PHP - Question Drop Menu

Okay, so, here's the scenario.

I have a form that is editing an item that is already in the database. The text fields fill in just fine with that info. However, the drop down menus don't retrieve that info, rather resorting to the defaults, which can be a problem if you don't remember what you originally had.

Is there anyway to make the dropdown menus pull the info from the table and use that rather than resorting back to the default?

I tried using this:

Code: [Select]
<tr><td width="20%">Bonus:</td><td><select name="bonus" value="{{bonus}}">
<option value="Attack" {{bonus1select}}>Add to the attack power of weapon</option>
<option value="Defense" {{bonus2select}}>Add to the defensive power of armor</option>
<option value="None" {{bonus3select}}>No effect</option>
</select><br /></td></tr>

So, it's obvious the "value" portion not working. Any help would be great!!

helo
i a newbie in php. i need some help regarding my system. i want my data from database to display in drop down menu form.anyone have idea how to do that? i really need some help. i have search all over and couldn't find any solution. any suggestions would be helpful 

I'm trying to code a drop-down menu that has four options; one for pie,exp, root 3, and the golden ratio.(all math values)  Beside the drop-down menu, there is an option for user-inputted data, they must input only positive numbers. I must take there selected drop-down menu option and times it by the user inputted number. I'm not sure how to check which option the user chose.

    <form id="s" method="post">
      <select id="math" name="options">
        <option value="nothing">Please select a constant</option>
        <option value="pie">Pi</option>
    <option value="econs">e</option>
    <option value="ratio">Golden Ratio</option>
    <option value="root">Root 3</option>
  

      </select>

    <input type="submit" name="Submit" value="Send">

Everything That ive tried has failed

Hi, i wonder whether someone may be able to help me please.

I am using a combination of PHP and AJAX to create two drop down menus on a HTML form. The data is being pulled from a mySQL database with the options available in the second drop down dependent on the value selected in the first.

The initial drop down menu called 'detectors' and the behaviours for the second drop down menu, 'searchheads' are created with the following AJAX code:

Code: [Select]
Function AjaxFunction(detectorid)
            {
            var httpxml;
            try
            {
            // Firefox, Opera 8.0+, Safari
            httpxml=new XMLHttpRequest();
            }
            catch (e)
            {
            // Internet Explorer
            try
            {
            httpxml=new ActiveXObject("Msxml2.XMLHTTP");
            }
            catch (e)
            {
            try
            {
            httpxml=new ActiveXObject("Microsoft.XMLHTTP");
            }
            catch (e)
            {
            alert("Your browser does not support AJAX!");
            return false;
            }
            }
            }
            function stateck()
            {
            if(httpxml.readyState==4)
            {
           
            var myarray=eval(httpxml.responseText);
            // Before adding new we must remove previously loaded elements
            for(j=document.addfindstolocation.searchheads.options.length-1;j>=0;j--)
            {
            document.addfindstolocation.searchheads.remove(j);
            }
           
           
            for (i=0;i<myarray.length;i++)
            {
            var optn = document.createElement("OPTION");
            optn.text = myarray[i];
            optn.value = myarray[i];
            document.addfindstolocation.searchheads.options.add(optn);
           
            }
            }
            }
            var url="searchheaddetails.php";
            url=url+"?detectorid="+detectorid;
            url=url+"&sid="+Math.random();
            httpxml.onreadystatechange=stateck;
            httpxml.open("GET",url,true);
            httpxml.send(null);
            }
The following code is the file 'searchheaddetails.php' (as highlighted above) which populates the second drop down menu.

Code: [Select]
<?
$detectorid=$_GET['detectorid'];
require "config.php";
$q=mysql_query("SELECT * FROM searchheads WHERE detectorid='$detectorid' ORDER BY 'searchheadname' ASC");
echo mysql_error();
$myarray=array();
$str="";
while($nt=mysql_fetch_array($q)){
$str=$str . "\"$nt[searchheadname]\",";
}
$str=substr($str,0,(strLen($str)-1)); // Removing the last char , from the string
echo "new Array($str)";

?> 
And this is the section of my form that pulls together the two drop down menus.

Code: [Select]
<form name="addfindstolocation" method="post" id="addfindstolocation">
<div align="left">
                                        <select name=detectors id="detectorid" onchange="AjaxFunction(this.value);">
                                            <option value=''>Select One</option>
                                            <?
                                            require "phpfile.php";// connection to database
                                            $q=mysql_query("SELECT * from detectors WHERE userid='1'ORDER BY 'detectorname' ASC");
                                            while($n=mysql_fetch_array($q)){
                                            echo "<option value=$n[detectorid]>$n[detectorname]</option>";
                                            }
                                           
                                            ?>
                                      </select>
                                    </div>
                                </div>
                                <p align="left">
                                    <label></label>
                                    <label>Search Head Used</label></p>
                                <div>
                                    <div align="left">
                                        <select name=searchheads id="searchheadid">
                                        </select>
                                    </div>
The drop down menus work fine, but I'm having a little difficulty with the data that is being saved. For the 'detectors' drop down menu the data being saved upon a selection being made is the 'id' pertient to the relevant detector e.g. 'Detector1' is selected and the 'id' value of '1' is saved which is exactly what I want. However when it comes to the second drop down menu, the value saved is the text value that the user selects, rather than the 'id'. e.g. 'Deep Search Head ' rather than an 'id' of '1'. Could someone perhaps tell me please what I need to change so that the 'id' value is saved rather than the text value.

If it helps, the coding is taken from the following http://www.plus2net.com/php_tutorial/ajax_drop_down_list.php.

Many thanks and kind regards.

Chris

Hey guys im looking to do a drop down menu which is pretty basic however, in this drop down menu it uses an array of 4 values eg Orange, Apples, Bananas, Strawberry's. For this menu if a user has previously chosen an Orange which is stored on the db then then when the drop down menu is loaded it should move Orange to the top of the list, the same applies for the other values. How could I do this? I know how to do the select statement and place the values in an array. from there I dont know quite how to approach it.


()
Thank you.

Hi, I'm no pro at PHP but I am trying to get a drop down menu to a authenticate before moving to the next part of the form.

What I want is once a selection has been made, ONLY THEN can the user move on, OTHERWISE a message echo appears.

This is the html menu box

<select size="1" name="title">
                    <option>Please Select</option>
                    <option value="Mr">Mr</option>
                    <option value="Mrs">Mrs</option>
                    <option value="Miss">Miss</option>
                    <option value="Ms">Ms</option>
                    <option value="Dr">Dr</option>
                  </select>   

Then this is what I have in the form PHP:

$visitortitle = $_POST['visitortitle'];

if ( HOW DO I GET THIS PART TO AUTHENTICATE AN OPTION HAS BEEN SELECTED? ) {
echo "<p>Please enter a title correctly<br />before you try submitting the form again.</p>\n";
die ( '<a href="pef.html">click here go back and try again</a>' );
echo $id;}

If anyone can help me sort out this part of the form I can move on as the rest is working fine?

Thanks
Gary
 

I am trying to get the drop downs to work in this order on the second <li> 903,896,898,1513 but they won't. I think it's picking up ascending order maybe? Is there another php/include I should use? It's a WordPress site.

Code: [Select]
<div id="navigation">
<div id="navbar_menu">
<ul id="nav_menu">
<li><?php wp_list_pages('include=3&title_li=&depth=2');?></li>
<li><?php wp_list_pages('include=2,903,896,898,1513&title_li=&depth=2');?></li> 
<li><?php wp_list_pages('include=23,911,913&title_li=&depth=1');?></li>
<li><?php wp_list_pages('include=25,921,925,927&title_li=&depth=1');?></li>             
<li><?php wp_list_pages('include=27&title_li=&depth=2');?></li>
<li><?php wp_list_categories('include=3&title_li=&depth=2');?></li>
<li><?php wp_list_pages('include=407&title_li=&depth=2');?></li> 
</ul>

</div>
</div>
Thanks in advance!

[attachment deleted by admin]

Hello!

This is my first post here.

I'm usually do programming in Ruby (Rails) but for a special occasion I need PHP.

So hi PHP people! I come in peace   

My question is quite simple. How do you make a drop down menu? Most preferably with database populated fields in it.

In the Rails community, this really helped me make these menus in Rails: http://www.kahfei.com/?p=23

Could someone briefly write something like that too? (The place holders)

Orrr you guys can just explain how to do it. The ones I found on the internet were quite confusing and not explained enough.

Help anyone? Thank you

I'm having a problem with populating a second drop down menu that needs to be identical to the first from a mysql database.
I have about 20 items in the DB & I only want to show the first 9.
My code is: Code: [Select]
      <select name="a_type" style="width: 175px;"> <option value="" selected="selected"> None selected</option>
      <?php
$i=1;
while (($_produce = mysql_fetch_array($result_produce)) && ($i <=9 ))
{
$produce_type=$_produce["type"];
$produce_id=$_produce["id"];
$i++;
?>

<option value="<?php echo $produce_id; ?>"><?php echo $produce_id; ?> - <?php echo $produce_type; ?> </option>
<?php
}
?>
        </select>

That works fine in my first drop down menu but the second drop down menu with the same exact code will show items starting at 11 & higher
instead of 1 to 9. I haven't tried closing the DB & the reopening it which I assume would work but there are a lot of other variables & things I'd
have to check to make sure they don't get effected so I'd rather not.   Is there a simple way to solve this?   

hii,everyone...
 i am new to the forum ..so forgive me if i make some errors in describing my problem

i want to create a drop down menu...the values of which i want to retrieve from a table i have created in my database...
i have been able to acheive this...but i want a second drop down menu...in which i want the values to get filtered according to the selection of the first drop down menu...
i hav a table with the name test...which i am connecting to the first drop down menu
i hav created another table with the name testtype...for second menu..
the id of first table is the testid of second...
plzz help me with creating the second menu...

my code is
Code: [Select]
   <?php
$sql_result = "select test_series FROM test";
$result = mysql_query($sql_result); 
?>
<td>Select Test Series </td>
         <td>
        <select name="testseries" id="testseries" )">
        <option>Select</option>
        <?php  if(mysql_num_rows($result)) 
    { 
    while($row = mysql_fetch_assoc($result)) 
    { 
$test = $row['test_series'];?>
    <option><?php echo $test ?></option>
    <?php } 
 
    } 
    else {?>
    <option>No Names Present</option>"; 
   <?php } 
    ?>
    </select>
    </td>
    </tr>
   
     <tr>
<?php
$testseries = $_POST['testseries'];
 $query = "select id from test where test_series = '$testseries'";
 $resultt = mysql_query($query) or die(mysql_error());
 $row = mysql_fetch_array($resultt);
 $id = $row['id'];
   
$sql_result = "select test_name FROM testtype where testid = $id";
$result = mysql_query($sql_result); 
?>
<td>Select Test Name </td>
         <td>
        <select name="testtype">
        <option>Select</option>
        <?php  if(mysql_num_rows($result)) 
    { 
    while($row = mysql_fetch_assoc($result)) 
    { 
$test2 = $row['test_name'];?>
    <option><?php echo $test2 ?></option>
    <?php } 
 
    } 
    else {?>
    <option>No Names Present</option>"; 
   <?php } 
    ?>
    </select>
    </td>
    </tr>






IN THIS CODE I AM GETTIN AN ERROR Undefined index: testseries in ..addques.php on line 155 WHICH IS THIS ONE $testseries = $_POST['testseries'];

Hi I'm new here, also I'm new in php. What I need to do is to write registration form with drop down menu. Actually, I need to have drop down menu which would write data into database when someone fill registration form. Also I need when someone choose one of the city name in drop down menu, to display certain image in specific php page. Can anyone help me.
Here is my code for registration form :
register.php
<?PHP

//Database Information

$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";

//Connect to database

mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

   
$name = $_POST['name'];
$email = $_POST['email'];   
$username = $_POST['username'];
$password = md5($_POST['password']);
$grad = $_POST['grad'];

// lets check to see if the username already exists

$checkuser = mysql_query("SELECT username FROM users WHERE username='$username'");

$username_exist = mysql_num_rows($checkuser);
if($username_exist > 0){
    echo "I'm sorry but the username you specified has already been taken.  Please pick another one.";
    unset($username);
    include 'register.html';
    exit();
}
$checkemail = mysql_query("SELECT email FROM users WHERE email='$email'");
$email_exist = mysql_num_rows($checkemail);
if($email_exist > 0){
    echo "Email koji ste izabrali vec postoji.";
    unset($email);
    include 'register.html';
    exit();
}
$checkname = mysql_query("SELECT name FROM users WHERE name='$name'");
$name_exist = mysql_num_rows($checkname);
if($name_exist > 0){
    echo "Ime koje ste izabrali vec postoji.";
    unset($name);
    include 'register.html';
    exit();
}
// lf no errors present with the username
// use a query to insert the data into the database.

$query = "INSERT INTO users (name, email, username, password, grad) VALUES('$name', '$email', '$username', '$password','$grad')";

mysql_query($query) or die(mysql_error());
mysql_close(); 
if(empty($name) || empty($email) || empty($username))
{
   echo "Nisu uneseni svi podaci.";
}
/*else if(empty(($password))) //ovaj uslov ne prolazi !!!
{
   echo "Niste uneli lozinku.";
}*/
else
{
   echo "Uspesno ste se registrovali";
}
?>

and for form :
register.html
<form name="login" method="post" action="register.php">
<table border="0" width="225" align="center">
    <tr>
        <td width="219" bgcolor="#ff6600">
            <p align="center"><font color="white"><span style="font-size:14pt;"><b>Registration</b></span></font></p>
        </td>
    </tr>
    <tr>
        <td width="555">
            <table border="0" width="555" align="center">
                    <tr>
                        <td width="116"><span style="font-size:12pt;">Name:</span></td>
                        <td width="156"><input type="text" name="name" maxlength="100"></td>   
                  <td width="200"><p align="left"><span style="font-size:11pt;"><i>*Ukucajte vase puno ime i prezime*</i></span></p></td>
                    </tr>
               
               <tr>
                        <td width="116"><span style="font-size:12pt;">Email:</span></td>
                        <td width="156"><input type="text" name="email" maxlength="100"></td><br/>
                  
                    </tr>
                <tr>
                    <td width="116"><span style="font-size:12pt;">Username:</span></td>
                    <td width="156"><input type="text" name="username"></td><br/>
               
                </tr>
                <tr>
                    <td width="116"><span style="font-size:12pt;">Password:</span></td>
                    <td width="156"><input type="password" name="password"></td><br/>
               
                </tr>
                <tr>
                    <td width="116">&nbsp;</td>
                        <td width="156">
                            <p align="right"><input type="submit" name="submit" value="Submit"></p>
                        </td>
                </tr>
            <tr>
               <td>
               <select name = "dropdown" value="options">
               <option value "register.php">Izaberi grad</option>
               <option value = "Beograd">Beograd</option>
               <option value = "Nis">Nis</option>
               <option value = "Novi Sad">Novi Sad</option>
               </select>
               </td>
            </tr>
            </table>
        </td>
    </tr>
    <tr>
        <td width="219" bgcolor="#ff6600">&nbsp;</td>
    </tr>
   
</table>
</form>

Hi

I have coded a drop down menu with php and i am trying to retrieve the data when a user select a option from the menu and the data is retrieved from the database. So far i have tried and nothing is displaying when i tried to process the php form.

Sales.php Page

<form action="saleprocess.php" method="GET">
<?php
echo 'Product Model:';
$query="SELECT * FROM products";
/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */
$result = mysql_query ($query);
echo "<select name=product_model value=Select>Product Model</option>";
// printing the list box select command
while($rows=mysql_fetch_array($result)){//Array or records stored in $nt

echo "<option name=product_model value='.$rows[product_id].'>$rows[product_model]</option>";
/* Option values are added by looping through the array */
}
echo "</select><br>";
?>
<input type='submit' name='submit' value='Create'></input> <br>
</form>

********************************************************************************
Salesprocess.php page

<?php
include("connect.php");

if(isset($_GET['product_id'])){

$product_id = $_GET['product_id'];


$query = mysql_query("SELECT * FROM products WHERE product_id= $product_id");

while($rows = mysql_fetch_assoc($query))
{
echo 'Product Model<br>';
echo $rows['product_id'];
echo $rows['product_model'];


}
}
?>

Muchly appreciated if someone can help me

Hi all,

I'm just wondering if there's an easier way of doing what accomplishing the following:

I have a value in my database which represents a selection in a drop down menu, i want to read it from the database and have it automatically selected depending on the stored data.

I have the following working but just wondered if there was an easier way to get the same result:

Code: [Select]
<?php
//database connection

$query = "SELECT id
  FROM  `tablename`
  WHERE  username='$username'";
$result = mysql_query($query);
$row = mysql_fetch_object($result);

$id = $row->id;
$id = (int)$id;
?>

<select name="id">
<option value="">Select your option...</option>
<option value="1" <?php if (($id - 1) === 0) { echo 'selected="selected"'; }?>>Selection 1</option>
<option value="2"<?php if (($id - 2) === 0) { echo 'selected="selected"'; }?>>Selection 2</option>
</select>

Sorry if it's not very clear, i'll explain best i can if anyone can help.

Thanks

I am having issues with my page. I am trying to send a email to a different address based on the entry that my user submits.

Here is my form on my test page.
Code: [Select]
<form method=POST action=formdata.php>

  <table width="640" border=0 align="center">
    <tr>
      <td align=right><b>Test</b></td>
      <td><input type=text name=test size=25></td>
      <td>&nbsp;</td>
      <td>&nbsp;</td>
    </tr>
  </table>
  </dl>
  <div align="center">
    <p>
      <input type=hidden name=FA value=SendMail>
    </p>
      <input type=submit value="Submit Form">
    </p>
  </div>
</form>
here is the php code for my formdata page

Code: [Select]

<?php
$con = mysql_connect("localhost","test","test");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
 
mysql_select_db("test", $con);
 
$sql="INSERT INTO formdata (test)
VALUES ('$_POST[test]')";
 
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Your Information Was Successfully Posted";
 
mysql_close($con);

if ($_POST['test'] =="A"){

$to = "email@sample.com";
$subject = "Custom Form";
$message = $_POST['test'] ;
$headers = "From: email@sample.com";
$sent = mail($to, $subject, $message, $headers) ; 
} else if ($_POST['test'] =="B"){

$to = "email2@sample.com";
$subject = "Custom Form";
$message = $_POST['test'] ;
$headers = "From: email@sample.com";
$sent = mail($to, $subject, $message, $headers) ; 
}
if($sent) 
{print "<p>Your mail was sent successfully"; }
else 
{print "<p>We encountered an error sending your mail"; }



?>

Hey guys
I wasnt sure if this is a php topic or java script
im trying to make my drop down menu's activate onchange. what confuses me is how do I make php know that the java script is being run. Or have the java script trip my php code.

To know a button has been pressed i use "if (isset['_POST'])"


 echo '<select name="siege_list" id="siege_list">';

foreach($name as $key => $value) {        

echo '<option value="' . $value['name'] . '" ' . ($value['name'] == $current1
['name'] ? 'selected="selected"' : '') . '> ' . $value['name'] . '</options>';
}                           
echo '</select><input type="submit" id="siege_planet" onchange="this.form.submit()">';