PHP - Select Dropdown List

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!

I'm trying to pull up an identical form after submission by using its "id" and ECHO for the form lines.

Question: What is the correct approach for ECHOing an HTML dropdown that is written as:
<select name='chixcutlet' value='' >
        <option value='0.00' selected>  --- </option>
        <option value='1.00'>  1  </option>
        <option value='2.00'>  2  </option>
        <option value='3.00'>  3  </option>
</select>

 I want it to get the info from the database so that the option that is saved in the database will be new default when the page is loaded.

If I don't change it to the previous info it will update the database with the default option rather then the actual option.

This topic has been moved to Ajax Help.

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

Hi i'm new to php.

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

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>

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);

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.

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++;

}


?>

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.

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!

Hi,

I am doing an EDIT user page and would like to check records with a dropdown and then set as selected, please help!


<select name="BrokerID" class="small-input">

<option value="Please select an option">Please select a Broker</option> 

<?php 

while($row2 = mysql_fetch_array($broker))

{

echo '<option name="BrokerID" value="'.$row2['BrokerID'].'">'.$row2['BrokerName'].'</option>';

}

?>

</select>

Hi,

I have a search form where users can search by age and country. Users can also save their search so they can go back and do it again but the problem is when they reload their saved search, I need it to select the country that has been saved in the drop down. So if someone searched the United Kingdom, I need it to show United Kingdom in the select drop down instead of (Select Country). Is there any easy and quick way around this?

Many Thanks

I'm trying to figure out why the options aren't appearing inside the select dropdown. Any ideas why?

Code: [Select]
echo "<label for=" . $row2['fullName'] . ">" . $row2['fullName'] . "</label>";
                       echo "<select name=" . $row2['fullName'] . " id=" . $row2['fullName'] . " class=dropdown title=" . $row2['fullName'] . " />";
                       if ($styleID == 1 || $styleID == 2 || $styleID == 6) {
                         $charactersQuery = "
                                SELECT
                                    characters.ID,
                                    characters.characterName
                                FROM
                                    characters
                                WHERE
                                    characters.styleID = 3
                                ORDER BY
                                    characters.characterName";
                         $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query
                         while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) {
                            print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r";
                         }   
                       } else {
                         $charactersQuery = "
                                SELECT
                                    characters.ID,
                                    characters.characterName
                                FROM
                                    characters
                                WHERE
                                    characters.styleID IN (1,2,6)
                                ORDER BY
                                    characters.characterName";
                         $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query
                         while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) {
                            print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r";
                         }
                       }
                       echo "</select>";
                   }

Hi,

I've got a basic sign up form but I want a drop down list which will list different catergories that relate to different tables which when selected will input the sign up information into that table which was selected from the catergory drop down.

This is the signup form

<html><head><title>Birthdays Insert Form</title>
<style type="text/css">
td {font-family: tahoma, arial, verdana; font-size: 10pt }
</style>
</head>
<body>
<table width="300" cellpadding="5" cellspacing="0" border="2">
<tr align="center" valign="top">
<td align="left" colspan="1" rowspan="1" bgcolor="64b1ff">
<h3>Insert Record</h3>
<form method="POST" action="test.php">
<?
print "Enter Company Name: <input type=text name=company_name size=30><br>";
print "Enter Contact Name: <input type=text name=contact_name size=30><br>";
print "Enter Telephone: <input type=text name=telephone size=20><br>";
print "Enter Fax: <input type=text name=fax size=30><br>";
print "Enter Email: <input type=text name=email size=30><br>";
print "Enter Address: <input type=text name=address1 size=20><br>";
print "Enter Address: <input type=text name=address2 size=30><br>";
print "Enter Postcode: <input type=text name=postcode size=30><br>";
print "Enter Town / City: <input type=text name=town_city size=20><br>";
print "Enter Website: <input type=text name=website size=30><br>";
print "Enter Company Type: <select name='table'> <option>stationary</option><option>reception</option></select><br>";
print "<br>";
print "<input type=submit value=Submit><input type=reset>";
?>
</form>
</td></tr></table>
</body>
</html>

This is the part which I can't figure out and is probably totally wrong! Im trying to use this script to sort the drop down list to then run the correct script to insert the form data.

<html xmlns="http://www.w3.org/1999/xhtml">

<head>
  <title>Hello!</title>
</head>

<body>

<?php
if($_POST['table']=='stationary' 'birthdays_insert_record.php')

else

if($_POST['table']=='reception' 'insert_reception.php')
?>

</body>

</html>

This is the script which works! that inserts the form data into a specific table

<html><head><title>Birthdays Insert Record</title></head>
<body>
<?
/* Change db and connect values if using online */
$company_name=$_POST['company_name'];
$contact_name=$_POST['contact_name'];
$telephone=$_POST['telephone'];
$fax=$_POST['fax'];
$email=$_POST['email'];
$address1=$_POST['address1'];
$address2=$_POST['address2'];
$postcode=$_POST['postcode'];
$town_city=$_POST['town_city'];
$website=$_POST['website'];
$db="myflawlesswedding";
$link = mysql_connect('localhost', 'root' , '');
if (! $link)
die(mysql_error());
mysql_select_db($db , $link) or die("Select Error: ".mysql_error());
$result=mysql_query("INSERT INTO reception (company_name, contact_name, telephone, fax, email, address1, address2, postcode, town_city, website) VALUES (
'$company_name',
'$contact_name',
'$telephone',
'$fax',
'$email',
'$address1',
'$address2',
'$postcode',
'$town_city',
'$website')") or die("Insert Error: ".mysql_error());
mysql_close($link);
print "Record added";
?>
<form method="POST" action="birthdays_insert_form.php">
<input type="submit" value="Insert Another Record">
</form>
<br>

<form method="POST" action="birthdays_dbase_interface.php">
<input type="submit" value="Dbase Interface">
</form>



</body>
</html>

I hope somebody can help me out here! or can point me in a better way to sort this problem!

Thanks for any advice!