PHP - Getting All Dropdown List Values Using Php

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.

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 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);

This topic has been moved to Ajax Help.

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

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++;

}


?>

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>

I have a sql query similar to below in  a .php file.       $sql = "select id,organisation,price from table where category = '$category';     These are the four unique values of the category column for your reference.   "A-t1" "B-t1" "C-t1" "D-t1"       Now I want to create a dropdown list or listbox in .php file with option to select multiple values. If customer selects multiple values, it should fetch query for the selected categories.   E.g. dropdown list should be similar to below : "A-t1" "B-t1" "C-t1" "D-t1"   If an user selects "A-t1"  and "C-t1", it should give output/query for the selected categories.    

i have the following code:
<td width='100px'>Suppliers
                <select name="supplier">
                    <?php
                       $catcher_id = $service->getCatcherId();
                       $supplier_names = LpmAdnetworkPeer::getByName($catcher_id);
                       foreach($supplier_names as $row)
                       {
                           ?>
                           <option><?php echo $row->getName();
                           ?></option>
                    <?php     
                       }
                    ?>
                </select>
                </td>   

then on the same form i have a submit button that takes me to the next form..the problem now is how can i access the ID of the item seleted in the dropdown on the NEXT form please?
i can get the name from the list by $_POST['supplier'] on the next form
thanks

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!

how come the lines:
PHP Code:
                 echo $form['catcher_id'];
                 
                  $catcher_names =  $form['catcher_id']                   
                  foreach($catcher_names as $val)
                  {
                      echo "testing loop";
                  }
produce nothing?? its not doing the foreach.
$form['catcher_id']; is a dropdown list containing catcher names
please help?
thanks

Hello everyone

I have two dropdown lists. First one you can see the tables of my db and the second one the files inside a folder. The problem is that i can't send my selections from the dropdown list to an other file.
My project is to choose a table, then a CSV file and then load the file in db.

index.php:
Code: [Select]
<?php //EMFANISH PINAKWN
include("../forma/dbCon.php");

echo "<form action='load_table_test.php' method='POST'>";
$sql="SHOW TABLES FROM misthodosia";
$result=mysql_query($sql);

if (!$result) {
echo "&#933;&#960;&#940;&#961;&#967;&#949;&#953; &#960;&#961;&#972;&#946;&#955;&#951;&#956;&#945; &#956;&#949; &#964;&#951;&#957; &#946;&#940;&#963;&#951;\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}

echo"<select>";
while ($row = mysql_fetch_row($result)) {
echo "<option name='{row[0]}' value='{row[0]}'>{$row[0]}</option>";
}
echo"</select>";
mysql_free_result($result);

//==========================EMFANISH ARXEIWN==========================

function populateDropdown(){
$path = glob("C:\\xampp\\htdocs\\dbadmin\\upload_files\\*"); //the double \\ is not a typo you

//have to escape it
//if the path cannot be found
if(!($path)){
echo("&#916;&#949;&#957; &#965;&#960;&#940;&#961;&#967;&#949;&#953; &#959; &#966;&#940;&#954;&#949;&#955;&#959;&#962;");
return;
}//if

//the path is valid
else{
foreach($path as $k){
$i = basename($k);

echo "<option name='$i'>".$i."</option>";

}//foreach
}//else
}//populateDropdown






echo"<select>";
populateDropdown();
echo"</select> ";


echo"<input type='submit' name='submit' value='&#934;&#972;&#961;&#964;&#969;&#963;&#949;1111111' />";
echo "</form>";


$hello="hello";

?>

load_table_test.php:
Code: [Select]
<?php
include("../forma/dbCon.php");



echo $_POST['hello'];

echo $_POST['{row[0]}'];
echo $i;


//$pinakas=$_POST['pinakas']
//$arxeio=$_POST['arxeio']

//$pinakas=$_POST['row']
//$arxeio=$_POST['diadromi']


//mysql_query("LOAD DATA LOCAL INFILE "$arxeio" INTO TABLE $pinakas
//FIELDS
//TERMINATED BY ','
//ENCLOSED BY '\"'
//LINES TERMINATED BY '\n')



echo "$arxeio";
echo "$pinakas";



    
echo "<br/><br/><br/><br/><br/><a href='index.php'>&#917;&#960;&#953;&#963;&#964;&#961;&#959;&#966;&#942; &#963;&#964;&#951;&#957; &#960;&#961;&#959;&#951;&#947;&#959;&#973;&#956;&#949;&#957;&#951; &#963;&#949;&#955;&#943;&#948;&#945;.</a>";                                     

?>

errors:
Code: [Select]
Notice: Undefined index: hello in C:\xampp\htdocs\dbadmin\load_table_test.php on line 6

Notice: Undefined index: {row[0]} in C:\xampp\htdocs\dbadmin\load_table_test.php on line 8

Notice: Undefined variable: i in C:\xampp\htdocs\dbadmin\load_table_test.php on line 9

Notice: Undefined variable: arxeio in C:\xampp\htdocs\dbadmin\load_table_test.php on line 27

Notice: Undefined variable: pinakas in C:\xampp\htdocs\dbadmin\load_table_test.php on line 28

i skipped solving this the other day with an easier way but now i am stuck with this stumble again in another area and i have no way out....
so here it goes,
i have an ajax dropdown box...i need to get the value that is selected by the user when it is clicked and then pass this value to a new pop window by appending to its url....any suggestions?

hi all,i have a php form that i would like to populate from a drop down list,my problem is that i cant place the drop down list at the correct place.what i want is that where the franchisee is the input should be the drop down,please help....this is the code that i am using

Code: [Select]
     <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
<p>Add a Bus in the Bus table.</p>
<p>
  Fleet Number:<br />
  <input type="text" name="fleet_number" size="6" maxlength="10" value="" />
  </p>
  <p>
  Registration Number:<br />
  <input type="text" name="registration_number"  size="10" maxlength="20" value="" />
  </p>
  <p>
  Model:<br />
  <input type="text" name="model"  size="10" max length="15" value="" />
  </p>
  <p>
  Franchisee :<br />
  <input type="select" name="franchisee_id" size="10" max length="15" value="" />
 
  <select name=dropdown_list>

    <?php   
  
    while($row = oci_fetch_array($stid))  
  
    {  
             
echo "<option value='".$row['FRANCHISEE_NAME']."'>";
echo $row['FRANCHISEE_NAME'];
echo "</option> ";
  
    }   
    
  ?> 
 
  </select>
</p> 
  <p>
  <input type="Submit" name="submit" value="Submit !" />
  </p>
</form>