jofftan Posted September 4, 2006 Share Posted September 4, 2006 Hi, I'm looking for a way to choose data content from a dropdown menu, linked to a MySQL DB, and echo it on that same page.The problem is when you get to that page for the first time I need to display the first entry of the data by default.Now I have it but with two php files, I want this on the same page.The way it is right now is Page 1 ;[code] <?php $c = mysql_connect ($host,$usager,$Mpasse); mysql_select_db ('db', $c); if (!$c) { die('Impossible de se connecter : ' . error_reporting(E_ALL)); } $req = "SELECT * FROM Team"; $resultat = mysql_query($req); echo '<ul id="subnavlist">'; while ($record = mysql_fetch_assoc($resultat)) { echo '<li id="subactive"><a href="corpoTeam.php?id_team='.$record["id_team"].'" id="current">'.$record["Titre"].'</a></li>'; } echo "</li>"; ?>[/code]Page 2;[code] <?php $_POST['id_team']; include 'connexion.php'; $c = mysql_connect ($host,$usager,$Mpasse); mysql_select_db ('db', $c); if (!$c) { die('Impossible de se connecter : ' . mysql_error()); } $req = "SELECT * FROM Team"; $resultat = mysql_query($req); echo '<ul id="subnavlist">'; while ($record = mysql_fetch_assoc($resultat)) { echo '<li id="subactive"><a href="corpoTeam.php?id_team='.$record["id_team"].'" id="current">'.$record["Titre"].'</a></li>'; } echo "</li>"; ?> </li> </ul> </div> <div> <?php ini_set('display_errors', 1); error_reporting(E_ALL); $req = "SELECT * FROM Team WHERE id_team = ".$id_team;// $resultat = mysql_query ("SELECT Titre, description, name FROM Team WHERE id_team = '".urldecode($_GET['id_team'])."'"); $resultat = mysql_query($req); $row=mysql_fetch_array($resultat); $name = $row['name']; $Titre = $row['Titre']; $description = nl2br($row['description']); echo "<br>"; echo "<br>"; echo '<h4>'; echo $name; echo '<br>'; echo $Titre; echo '</h4>'; echo '<p>'; echo $description; echo '</p>'; ?>[/code]Thank you Quote Link to comment Share on other sites More sharing options...
marker5a Posted September 4, 2006 Share Posted September 4, 2006 It sounds like to me you are talking about implementing Javascript... but your question is not quite to clear. COuld explain in clearer terms what you want...Thanks Quote Link to comment Share on other sites More sharing options...
jofftan Posted September 4, 2006 Author Share Posted September 4, 2006 Here's the link[url=http://www.virochempharma.com/corpo.php]http://www.virochempharma.com/corpo.php[/url]When you get to this page I'm echoing the first entry of the DB.If you select another entry from the dropdown it call a second page.I want all that operation on a single page.I don't know if it's possible ? Quote Link to comment Share on other sites More sharing options...
marker5a Posted September 4, 2006 Share Posted September 4, 2006 Yea, I understand what you are saying now... instead of these links on the left changing the page (corpo_mission.php, corpo_history.php, etc...) you want to only change the text.This cannot all be done with PHP. You will need to change the hyperlinks to call a javscript function that changes the text in that box. However, within this Javascript function you create, you will need to use PHP to pull the data from the DB.I cannot help you easily with the javascript aspect... I would post this in the Javascript help forum. Quote Link to comment Share on other sites More sharing options...
jofftan Posted September 4, 2006 Author Share Posted September 4, 2006 No I'm sorry I wasn't clear enough.The links are in the dropdown of team members in the middle of the page. Quote Link to comment Share on other sites More sharing options...
.josh Posted September 4, 2006 Share Posted September 4, 2006 if you are asking to have a dropdown populated by entries in your db, and info changing on your page, like say, info in another dropdown, when you select a different dropdown value, before you have to go and click submit, then you are going to need a javascript method called AJAX. go to ajaxfreaks.com or google it. Quote Link to comment Share on other sites More sharing options...
jofftan Posted September 4, 2006 Author Share Posted September 4, 2006 Not exactly.Ok now it's calling the same page within the link so it work I don't know if it's secure but it work.[code] while ($record = mysql_fetch_assoc($resultat)) { echo '<li id="subactive"><a href="corpo.php?id_team='.$record["id_team"].'" id="current">'.$record["Titre"].'</a></li>'; }[/code]But now I got this error to fix[quote]Notice: Undefined variable: id_team in /var/www/vhosts/virochempharma.com/httpdocs/corpo.php on line 106Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/vhosts/virochempharma.com/httpdocs/corpo.php on line 109[/quote] Quote Link to comment Share on other sites More sharing options...
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