PHP - How To Create Dependent Dropdown List?

Hi all,
firstly apologies as this is a cross post from another forum and we have hit a block.. I am hoping that opening this up to another set of gurus we can get a resolution.

What I am trying to achieve is this...
I have 2 tables Main and FinancialYear. Main holds all data which I use a form to post the data to it..(all works fine). I use this code to create a drop down in the insert.php form. again this works.
Code: [Select]
<tr><td>Financial Year: xxxx/xxxx</td><td>
<!-- pulls the data from the table variable to populate the dropdown menu -->
   <?php
      $database = 'Projects_Main';
      $fintable = 'FinancialYear';
         if (!mysql_connect($db_host, $db_user, $db_pwd))
               die("Can't connect to database 'cos somethin' is wrong");
         if (!mysql_select_db($database))
             die("Can't select database");
      $result = mysql_query("SELECT FinancialYear_id, FinancialYear FROM {$fintable} order by FinancialYear");
      $options="";
         while ($row=mysql_fetch_array($result)) 
            {
               //$id=$row["FinancialYear_id"];
               $thing=$row["FinancialYear"];
               $options.="<OPTION VALUE=\"$thing\">".$thing.'</option>';
            }
   ?>
      <SELECT NAME="FinancialYear">
      <OPTION VALUE=0>Choose</OPTION>
      <?=$options?>
      </SELECT>
   </td></tr>

What I have done is built another form which list all records in the database and creates an update url for every record that passes the field Project_id where i use $_get to retrieve the Project_id to retrieve the relevant data into the update.php form.

I am able to populate the form with all the correct information BUT I am looking to introduce some dropdowns to aid updating the data and provide consistency to the data. .

Code: [Select]
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db('Projects_Main')or die("cannot select DB");

// get value of id that sent from address bar
$Project_id=$_GET['Project_id'];
//define vars
$FinancialYear=$_POST['FinancialYear'];
// other vars defined here also.. about 30

// Retrieve data from database
$sql="SELECT * FROM Main WHERE Project_id='$Project_id'";
$result=mysql_query($sql);

$rows=mysql_fetch_array($result);
?>
//update_record_ac.php posts the data to the dbase.
<form name="form1" method="post" action="update_record_ac.php">
<center>
<table>
<tr><td><b>Store Details<b></td></tr>
<tr><td>Financial Year:</td><td>
// takes the data from $rows and present to form
<input name="FinancialYear" type="text" id="FinancialYear" value="<?php echo $rows['FinancialYear']; ?>">

// this is where I need to create the drop down.. see my other comments in the post.....

</td></tr>


the financialYear table consists if the following;
financialyear_id - pri, auto inc. ---- data format is 2010/2011, 2011/2012....
financialyear

the main table contains 30 fields .. won't list em all...
Project_id - pri, auto inc
financialyear

I need the drop down to pull the data from the financialyear table and then to present or focus on the currently stored data...
so if the store value in the table Main is 2010/2011 if Ii was to select the update url in the list_record.php it will pull all the relevant data into update_record.php form. the financialyear field in the form should be a dropdown with all the financial years listed but the 2010/2011 is selected or focused. I still need to be able to change the entry and post this back to the table Main.....

So the dropdown contains the list of years from the financialyear table but when the record is pulled from table main the year that is stored in table Main should be highlighted in the dropdown and I should be able to select a new record and post back to the table Main..


any thoughts...

please don't slate for the cross post, I haven't sanatised the data at any stage. I know i'm open to injection attacks. and yes my code is a little dirty... all these will be rectified as i finalise the process and ensure the consept works.

Thanks for taking the time to read and hopefully you are able to understand the requirement and are able to assist.

thanks
Balgrath

Hi,

I am trying to call the data from Mysql but I am getting an empty drop down list, this is the code:


mysql:

Code: [Select]
create table years
(  yearID integer auto_increment,
   year varchar(30),
   primary key (yearID)
);

insert into years (yearID, year) values ('1', '2007-2008');
insert into years (yearID, year) values ('2', '2008-2009');
insert into years (yearID, year) values ('3', '2009-2010');
insert into years (yearID, year) values ('4', '2010-2011');
insert into years (yearID, year) values ('5', '2011-2012');
insert into years (yearID, year) values ('6', '2012-2013');

PHP:

Code: [Select]
<?php require_once('../Connections/connection.php'); ?>
<?php
 
$result = @mysql_query( "select yearID, year, from sss.years");

print "<p>Select a year:\n";
print "<select name=\"yearID\">\n";
while ($row = mysql_fetch_assoc($result)){
$yearID = $row[ 'yearID' ];
$year = $row[ 'year' ];
print "<option value=$yearID>$year\n";
}
print "</select>\n";
print "</p>\n";
?>

Thank you!

Hi i'm new to php.

I want to get all the values of dropdown list on another page wether selected or not.

Hey,

I have the following coding:
Quote

<?
$dbuser="*******";
$dbpass="*******";
$dbname="virtuda_db";  //the name of the database
$chandle = mysql_connect("localhost", $dbuser, $dbpass)
    or die("Connection Failure to Database");
mysql_select_db($dbname, $chandle) or die ($dbname . " Database not found. " . $dbuser);

$mainsection="license";

$query1="select name from license";
$result = mysql_db_query($dbname, $query1) or die("Failed Query of " . $query1);  //do the query
while($thisrow=mysql_fetch_row($result))
{
  $i=0;
  while ($i < mysql_num_fields($result))
  {
    $field_name=mysql_fetch_field($result, $i);
    echo $thisrow[$i] . " ";  //Display all the fields on one line
    $i++;
  }
echo "<br>";  //put a break after each database entry
}
?>



How would I set up this so that instead of just "listing" them out on new lines, it would list the results into a drop down list? Thanks!

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=316599.0

i have been trying to get this code to get a list of usernames from a database and i have now got that to work but when i try and save it it saves all the usernames from the drop down list and not just the one i have selected

how can i get it to just use the one i have selected

Code: [Select]
<?php

include "connect.php"; //connection string
include("include/session.php");

print "<link rel='stylesheet' href='style.css' type='text/css'>";

print "<table class='maintables'>";

print "<tr class='headline'><td>Post a message</td></tr>";

print "<tr class='maintables'><td>";

// Write out our query.
$query = "SELECT username FROM users";
// Execute it, or return the error message if there's a problem.
$result = mysql_query($query) or die(mysql_error());

$dropdown = "<select name='username'>";
while($row = mysql_fetch_assoc($result)) {
$dropdown .= "\r\n<option value='{$row['username']}'>{$row['username']}</option>";
}
$dropdown .= "\r\n</select>";

if(isset($_POST['submit']))

{

   $name=$session->username;

   $yourpost=$_POST['yourpost'];

   $subject=$_POST['subject'];

$to=$dropdown;

   if(strlen($name)<1)

   {

      print "You did not type in a name."; //no name entered

   }

   else if(strlen($yourpost)<1)

   {

      print "You did not type in a post."; //no post entered

   }

   else if(strlen($subject)<1)

   {

      print "You did not enter a subject."; //no subject entered

   }

   else

   {

      $thedate=date("U"); //get unix timestamp

      $displaytime=date("F j, Y, g:i a");

      //we now strip HTML injections

      $subject=strip_tags($subject);

      $name=strip_tags($name);

      $yourpost=strip_tags($yourpost); 
$to=strip_tags($to);

      $insertpost="INSERT INTO forumtutorial_posts(author,title,post,showtime,realtime,lastposter,name) values('$name','$subject','$yourpost','$displaytime','$thedate','$name','$to')";

      mysql_query($insertpost) or die("Could not insert post"); //insert post

      print "Message posted, go back to <A href='forum.php'>Forum</a>.";

   }



}

else

{

   print "<form action='newtopic.php' method='post'>";

   print "Your name:<br>";

   print "$session->username<br>";

   print "User to send to:<br>";

print "$dropdown";

   print "Subject:<br>";

   print "<input type='text' name='subject' size='20'><br>";

   print "Your message:<br>";

   print "<textarea name='yourpost' rows='5' cols='40'></textarea><br>";

   print "<input type='submit' name='submit' value='submit'></form>";



}

print "</td></tr></table>";

?>

MOD EDIT: Changed PHP manual link [m] . . . [/m] tags to [code] . . . [/code] tags.

I have the following code currently:
Code: [Select]
<?php foreach ((array)$node->field_buy_at as $item) { ?>
<?php print $item['view'] ?>
<?php } ?>
I would like to make the list a drop down with a link so that when a user selects, he goes to a new page. I tried the following:
Code: [Select]
        <select name="select">
      <?php foreach ((array)$node->field_buy_at as $item) { ?>
<?php
            $url = $node->field_buy_at[0]['url'];
            $store = $item['view'];
         ?>
        <? echo "<option value='$url'>$store</option>";?>
<?php } ?>
  </select>
I'm pretty sure it's this "$url = $node->field_buy_at[0]['url'];" that I don't have correct.

Hey Guys,

I know it may seem pretty simple, but im having trouble populating a drop down list. Here is my code at the moment, but what it's doing is displaying the names all in one value, where it should be in separate select values. *Note that i have only done it to the first one.

See attachment. 'AntonMatt' are next to each other, they should be separate select values.

Code: [Select]
<?

$id = $_GET['id'];

$selectplayers="SELECT * FROM players WHERE club='$club' AND team='$team'";
$player=mysql_query($selectplayers);


?>

<table class='lineups' width="560" cellpadding="5">
  <tr>
    <td colspan="2">Starting Lineup</td>
    <td colspan="2">On the Bench</td>
  </tr>
  <tr>
    <td width="119">&nbsp;</td>
    <td width="160">&nbsp;</td>
    <td width="69">&nbsp;</td>
    <td width="160">&nbsp;</td>
  </tr>
  <tr>
    <td>Prop</td>
    <td><select name="secondary" style="width: 150px">
      <option value='' selected="selected"><? while($rowplayer = mysql_fetch_array($player)) { echo $rowplayer['fname']; } ?></option>
    </select></td>
    <td>16.</td>
    <td><select name="secondary16" style="width: 150px">
      <option value='' selected="selected">Secondary Position</option>
    </select></td>
  </tr>
  <tr>
    <td style="padding-top: 8px;">Hooker</td>
    <td style="padding-top: 8px;"><select name="secondary2" style="width: 150px">
      <option value='' selected="selected">Secondary Position</option>
    </select></td>
    <td style="padding-top: 8px;">17.</td>
    <td style="padding-top: 8px;"><select name="secondary17" style="width: 150px">
      <option value='' selected="selected">Secondary Position</option>
    </select></td>
  </tr>
</table>

    </div>

</div>

Alright, so I have an xml file differences.xml that is being parsed in XML. This is what the xml looks like:

<item  code="lM" name="dog">
<cost>5000</cost>
<Start>12/15/2010</Start>
<End>01/13/2011</End>
</item>
<item  code="lF" name="cat">
<cost>5000</cost>
<Start>04/15/2010</Start>
<End>04/23/2011</End>
</item>[/

I want to have the item names (dog, cat) show in a dropdown menu so that I can select these items for editing before storing in my mysql database.

This is the php code I have so far:
<?PHP

$xml = simplexml_load_file("differences.xml");

$object = $xml->xpath("//item");

echo '<SELECT name=object>';

foreach ($object['name'] as $key => $value)

{

echo '<OPTION value='.$value.'> '.$value;

}

echo '</select>';
?>

I do have a dropdown list but there are no values inside it (it is empty). Can anyone help me figure out why?

I do have this code that does work which lists the items in plaintext (not in a dropdown) so hopefull this will help us out:

<?PHP

$xml = simplexml_load_file("differences.xml");

$object = $xml->xpath("//item");

$count = count($object);

$i = 0;

while($i < $count)
{

echo '<h1>'.$object[$i]['name'].'</h1>';

$i++;

}


?>

Hi.

I am using this script to populate a dropdown list box from sql, it works but does anyone know how to sort the list in alphabetical order?


$sql="SELECT * FROM Fish WHERE ***** = '".$_GET['stocktype']."'"; 
$result=mysql_query($sql); 

$options=""; 

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

    $ID=$row["ID"]; 
    $Stock=$row["Commonn"]; 
    $Options.="<OPTION VALUE=\"$ID\">".$Stock; 
} 


<SELECT NAME='stock1'> 
<OPTION VALUE='$Options'>$Options</option>

</SELECT>

I'm trying to sort this dropdown box. It reads from a directory, and lists the file name in the dropdown box. Here's the tricky part...

the filename is listed differently in the dropdown than in the directory by using explode(). I want to sort it though since it's still being sorted by the directory listings...

For example:
Filename starts out as: 123_abc_567.pdf then gets listed as abc_123_567.pdf in the dropdown, but it's still getting sorted as if it were 123_abc_567.pdf

How can I do that?

Here's my code: // Define the full path to folder from root
    $path = "C:/Work_Orders/";

    // Open the folder
    $dir_handle = @opendir($path) or die("Unable to open $path");
    
echo "<form method=\"POST\" action='".$_SERVER['PHP_SELF']."' name='selectworkorder'><select name='ordernumber2'>";

    // Loop through the files
    while ($file = readdir($dir_handle)) {
    
            //Remove file extension
            $ext = strrchr($file, '.'); 
             if($ext !== false) 
             { 
                 $file = substr($file, 0, -strlen($ext)); 
             } 
    
    if($file == "." || $file == ".." || $file == "index.php" )
            
    continue;

//explode file name        
$changedordernumber = explode("_",$file);
//put in new order
$changedordernumber = $changedordernumber[1]."_".$changedordernumber[0]."_".$changedordernumber[2];
        $changedordernumber=trim($changedordernumber,"_");
        
//list options
echo "<option name='$file' value='$file'>$changedordernumber</option>\n";

    }
    echo "</select><input type='submit' value='Change' name='submit'/></form></div>";
    // Close
    closedir($dir_handle);

If I can get this fixed, I will have completed all but the admin login for this project - my first php/mysql project.

Here is what I need.  I have a list_records.php that list all the records in the table 'links' and the category each entry is in from the table 'categories'.

Here are my table structures.
Code: [Select]
-- Table structure for table `categories`
--

DROP TABLE IF EXISTS `categories`;
CREATE TABLE IF NOT EXISTS `categories` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `categories` varchar(37) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ;

-- --------------------------------------------------------

--
-- Table structure for table `links`
--

DROP TABLE IF EXISTS `links`;
CREATE TABLE IF NOT EXISTS `links` (
  `id` int(4) NOT NULL AUTO_INCREMENT,
  `catid` int(11) DEFAULT NULL,
  `name` varchar(255) NOT NULL DEFAULT '',
  `url` varchar(255) NOT NULL DEFAULT '',
  `content` varchar(255) NOT NULL DEFAULT '',
  PRIMARY KEY (`id`),
  KEY `catid` (`catid`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ;

On the update.php file, I have a form that lets me make changes to the record. 

Here is the codes for update.php

Code: [Select]
<? include "menu.php" ?>

<? include "db.php" ?>
<?php
$id=$_GET['id'];

$sql = "select * from links where id =$id";
$query = mysql_query($sql);
 
while ($row = mysql_fetch_array($query)){
 
    $id = $row['id'];
    $catid = $row['catid'];
    $name = $row['name'];
    $url = $row['url'];
    $content = $row['content'];
 
    //we will echo these into the proper fields
 
}
mysql_free_result($query);
?>
 
<table width="65%" align="center">
<tr><td align="left">

<form action="updated.php" method="post">

<input type="hidden" value="<?php echo $id; ?>" name="id"/>
 
<br>

 
<b>Website Name:</B><br> Change the name of the website listing.<br>
<input type="text" value="<?php echo $name; ?>" name="name"/>
 
<br>
 <br>
<b>URL:</b><br> Change the URL of the website listing.<br>
<input type="text" value="<?php echo $url; ?>" name="url"/>
 
<br>
<br>
 
<b>Description:</b><br> Change the description of the website listing.<br> Limit 255 characters.<br/>
<textarea name="content" cols="45" rows="4" wrap="soft"><?php echo($content);?></textarea>

<br>
 <?php
$result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); 
while ($row = mysql_fetch_array($result)) {
$id = $rows["id"]; 
$categories=$row["categories"];
$options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>';
};


?>


<SELECT NAME=catid>
<OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION>
</SELECT>

<?php
mysql_close();
?>

<div align="center">
<input type="submit" value="submit changes"/>
 </div>
</form>

<br>
</td></tr></table>


The part of he code I need help with is

Code: [Select]
<?php
$result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); 
while ($row = mysql_fetch_array($result)) {
$id = $rows["id"]; 
$categories=$row["categories"];
$options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>';
};


?>


<SELECT NAME=catid>
<OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION>
</SELECT>

I want it to default to the category that the entry is in.  If you look, you will see in the select portion that I I have Code: [Select]
<$ echo $catid; ?> which echos the proper category ID, but if I use Code: [Select]
<? echo $categories; ?> it echos Writing, which is the last category in the list.  Yet, the $options echo the catid and it corresponding category.  How can I get the default option to echo BOTH the catid and category name while also listing all the other categories so that the records can be moved to a new category is needed?

Any help will be appreciated. Thank you in advance.

Hi, I have a chunk of text I am pulling from a mysql db like below.
                           
Air conditioning in rooms
24 Hour Reception
Smoking rooms available

How can I write this into -

<ul>                           
<li>Air conditioning in rooms</li>
<li>24 Hour Reception</li>
<li>Smoking rooms available</li>
</ul>

Many thanks.

Hey everyone,

I am trying to set up a thing where i will have a list of states in a drop down list. After a state is select i would like a section of check boxes to appear with cities that a person can choose from.

Can anyone help me out with this i have looked on the internet and been trying to code this my self but with progress.

I can work with php or java script all i need is some help on getting started and maybe an example.

Thanks

E