafi2012 · April 12, 2012

The following is the code for the application i am trying to develop, the selected values from the three drop down menus doesn't show the content from the database!

<html>

<head>

<title> Interactive AGE+ </title>

</head>

<body>

<p> Interactive AGE+ </p>

</div>

<?php

include('connectland.php');

?>

</div>

<ul>

<li><a href='#'>Aardrijkskunde</a></li>

<li><a href='#'>Economie</a></li>

<li><a href='#'>Techniek</a></li>

</ul>

</div>

<h1 class='tekst'>Land selecteren</h1><br />

<?php

$query = "SELECT landID ,landnaam FROM landen";

$result = mysql_query ($query);

$query2 = "SELECT PerID ,Periode FROM periode";

$result2 = mysql_query ($query2);

$query3 = "SELECT CatID ,categorie FROM categorien";

$result3 = mysql_query ($query3);

echo '<form action="members.php" method="POST">';

echo '<select name="landname" id="landnaam">';

echo '<optgroup label="Landen">';

while($row=mysql_fetch_assoc($result)){ //Array or records stored in $nt

$countryname = $row['landnaam'];

echo '<option value='. $countryname .'> '. $countryname .' </option>';

}

echo '</optgroup>';

echo '</select></td>';

echo '<br />';

echo '<select name="pers" id="perioden">';

echo '<optgroup label="Periodes">';

while($row2=mysql_fetch_assoc($result2)){ //Array or records stored in $nt

$periodenaam = $row2['Periode'];

echo '<option value='. $periodenaam .'> '. $periodenaam .' </option>';

}

echo '</optgroup>';

echo '</select></td>';

echo '<br />';

echo '<select name="cats" id="categorien">';

echo '<optgroup label="Categorie">';

while($row3=mysql_fetch_assoc($result3)){ //Array or records stored in $nt

$categorienaam = $row3['categorie'];

echo '<option value='. $categorienaam .'> '. $categorienaam .' </option>';

}

echo '</optgroup>';

echo '</select></td>';

echo '<br /><br />';

echo '<input class="Knoppen" name="verder" type="submit" align="center" value="Ga verder"/>';

echo '</form>';

?>

</div>

<?php

if (isset($_POST['verder'])){

$land = $_POST['landname'];

echo "<h1>$land</h1><br />";

$query = "SELECT informatie FROM landen WHERE landnaam = '$land'";

$result = mysql_query ($query);

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

$info = $row['informatie'];

echo nl2br($info);

}

?>

</div>

</div>

</body>

</html>

Muddy_Funster · April 12, 2012

Try using this, I have only sampled the first select group, but it should be easy enough to work out from that

<?php

function buildMenu($value){//going to use this function to build the option list
if(!empty($value)){
$output = "<option value =\"$value\">$value</option>\n";
return $output;
}
}

$query = "SELECT landen.landnaam, periode.Periode, categorien.categorie FROM landen, periode, categorien";
$reusult = mysql_query($query) or die (mysql_error())."<br>------------------<br>$query");
while ($row = mysql_fetch_assoc($result)){
$landMenu[] = buildMenu($row['landen']);
$periodMenu[] = buildMenu($row['periode']);
$catagorieMenu[] = buildMenu($row['categorie']);
}
//.....
echo "<select name=\"landname\" id=\"landnaam\">";
echo "<optgroup label=\"Landen\">";
foreach($landMenu as $key){
echo $landMenu[$key];
}
echo "</optgroup></select>";
//.......

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Dynamic dropdown menu!

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