PHP - Php/mysql Drop Down Error

Hi,
At present the PHP page reads the MySQL data and prints on screen and if the result if 'FREE' it shows a clickable image for the user to book. I am looking for something that allows the user to drag the result they fund to another location with in the table. I.e. user has box 1 filled with their name, they can drag their name to box 9 and it saves it, so they can go back on the it will be their.

If anyone knows where I can firn maybe a tutorial, or guide to get started with this, as I have looked and cannot see anything.

Cheers.
Stu

Hello Friends!.... here is the big idea. I am trying to make a form in which there will be 2 drop down lists which will be populating directly from MySQL DB ....
Actually i am developing a student management system as my first PHP project... here i want to have 2 drop down lists first is roll number and second is student name. i want that when some one select the roll number in the first drop down the student name against it is automatically populated in the next drop down.


please help me friends i am new to php and dont know soo much.! any help PLZZZZZZZZZZ

hey all
so I have this bit down:
Code: [Select]
$query="SELECT `2010 Region Code` AS codes FROM locations";
 
$results = mysql_query($query);


$options="";

$options = "<select location='codes'>";
while($nt=mysql_fetch_assoc($results))
{

$thing=$nt["codes"];

$options.="\r\n<option value ='{$nt['codes']}'>
{$nt['codes']}</option>";
}
$options .="\r\n</select>";
echo $options;


what I'm trying to do is grab the selection from the drop down and display it as a table (the sql query would be extended should we manage to figure this one out

I've tried
Code: [Select]
echo"<form name='LOCATIONS' action='".$_SERVER['PHP_SELF']."' target='iframe' method='post'>";
any ideas?

I've got a HTML drop down box as similar to this:

<select name="dropdown">
<option>Option 1</option>
<option>Option 2</option>
<option>Option 3</option>
</select>

If I run a mysql query and get a result of "Option 3", is there anyway using PHP to give Option 3 the selected value?

I am trying to insert a date into a mySQL table from html drop downs with php. Right now when I enter the date it goes into the database as 0000-00-00. Can anybody see why it might be doing this?

My html:
Code: [Select]
<label for='birthdate' >Birthdate (Optional):</label><br/>
    <select name='month' id='month' value='<?php echo $fgmembersite->SafeDisplay('month') ?>'>
<option value="01" selected="January">January</option>
<option value="02">February</option>
<option value="03">March</option>
        etc...
    </select>&nbsp;&nbsp;
    <select name='day' id='day' value='<?php echo $fgmembersite->SafeDisplay('day') ?>'>
<option value="01" selected="1">1</option>
<option value="02">2</option>
<option value="03">3</option>
etc...
    </select>&nbsp;&nbsp;
    <select name='year' id='year' value='<?php echo $fgmembersite->SafeDisplay('year') ?>'>
<option value="2010" selected="2010">2010</option>
<option value="2009">2009</option>
<option value="2008">2008</option>
etc...
    </select>

And my php to collect info:
Code: [Select]
$formvars['birthdate'] = $this->Sanitize($_POST['year'], $_POST['month'], $_POST['day']);
php where I make table:
Code: [Select]
"birthdate DATE NOT NULL ,".

php to insert into mysql:
Code: [Select]
$insert_query = 'insert into '.$this->tablename.'(
                name,
address,
birthdate,
sex,
program,
guide,
                email,
                username,
                password,
                confirmcode
                )
                values
                (
                "' . $this->SanitizeForSQL($formvars['name']) . '",
"' . $this->SanitizeForSQL($formvars['address']) . '",
"' . $this->SanitizeForSQL($formvars['birthdate']) . '",
"' . $this->SanitizeForSQL($formvars['sex']) . '",
"' . $this->SanitizeForSQL($formvars['program']) . '",
"' . $this->SanitizeForSQL($formvars['guide']) . '",
                "' . $this->SanitizeForSQL($formvars['email']) . '",
                "' . $this->SanitizeForSQL($formvars['username']) . '",
                "' . md5($formvars['password']) . '",
                "' . $confirmcode . '"
                )'; 

Thank you!

your MySQL server version:  5.1.36

Code: [Select]
SELECT * FROM game_weapons

Table: Picture Attached

the EXPLAIN output for your query, if applicable:
I wish to connect a drop down menu's selection of weapons with the correlated table information.

What do I want to happen:
Click on a drop down Menu and have a list of weapons, these weapons are associated with a set number in the database and passed onto the next screen.   (The larger number wins, this part I have figured out).

               <p>
               <select name="weapon2" style="font-size:20px;font-family:Arial;width:275px">
               <option value="power">Power</option>
               <option value="intelligence">Intelligence</option>
               <option value="speed">Speed</option>
               <option value="reserve">Reserve</option>
               </select>
               </p>


I do have the battle code figured out!  (This should be the last step).

Hi all,

I am currently learning PHP and have the homework to produce a function that can delete a row in a MySQL database table by clicking on an item in a drop-down menu in a web page.

The code I have produced up until now is this:

<!DOCTYPE HTML>
<html lang="de">
<head>
	<meta charset="utf-8" />
	<title>E3_Artikel_Löschen</title>

</head>
<body>
<form method = "GET">

<?php
	$anr="";
	
	try {
	$pdo = new PDO ('mysql:dbname=bestelldatenbank;host=localhost;charset=utf8', 'root', '');
	} catch (PDOException $error){
		die ($error->getMessage());
		}		
?>

	<div>
		<p>
			<label for="artikel">Artikel: </label>
				<select id="artikel" name="artikel">
					<?php
						$sqlSelect = "SELECT anr, name FROM artikel ORDER BY anr ASC";
						foreach ($pdo->query($sqlSelect) as $row) {
							echo "<option value=$row[0]>$row[0] | $row[1]</option>\n";
							$anr = $row[0];

						}
					?>
				</select>
			
			<input type = "submit" value = "Delete row" />
			
		</p>
	</div>

<?php
	function artLoeschen($anr) {
		echo "Function called $anr";
		if(isset($_GET[$anr])) {
			$anr = $_GET[$anr];
			$sqlDelete = $pdo->query("DELETE FROM artikel WHERE anr = :anr");
			if ($stmt = $pdo->prepare($sqlDelete)) {
				$stmt->bindParam(':anr', $anr);
				$stmt->execute();
			}
			echo "<h2><b>Artikel gelöscht!</b></h2>";
		}
	}
?>

</form>
</body>
</html>

So, I have observed the following when I run the script in a browser:

1.  The HTML works as expected and I get a drop-down list with the article number and description of each item in the affected table.

2.  I can click on an item in the list and it populates the top item in the drop-down list.

3.  When I click delete row, the selected item is not deleted.

4.  There are no error messages returned but the function is not executed (at least not as I would like to expect).

 

I have obviously missed something or made a mistake in my code. 

I would be very grateful for any help...this is driving me mad! :)

Regards,

Kevin

I am trying to update a mysql table called AvItems with the value 'Torso' in the Equip "section?" I have been through the forums and cannot see anything to match.
I dont mind if the page looses the onsubmit() and has a button instead.
Though I would like to update the database and link back to the same page: There is a display that shows the item that is currently equiped, I have put this in to show it works, or doesn't as the case may be.
Hope I got the code /code right this time.

many thanks in advance

Andy Curtis

Code: [Select]
create table Items(
ItemID integer unsigned auto_increment primary key,
ItemName varchar(20) not null,
Type varchar(10),
UsedOn varchar(10),
);

create table AvItems(
AvItemID integer unsigned auto_increment primary key,
AvID integer unsigned,
ItemID integer unsigned,
Equip varchar(8));


<?php

$username="root";
$password="MyPassword";
$database="MyDataBase";
$AvName = "AndyJCurtis";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$AvAccR = mysql_query( " SELECT AvID FROM AvAcc WHERE AvName = '$AvName' " );
$AvID = mysql_result($AvAccR, 0, 'AvID');
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
$Torso = mysql_query(" select ItemName from AvItems, Items where AvItems.itemID = Items.itemID AND AvItems.AvID = '$AvID' AND UsedON = 'Body' ");
$TorsoE = mysql_query(" select ItemName from AvItems, Items where AvItems.itemID = Items.itemID AND AvItems.AvID = '$AvID' And UsedON = 'Body' AND Equip = 'Body' ");
if(mysql_num_rows($TorsoE) != 0)
{
$TorsoItem = mysql_result($TorsoE ,0,"ItemName");
//mysql_close();
?>

<title></title>
<head></head>
<body>
<form action="http://localhost/CI/Equip2.php" method="post">

<table border=1>
<tbody>
<tr>
<td>Torso<BR>
<?PHP
echo "$TorsoItem <BR>";
?>
<select name="Torso" onchange="submit();" value =" Update">
<?PHP
while($TorsoRow = mysql_fetch_array($Torso))
{
echo "<option value=\"".$TorsoRow['ItemName']."\">".$TorsoRow['ItemName']."\n </option>";
}
?>
</select>
</td>
</tr>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
<?php
if($_POST['Torso'] == 'Update')
{
mysql_query("update AvItems set Equip = '' where Equip='Torso'") or die("cant update unequip"); 
mysql_query("update AvItems set Equip = 'Torso' where ItemID='{$_POST['ItemName']}'") or die("cant update equip"); 
}
?>
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

</tbody>
</table>
</.form>
</body>
</html>

I have created a drop down list and it does retrieve information from mysql but now I want to use what is been selected to retrieve information.

How Do I do this?

<?php 

MYSQL_CONNECT(localhost,'root','') OR DIE("Unable to connect to database"); 
@mysql_select_db(Examination) or die( "Unable to select database");

$query=("SELECT * FROM subject");

$result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() );
echo "<select name=myselect>";  

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

echo "<OPTION VALUE=".$row['Sub_ID'].">".$row['Sub_Name']."</OPTION>";
}

echo "</select>"; 


?>

Alright so I created a MySQL database that has 5 tables each named a brand of a dirt bike.

In each table their are 2 fields, one INDEX_ID and MODELS.

Under that for rows I have every model bike named.

A quick question before I get onto what I want to do:

Can data be added underneath the rows?

For example, users will be able to submit information about each bike model. If each bike model is a row, can there be a category past that row or do I need to make each field a model name and just have a ton of fields and have the rows be the information users submit.

To make it easier to understand I'll post the SQL code for the brand Honda:

CREATE TABLE `Honda` (
  `INDEX_ID` int(3) NOT NULL auto_increment,
  `MODELS` varchar(20) collate latin1_general_ci NOT NULL,
  PRIMARY KEY  (`INDEX_ID`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=23 ;

--
-- Dumping data for table `Honda`
--

INSERT INTO `Honda` VALUES(1, 'CR85');
INSERT INTO `Honda` VALUES(2, 'CR125');
INSERT INTO `Honda` VALUES(3, 'CR250');
INSERT INTO `Honda` VALUES(4, 'CRF100');
INSERT INTO `Honda` VALUES(5, 'CRF150');
INSERT INTO `Honda` VALUES(6, 'CRF230');
INSERT INTO `Honda` VALUES(7, 'CRF250X');
INSERT INTO `Honda` VALUES(8, 'CRF250R');
INSERT INTO `Honda` VALUES(9, 'CRF450X');
INSERT INTO `Honda` VALUES(10, 'CRF450R');
INSERT INTO `Honda` VALUES(11, 'CRF50');
INSERT INTO `Honda` VALUES(12, 'CRF70');
INSERT INTO `Honda` VALUES(13, 'CRF80');
INSERT INTO `Honda` VALUES(14, 'XR650');
INSERT INTO `Honda` VALUES(15, 'CR500');
INSERT INTO `Honda` VALUES(16, 'XR100');
INSERT INTO `Honda` VALUES(17, 'XR200');
INSERT INTO `Honda` VALUES(18, 'XR250');
INSERT INTO `Honda` VALUES(19, 'XR400');
INSERT INTO `Honda` VALUES(20, 'XR50');
INSERT INTO `Honda` VALUES(21, 'XR70');
INSERT INTO `Honda` VALUES(22, 'XR80');

Anyway I still need to figure out how to have this under a form that a user can use to select the bike they want to submit information about. A code like this perhaps?:


<?
$connection = mysql_connect("localhost","user","pass");
$fields = mysql_list_fields("dbname", "table", $connection);
$columns = mysql_num_fields($fields);
echo "<form action=page_to_post_to.php method=POST><select name=Field>";
for ($i = 0; $i < $columns; $i++) {
echo "<option value=$i>";
echo mysql_field_name($fields, $i);
}
echo "</select></form>";
?>


Thanks.

I am simply trying to insert a value generated from an array in a while loop, but it seems the value is not global and I can't pass it as I like. I did some research on this, but could not find an answer that solved my issue...

Here is my select box code which is working perfect:

<select name="city">
<?php
$sql = "SELECT id, city_name FROM cities ".
"ORDER BY city_name";
$results_set = (mysqli_query($cxn, $sql)) or die("Was not able to produce the result set!");
while($row = mysqli_fetch_array($results_set)) {
echo "<option value=$row[id]>$row[city_name]</option>";
}
?>
</select>

Are any variables defined in a while loop global to the while loop only?

Here is my SQL which you can see my $row[id] being passed thru field city_id... The value is being generated as supposed to based on value like: value="1", value="2" etc.. for the select options for each city name. So the values are there... But I CANNOT get that numerical id to pass to the database when submitting my form.

Any ideas for a workaround to get this value passing as normal?

if (isset($_POST['addPosting'])) {
      $query = "INSERT INTO Postings (id, city_id, title, description) VALUES
('','$row[id]','$_POST[title]','$_POST[description]')";

Hi all

I need to combine these two scripts:

Firstly, the following decides which out of the following list is selected based on its value in the mySQL table:

<select name="pack_choice">
<option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option>
<option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option>
<option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option>
</select>

Although this works OK, I need it also to show dynamic values like this:

select name="category">
<?php
$listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC");
while($categoryReturned=mysql_fetch_array($listCategories)) {
echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>";
}
?>
</select>

I'm not sure if this is possible?

Many thanks for your help.

Pete

I am working on a project that uses a drop down list to chose the category when inserting new data into the database. What I want to do now is make the drop down list default to the chosen category on the list records page and the update page. I have read several tutorials, but they all say that I have to list the options and then select the default. But since it is possible to add and remove categories, this approch won't work. I need the code to chose the correct category on the fly.

There are two tables, one that has the category ID and category name.  The second table has the data and the catid which is referenced to the category id in the first table.

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 ;


Then is the list records file, I have

Code: [Select]
<?php
include ("db.php");
include ("menu.php");


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



$id = $_GET['id'];
$query="SELECT * FROM links ORDER BY catid ASC";
$result=mysql_query($query);

?>
<table width="65%" align="center" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>
<table width="100%" border="1" cellspacing="0" cellpadding="3">
<tr>
<td colspan="7"><strong>List data from mysql </strong> </td>
</tr>

<tr>
<td align="center"><strong>Category ID</strong></td>
<td align="center"><strong>Category ID</strong></td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>URL</strong></td>
<td align="center"><strong>Content</strong></td>
<td align="center"><strong>Update</strong></td>
<td align="center"><strong>Delete</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td>
<SELECT NAME=catid>
<OPTION>Categories</OPTION>
<?php echo $options; ?>
</SELECT>
</td>
<td><? echo $rows['catid']; ?></td>
<td><? echo $rows['name']; ?></td>
<td><a href="<? echo $rows['url']; ?>"><? echo $rows['url']; ?></a></td>
<td><? echo $rows['content']; ?></td>
<td align="center"><a href="update.php?id=<? echo $rows['id']; ?>">update</a></td>
<td align="center"><a href="delete.php?id=<? echo $rows['id']; ?>">delete</a></td>

</tr>
<?php
}
?>
</table>
</td>
</tr>
</table>
<?php
mysql_close();
?>

So, how do I get this code
Code: [Select]
$result = mysql_query("SELECT categories FROM categories") or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
$categories=$row["categories"];
$options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>';
};

<SELECT NAME=catid>
<OPTION>Categories</OPTION>
<?php echo $options; ?>
</SELECT>

to give me an output that will be

something like if catid exactly matches categories.id echo categories.categorie  ???

so far everything I have done produces either a default category of the last category,  the catid (which is a number), all of the categories (logical since catid will always be = id, or nothing.  How do I get just the category name? 

I will keep reading and try to figure this out, but any help would be greatly appreciated.

Thanks in advance

Hi I have tried the mysql forum but have had no joy with an answer to my problem so wondered if php would be better.

I want my users to be able to select from 5 different drop down lists where they can chose any combination from 1 up to all 5,  I have attached the front end.  These lists are being populated from mysql tables.

Code for the drop down lists is as follows
Code: [Select]
  <form action="horse-events-devon.php?url_countyid=<?php echo  urlencode ($url_countyid ['url_countyid'])  ; ?>&go" method="POST">
    <table id="searchtable">
      <tr>
        <th>Find By Discipline</th>
        <th>Find By Venue</th>
        <th>Find By Championship</th>
        <th>Find By Organiser</th>
        <th>Equine Association</th>
        <th>Submit Your Selections</th>
        </tr>
      <tr>
        <td><select name="dis_id">
                                    <?php 
                                        $upcomingdis = upcomingdis($url_countyid);
                                        $upcoming_dis_bycounty = mysql_fetch_assoc ($upcomingdis);
                                    ?>
                                    <?php 
                                    do { 
       ?>   <option value="<?php echo $upcoming_dis_bycounty ['dis_id']; ?>" >
<?php echo $upcoming_dis_bycounty ['dis_description']; ?></option>
<?php 
                                    } while ($upcoming_dis_bycounty = mysql_fetch_assoc ($upcomingdis)); 
                                    ?></select></td>
        <td><select name="ven_id">
                                    <?php 
                                        $upvenbycounty_set = upcoming_venevents_bycounty($url_countyid);
                                        $upcoming_ven_bycounty = mysql_fetch_assoc ($upvenbycounty_set);
                                    ?>
                                    <?php 
                                    do { 
       ?>   <option value="<?php echo $upcoming_ven_bycounty ['ven_id']; ?>" >
<?php echo $upcoming_ven_bycounty ['ven_name']; ?></option>
<?php 
                                    } while ($upcoming_ven_bycounty = mysql_fetch_assoc ($upvenbycounty_set)); 
                                    ?></select></td>
        <td><select name="champ_id">
                                    <?php 
                                        $championship_set = findchampionships();
                                        $champlist = mysql_fetch_assoc ($championship_set);
                                    ?>
                                    <?php 
do { 
?>   <option value="<?php echo $champlist ['champ_id']; ?>" >
<?php echo $champlist ['champ_description']; ?></option>
<?php 
} while ($champlist = mysql_fetch_assoc ($championship_set)); 
?></select></td>
        <td>
        <select name="org_id">
                        <?php 
                            $uporgbycounty_set = upcoming_organevents_bycounty($url_countyid);
                            $upcoming_org_bycounty = mysql_fetch_assoc ($uporgbycounty_set);
                        ?>
                        <?php 
                        do { 
                        ?>   <option value="<?php echo $upcoming_org_bycounty ['org_id']; ?>" ><?php echo $upcoming_org_bycounty ['org_name']; ?></option>
                        <?php 
                        } while ($upcoming_org_bycounty = mysql_fetch_assoc ($uporgbycounty_set)); 
                        ?></select>
       
        </td>
        <td><select name="ass_id">
                                    <?php
                                        $upassbycounty_set = upcoming_assevents_bycounty($url_countyid);
                                        $upcoming_assbycounty = mysql_fetch_assoc ($upassbycounty_set);
                                    ?>
                                    <?php 
                                    do { 
                                    ?>   <option value="<?php echo $upcoming_assbycounty ['ass_id']; ?>" >
<?php echo $upcoming_assbycounty ['ass_description']; ?></option>
                                    <?php 
                                    } while ($upcoming_assbycounty = mysql_fetch_assoc ($upassbycounty_set)); 
                                    ?></select></td>
      <td><input name="submit" type="submit" /><input name="countyid" type="hidden" value="<?php echo $url_countyid ['url_countyid']; ?>" /></td>
        </tr>
      </table>
  </form>
My search processing is as follows

Code: [Select]
  <?php
if (isset($_POST['submit'])){
if (isset($_GET['go'])){
$countyid = $_POST['countyid'];
$ven_id = $_POST['ven_id'];
$dis_id = $_POST['dis_id'];
$champ_id = $_POST['champ_id'];
$org_id = $_POST['org_id'];
$event_id = $row['event_id'];

$sql = "SELECT DATE_FORMAT (events.startdate, '%a, %d, %b') 
   AS stdate, events.event_id, events.title, events.ven_id, events.org_id,
   venue.county_id, venue.ven_id, eventdisciplines.event_id, eventdisciplines.dis_id, county.county_id, discipline.dis_id \n"
. "FROM events \n"
. "LEFT OUTER JOIN eventdisciplines \n"
. "ON events.event_id = eventdisciplines.event_id \n"
. "LEFT OUTER JOIN discipline \n"
. "ON eventdisciplines.dis_id = discipline.dis_id \n"
. "LEFT OUTER JOIN venue \n"
. "ON events.ven_id = venue.ven_id \n"
. "LEFT OUTER JOIN county \n"
. "ON venue.county_id = county.county_id \n"
. "WHERE events.ven_id = ({$ven_id} OR events.org_id = {$org_id})\n"
. "AND events.startdate > NOW()\n"
. "AND venue.county_id = {$countyid}  \n"
. "ORDER BY startdate ASC";

$result = mysql_query ($sql, $connection);
?>
How am I best to do this please?  my OR within the mysql does not work, should I not be doing this with php in the search processing?  someones help would really be appreciated, just to point me in the right direction.

[attachment deleted by admin]

I am getting this error:
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/content/20/8917120/html/goldenfruit/test.php on line 11

for this code:


<?php

$connect = mysql_connect('DB credentials removed');

if (!$connect) {
   die('Could not connect: ' .mysql_error());
}


$data = mysql_query("SELECT * FROM fruits");
$row  = mysql_fetch_row($data);

echo $row[0];

echo hello;


?>


I cannot seem to find the error in this.
Can anyone help?