PHP - How To Create A Dropdown List In A Header Menu In Html And Css

I am very new new php (wrote my first PHP script 5 Days ago) and am trying to give myself a crash course but I have hit a pit-stop which is killing me a little!

I hope that title makes sense.... Basically I created PHP script to take data from a database and display in, I then wrote some code to use a drop down menu to order that data. That all worked ok until I tried to utilise some pagination. I can make the pagination work, and I can make the ordering work, but not at the same time! At the moment the code that I have will allow me to order the list and almost paginate it. There are 40 results and I want to display 10 at a time. When not using the ordering code I can paginate it perfectly but when I try to intergrate the two bits of code it will only display the first 10 results and not give me an option to go to the next page to see the rest!

The code:

if (!isset($_GET['start'])) {

$_GET['start'] = 0; }

$per_page = 10; $start = $_GET['start'];
if (!$start)
   $start = 0;
 
 
 
$sort = @$_POST['order'];   
if (!empty($sort)) {   
$get = mysql_query("SELECT bookname, bookauthor, bookpub, bookisbn
            FROM booktable
            ORDER BY ".mysql_real_escape_string($_POST['order'])." ASC
            LIMIT $start, $per_page");
}
 else { 
$get = mysql_query("SELECT bookname, bookauthor, bookpub, bookisbn 
                FROM booktable 
                ORDER BY bookname ASC
                LIMIT $start, $per_page"); 
}
 
$record_count = mysql_num_rows($get);
 
 
?>
<?php
if (isset($_GET['showerror']))
$errorcode = $_GET['showerror'];
else
$errorcode = 0;
?>

 wont include all the html rubbish and the ordering menu!
<div id="mid">
 
<?php
echo "<table>";
echo "<tr>";
echo "<th>";
 
echo "</th>";
echo "<th>";
echo "Book Title";
echo "</th>";
echo "<th>";
echo "Book Author";
echo "</th>";
echo "<th>";
echo "Book Publisher";
echo "</th>";
echo "<th>";
echo "Book ISBN";
echo "</th>";
echo "<th>";
 
echo "</th>";
echo "</tr>";
while ($row = mysql_fetch_assoc($get))
{
 // get data
 $bookname = $row['bookname'];
 $bookauthor = $row['bookauthor'];
 $bookpub = $row['bookpub'];
 $bookisbn = $row['bookisbn'];
 
 
 
    echo "<tr>";
    echo "<td>";
    echo "<a href='addtolist.php?bookname=".$bookname."&bookauthor=".$bookauthor."&bookpub=".$bookpub."&bookisbn=".$bookisbn."'>Add to basket</a>";
 
    echo "</td>";
 
    echo "<td>";
    echo $bookname;
    echo "</td>";
 
    echo "<td>";
    echo $bookauthor;
    echo "</td>";
 
    echo "<td>";
    echo $bookpub;
    echo "</td>";
 
    echo "<td>";
    echo $bookisbn;
    echo "</td>";
 
    echo "</tr>";
 
 
}
 
echo "</table>";
 
$prev = $start - $per_page;
$next = $start + $per_page;
 
if (!($start<=0))
       echo "<a href='products.php?start=$prev'>Prev</a> ";
 
       //set variable for first page number
$i=1;
 
//show page numbers
for ($x = 0; $x < $record_count; $x = $x + $per_page)
{
 if ($start != $x)
    echo "<a class='pagin' href='products.php?start=$x'> $i </a>";
 else
    echo "<a class='pagin' href='products.php?start=$x'><b> $i </b></a>";
 $i++;
}
 
//show next button
if (!($start >= $record_count - $per_page))
       echo "<a class='pagin' href='products.php?start=$next'> Next </a>";
?>


Thank you so much for reading!

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.

I am querying a database and trying to display the names of categories in a drop-down menu.  My problem is that the names are not visible.

I have the following code:

   $result = mysql_query('SELECT name FROM category') or die(mysql_error());
   echo('Choose a category');
   echo('<select name="category">');
   echo('<option value="general" >general</option>');
   while($row = mysql_fetch_array($result)){
      echo('<option style="color:black;" value = "'.$row['name'].'">'.$row['name'].'</option>');
   }
   echo('</select><br /><br />');

The query is good and executing the query retrieves the expected number of results.  My dropdown box is the proper length and width (showing that it is trying to print the names); however, nothing is displayed.  Any ideas?

Hi all, here's my code:

Code: [Select]
<?php
foreach ($_SESSION['topping'] as $value) {
    echo "<tr><td width='30%'>Topping</td><td width='50%'>$value</td><td width='20%'><select name='notopping'>";
foreach ($_SESSION['cupcake'] as $number) { '<option name="notoppings[]" value="'.$number.'">".$number."</option>'; }
echo "</select></td></tr>";
}
?>

$_SESSION['cupcake'] is a value from either 6, 12, 24 or 36.  What I want to do is put them into a drop down box (second foreach) as the value and the displayed value - counting up from 1 (so 1,2,3,4,5,6 or up to 12,24 etc).

Also by creating this as an array, does this mean than for each topping (say Vanilla and Chocolate) the value dynamically created can be used on the next page by using $_POST['notoppings'] to display each type (two different numbers - one for Vanilla and one for Chocolate).  Does that make sense?

Thanks!

Jason

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!

This topic has been moved to Ajax Help.

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

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'm new to php.

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

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++;

}


?>

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.

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'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);

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 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>