PHP - Dropdown 'selected' Not Showing Right

friend the code display all the managers name  

  <?php

        echo '<select name="Manager">';

        $query="SELECT * FROM managers";

        $result = mysql_query($query);

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

            echo "<option value='" . $row['manager_name'] . "'>" . $row['manager_name'] . "</option>";

        }

        echo '</select>';

        ?>

  and the variable below displays the result   $Manager= $row['Manager'] ;     but how do i do it on the above dropdown list the it shows which result is in the $Manager variable?

mysql.php
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("ckeditor",$con);
?>
---------------------------------------------------------
add.php
<?php
include("mysql.php");


if(isset($_POST["button2"]))
{
$sql="INSERT INTO cktext (section)
VALUES
('$_POST[select2]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }

?>

-----------------------------------------------------------------
home.php

<form id="form1" name="form1" method="post" action="add.php">
<tr>
    <td>Section:</td>
    <td><select name="select2" id="select2">
     <option selected="selected" value="MALE">Male</option>
  <option selected="" value="FEMAIL">FEMAIL</option>
    </select></td>
  </tr>
<input type="submit" name="button2" id="button2" value="Upload" />
  </form>

In DATABASE :- cktext table attribute "section" is varchar type.

BUT IT RETURN ME BLANK OUTPUT.

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?

Yesterday, I created a topic about how I could update records and I managed to achieve that successfully.

Now I have another dilemma.

When I have a specific record I want to update, I want to change a category ID of an product (e.g. change it from 1 to 2) but how do I go about doing this?

Here is my code thus far:

Code: [Select]
<?php

require_once ('./includes/config.inc.php');

require_once (MYSQL);

$id=$_GET['prodID'];

$results = mysqli_query($dbc, "SELECT * FROM product WHERE prodID=".$_GET['prodID']."");
$row = mysqli_fetch_assoc($results);

?>
<form action="" method='POST'>
Product ID: <input type="text" value="<?php echo $row['prodID'];?>" name="prodID" /> <br />
Product: <input type="text" value="<?php echo $row['product'] ;?>" name="product" /> <br />
Product Description: <input type="text" value="<?php echo $row['prod_descr'] ;?>" name="prod_descr" /> <br />
Category:
<select name="category">
<option value="<?php echo $row['catID'];?>"></option>
</select>
Price: <input type="text" value="<?php echo $row['price'] ;?>" name="price" /> <br />
In Stock: <input type="text" value="<?php echo $row['stock'] ;?>" name="stock" /> <br />
<br /><input type="submit" value="save" name="save">
</form>

<?php

if(isset($_POST['save']))
 {
    $id = $_POST['prodID'];
    $product = $_POST['product'];
$descr = $_POST['prod_descr'];
$price = $_POST['price'];
$stock = $_POST['stock'];

// Update data
    $update = mysqli_query($dbc, "UPDATE product SET product='$product', prod_descr='$descr', price='$price', stock='$stock' WHERE prodID=".$_GET['prodID']."");
header( 'Location: update_save.php' ) ;
}
?>

Hey Everyone...

First off, I am only a young web developer and i'm working on a school project and am making a text-based game online...

Now what i'm having trouble with...

I want a drop-down list that has a list of characters classes
Clubber
Mixer
Sauceror
Tamer

And I want whatever is selected to be placed into the database along with the username/password (THIS ALL WORKS FINE JUST NOT THE DROP DOWN LIST)

All help appreciated

I'm new to this form and php/mysql so sorry if this isn't the right place to post this.


This is what is in the first dropdown box.
Code: [Select]

$selValues['john'] = "a, b, c, d, e";

These are the different lists I want it to put in the second drop down box depending on what they choose in the first.
Code: [Select]
$selValues['list1']= " a1, a2, a3, a4, a5";
$selValues['list2']= " b1, b2, b3, b4, b5";

This is how I made an attempt to make it work before posting here. Along with plently of other ways.
Code: [Select]
if ($selValues['john']=a)
{
     $chan_input = selectBuilder($selValues['$list1'] };
else if ($selValues['john']=b)
 {
     $chan_input = selectBuilder($selValues['$list2'] };

Basicly how I need it to work is there are two drop down boxes. First they will chose between a,b,c,d,e. If they chose a then on the second drop down box I only want them to be able to select a1,a2,a3,a4,a5.

Folks,

I have a dropdown, once values are selcted, these values should be put in a URL structure so that it matches with the one MoD_rewrite rule in my .htaccess.

Here is my Mod_Rewrite Rule:

   RewriteRule ^(.*)/([^/]*)\.html$ search.php?q=$1&sc=$2 [QSA,L] 


Here is my Dropdown Code:

 <div id="search" >

<form id="searchform" method="get" action="search.php">
<label>Search By Brand/ Manufacturer: </label>

<select name="q">

<option value="SelectBrand">Select Brand</option>

<?php if(isset($this->search->options)): ?>
<?php foreach($mfg as $lolachild): ?>

  <option value="<?php echo $lolachild; ?>"><?php echo ucwords($lolachild); ?></option>

<?php endforeach; ?>
<?php endif; ?>

</select>

<select name="sc">

  <option value="All"><?php eprint(LangAll); ?></option>
<?php if(isset($this->search->options)): ?>
<?php foreach($this->search->options as $cat): ?>

  <option value="<?php echo $cat->value; ?>"><?php echo $cat->name; ?></option>
<?php endforeach; ?>
<?php endif; ?>

</select>

        <input type="submit" value="<?php eprint(LangSearch); ?>" />

</form>

 </div> 



Problem is, upon Submit, it goes to this link structu

http://mydomain.co.uk/search.php?q=fisher&sc=302

It should rather be:

http://au2.co.uk/fisher/302.html

What am i missing or doing wrong?

Cheer
Natasha

Hello, im trying to make a dropdown show up for every single row gotten from the database table. Instead it shows only one for the first row.

Here is the code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
  <title>Example Form</title>
  <link rel="stylesheet" type="text/css" href="dd.css" />
<script type="text/javascript" src="jquery-1.3.2.min.js"></script>
<script type="text/javascript" src="jquery.dd.js"></script>
</head>
<body>

<?php
mysql_connect("localhost", "", "")or die("cannot connect");
mysql_select_db("test")or die("cannot select DB");
$tbl_name="test_mysql";
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
$count=mysql_num_rows($result);
?>
<form name="form1" method="post" action="">
<tr>
<td>
<table width="100%" border="0" cellspacing="1" cellpadding="0">

<tr>
<td align="center"><strong>Row</strong></td>
<td align="center"><strong>Month  Date</strong></td>
<td align="center"><strong>Message</strong></td>
<td align="center"><strong>Title</strong></td>
<td align="center"><strong>Icon</strong></td>

</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td align="center"><?php $id[]=$rows['id']; ?><?php echo $rows['id']; ?></td>
<td align="center"><input name="month[]" MAXLENGTH="3" size="3" type="text" id="month" value="<?php echo $rows['month']; ?>">
                   <input name="date[]" MAXLENGTH="2" size="2" type="text" id="date" value="<?php echo $rows['date']; ?>">
</td>
<td align="center"><input name="message[]" size="50" type="text" id="message" value="<?php echo $rows['message']; ?>"></td>
<td align="center"><input name="title[]"  size="50"  type="text" id="title" value="<?php echo $rows['title']; ?>"></td>
<td align="center">
    <select name="test[]" style="display:none; width:200px" class="mydds">
      <option value="icon_phone.gif" title="icon/icon_phone.gif">Phone</option>
      <option value="icon_sales.gif" title="icon/icon_sales.gif">Graph</option>
    </select> 
</tr>
<?php
}
?>
<tr>

<td colspan="4" align="center"><br><input type="submit" name="Submit" value="Submit"></td>
</tr>
</table>
</td>
</tr>
</form>


<script language="javascript" type="text/javascript">
function showvalue(arg) {
    alert(arg);
    //arg.visible(false);
}
$(document).ready(function() {
try {
        oHandler = $(".mydds").msDropDown().data("dd");
        oHandler.visible(true);
        //alert($.msDropDown.version);
        //$.msDropDown.create("body select");
        $("#ver").html($.msDropDown.version);
        } catch(e) {
            alert("Error: "+e.message);
        }
})    
</script>


<hr>

<?php
if (isset($_POST['Submit'])) { 
for($i=0;$i<$count;$i++){

$month = $_POST['month'];
$date = $_POST['date'];
$message = $_POST['message'];
$title = $_POST['title'];
$monthday = $month[$i]."<br>".$date[$i];

$sql1="UPDATE $tbl_name SET monthday='$monthday', month='$month[$i]', date='$date[$i]', message='" . mysql_real_escape_string($message[$i]) . "', title='" . mysql_real_escape_string($title[$i]) . "' WHERE id='$id[$i]'";


//$sql1="UPDATE $tbl_name SET monthday='$monthday', month='$month[$i]', date='$date[$i]', message='$message[$i]', title='$title[$i]' WHERE id='$id[$i]'";
$result1 = mysql_query($sql1);
} 
header("location:update2.php");
}
?>    

</body>
</html>

<BR>



I have no idea if it has to do with the PHP, HTML, or JAVASCRIPT.

So if someone can tell me what is going on I would be happy.

Here is how it looks:



Youl see it shows the fields but not the dropdown... Please help and thank You!

Hello there,

I'm trying to show a picture based on the value that was chosen in the dropdown menu of the page before the submit.

Let's say i'm having a drop down form with 3 values: Porsche, BMW, Audi, the form also consists of a button to submit form and some other text fields but these aren't really relevant.

So what i want is, you choose whatever car, let's say BMW, fill in all the other data of the form, hit submit, and on the next page it should show a Picture which i define for each car.

I hope i explained that somehow understandable.

Thanks in advance,

Sabine

I am pretty new to PHP and am trying to create a simple (so I assumed) page to takes data from one html page(works fine) and updates a MYSQL Database.  I am getting no error message, but the connect string down to the end of the body section is showing up as plain text in my browser window.  I do not know how to correct this.  I have tried using two different types of connect strings and have verified my names from the HTML page are the same as listed within the php page.  Suggestions on what I need to look for to correct would be great.  I have looked online, but so far all I am getting is how to connect, or how to create a comment, so I thought I would try here. 
Thank you for any assistance I may get!!  - Amy -
 
Code: [Select]
<body><font color="006600">
 <div style="background-color:#f9f9dd;">
  <fieldset>
<h1>Asset Entry Results</h1>
<?php
  // create short variable names
  $tag=$_POST['tag'];
  $serial=$_POST['serial'];
  $category=$_POST['category'];
  $status=$_POST['status'];
  $branch=$_POST['branch'];
  $comments=$_POST['comments'];

  if (!$tag || !$serial || !$category || !$status || !$branch) {
     echo "You have not entered all the required details.<br />"
          ."Please go back and try again.";
     exit;
  }

  if (!get_magic_quotes_gpc()) {
    $tag = addslashes($tag);
    $serial = addslashes($serial);
    $category = addslashes($category);
    $status = addslashes($status);
    $branch = addslashes($branch);
    $comments = addslashes($comments);
  }

  //@ $db = new mysqli('localhost', 'id', 'pw', 'inventory');
  $db = DBI->connect("dbi:mysql:inventory:localhost","id","pw") or die("couldnt connect to database");

  $query = "insert into assets values
            ('".$serial."', '".$tag."', '".$branch."', '".$status."', '".$category."', '".$comments."')";
  $result = $db->query($query);

  if ($result) {
      echo  $db->affected_rows." asset inserted into Inventory.";
  } else {
     echo "An error has occurred.  The item was not added.";
  }

  $db->close();
?>
  </fieldset>
 </div>
</body>

I have this loop on a shorturl website i own that selects domains from my website and lists them in a dropdown list where users can select one domain to use as url shortener.

The problem is that when a users selects and entry from the dropdown list this created only the [[[main_url]]] entry from the database table bellow can be used. When someone selectes "coolartistname.site.com the submit button cannot be pressed. in other words it's not adding the SELECTED attribute other than the [[[main_url]]] domain.

Can anyone see on the code bellow why this would happen?

When i echo $shortUrlDomains it's just saying "Array" and $_REQUEST['shortUrlDomain'] shows nothing.

**Index.php**

    // get base urls
    $shortUrlDomains = getShortUrlDomains();
    
        <?php
        foreach ($shortUrlDomains AS $k => $shortUrlDomain)
        {
        
            // active domains only
            if (($shortUrlDomain['premium_only'] !== '1') && ($shortUrlDomain['premium_only'] !== $Auth->id))
            {
                continue;
            }
        
            echo '<option value="' . (int)$k . '"';
        
            if (isset($_REQUEST['shortUrlDomain']))
            {
                if ($k == (int)$_REQUEST['shortUrlDomain'])
                {
                    echo 'SELECTED';
                }
            }
        
            echo '>';
        
            echo $shortUrlDomain['domain'];
        
            if ($disabled == true)
            {
                echo ' (' . safeOutputToScreen(t('unavailable', 'unavailable')) . ')';
            }
        
            '</option>';
        }
        echo '</optgroup>';
        ?>

FUNCTIONS.PHP

    function getShortUrlDomains()
    {
        // if we already have it cached
        if(defined('_SETTINGS_SHORT_URL_DOMAINS'))
        {
            return unserialize(_SETTINGS_SHORT_URL_DOMAINS);
        }
        
        // get connection
        $db  = Database::getDatabase(true);
        
        // load from database
        $domains = $db->getRows("SELECT id, domain, premium_only, status FROM url_domain ORDER BY premium_only ASC, id ASC");
        
        // organise and replace site url
        $rs = array();
        foreach($domains AS $domain)
        {
            if($domain['domain'] == '[[[main_url]]]')
            {
                $domain['domain'] = _CONFIG_SITE_FULL_URL;
            }
            $rs[$domain{'id'}] = $domain;
        }

**MYSQL DATABASE**

    url_domain (id, domain, premium_only, status, date_created, owner) VALUES
    (1, '[[[main_url]]]', 0, 'enabled', '2019-03-02 08:13:00', 1) 
    (14, 'coolartistname.site.com', 6, 'enabled', '2020-04-18 19:21:59', 6);

 

I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time.

I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.

When i load ajax.php i get 2 error mesages: 
Notice: Undefined index: machinery in C:\wamp64\www\gecko\ajaxfile.php on line  *8*
Warning: sqlsrv_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp64\www\site\ajaxfile.php on line  *16*

This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID*  is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID.

Index.php

    <!doctype html>
      <?PHP
    $server = "server";
    $options = array(  "UID" => "user",  "PWD" => "pass",  "Database" => 
    "database");
    $conn2 = sqlsrv_connect($server, $options);
    if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
    echo " ";
    ?>
    <html>
    <head>
    <meta charset="utf-8">
    <title>Untitled Document</title>
    </head>
    
    <body>
                <section id="formaT2" class="formaT2 formContent">
                <div class="row">
                    <div class="col-md-2 col-3 row-color remove-mob"></div>
                    <div class="col-md-5 col-9 bg-img" style="padding-left: 0; 
    padding-right: 0;">
                        <h1>Form</h1>
                        <div class="rest-text">
                           <div class="contactFrm">
                              <p class="statusMsg <?php echo 
    !empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
                               
                                <form action="connection.php" method="post">
                                    <div>machinery</div>     
    <select id="machinery">
       <option value="0">--Please Select Machinery--</option>
       <?php 
       // Fetch Department
       $sql = "SELECT Machinery FROM T013";
       $machanery_data = sqlsrv_query($conn2,$sql);
       while($row = sqlsrv_fetch_array($machanery_data) ){
          $id = $row['Id'];
          $machinery = $row['Machinery'];
          
          // Option
          echo "<option value='".$id."' >".$machinery."</option>";
       }
       ?>
    </select>
    <div class="clear"></div>
                                    <div>Sub Machinery</div>
    <select id="sub_machinery">
       <option value="0">- Select -</option>
    </select>
                                    
                                    <input type="submit" name="submit" 
    id="submit" class="strelka-send" value="Insert">
                                    <div class="clear"> </div>
                                </form>
                             </div> 
                         </div>
                    </div>
                </div> 
            </section>
             </script>
          <script type="text/javascript">
          $(document).ready(function(){

        $("#machinery").change(function(){
            var machinery_id = $(this).val();
    
            $.ajax({
                url:'ajaxfile.php',
                type: 'post',
                data: {machinery:machinery_id},
                dataType: 'json',
                success:function(response){
    
                    var len = response.length;
     
                    $("#sub_machinery").empty();
                    for( var i = 0; i<len; i++){
                        var machinery_id = response[i]['machinery_id'];
                        var machinery = response[i]['machinery'];
                        
                        $("#sub_machinery").append("<option 
    value='"+machinery_id+"'>"+machinery+"</option>");
    
                    }
                }
            });
        });
    
    });
          </script>
    </body>
    </html>

Ajaxfile.php

<?php
    $server = "server";
    $options = array(  "UID" => "user",  "PWD" => "pass",  
    "Database" => "database");
    $conn2 = sqlsrv_connect($server, $options);
    if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
    echo " ";

    $machineryID = $_POST['machinery'];   // department id

    $sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
    
    $result = sqlsrv_query($conn2,$sql);
    
    $machinery_arr = array();
    
    while( $row = sqlsrv_fetch_array($result) ){
        $machinery_id = $row['ID'];
        $machinery = $row['MachineID'];
    
        $machinery_arr[] = array("ID" => $machinery_id, "MachineID" => 
    $machinery);
    }
    
    // encoding array to json format
    echo json_encode($machinery_arr);
    ?>

Edited May 6, 2019 by davidd

Hi all,

I have a html drop down menu and want to show the initially selected value as the one stored in the mysql database that a member selected.

I have 180+ rows in mysql that a member could have selected and would like to basically do this:

IF (value of option == $row['value'] ) { echo "selected=\"selected\"; }
I could put this in every one of the rows it would be much better to have an if statement. As there are so many values I do not know where to begin.

Can someone point me in the right direction?

Thanks