PHP - How To Add Selected Dropdown Option In Database With Php

Hi,

I wanted to make a dynamic php page. I want the user to choose an option you could select in an option box (html) and after that I wish that down the page some text would change. If I take option 1 -> text 1, option 2 -> text 2, ...
But this doesn't work, do I need a button or something to made my code running, and how do I have to use it on the same page?
Or is it possible to run it automaticly after a new choice?

Thanx, Kevin



My code:

<?php
/*
Template Name: template voor Ieder1Trainer #
*/
get_header(); 

$Tacktiek = "";

$Tacktiek1 = "3-5-2";
$Tacktiek2 = "3-4-3";
$Tacktiek3 = "4-5-1";
$Tacktiek4 = "4-4-2";
$Tacktiek5 = "4-3-3";
$Tacktiek6 = "5-4-1";
$Tacktiek7 = "5-3-2";

$aantalDef;
$aantalMid;
$aantalAtt;

?>

<div id="newsfeed">

<div class="banderol">
<div class="banderol_left"></div>
<h1>Dynamic page</h1>
</div>

<?php if(is_user_logged_in()) { 
get_currentuserinfo();
//echo 'Username: ' . $current_user->user_login . "\n";
//echo 'User email: ' . $current_user->user_email . "\n";
?>

<!-- 
Here you choose a tactiek; 1,2,3, .. after that has been choosen I whise the next php-code would be run (the if-then-elseif-...)
-->

<tr>
<td><p class="formulier">Kies je Tacktiek:</p></td>
<td>
<select size="1" name="$Tacktiek">
<option selected><?php echo $Tacktiek1;?></option>
<option><?php echo $Tacktiek2;?></option>
<option><?php echo $Tacktiek3;?></option>
<option><?php echo $Tacktiek4;?></option>
<option><?php echo $Tacktiek5;?></option>
<option><?php echo $Tacktiek6;?></option>
<option><?php echo $Tacktiek7;?></option>
</select>
</td>
</tr>

<!--
This must be run after choosen a tactiek.
-->

<?php
if($Tacktiek == '3-5-2') {
$aantalDef = 3;
$aantalMid = 5;
$aantalAtt = 2;
} elseif($Tacktiek == '3-4-3') {
$aantalDef = 3;
$aantalMid = 4;
$aantalAtt = 3;
} elseif($Tacktiek == '4-5-1') {
$aantalDef = 4;
$aantalMid = 5;
$aantalAtt = 1;
} elseif($Tacktiek == '4-4-2') {
$aantalDef = 4;
$aantalMid = 4;
$aantalAtt = 2;
} elseif($Tacktiek == '4-3-3') {
$aantalDef = 4;
$aantalMid = 3;
$aantalAtt = 3;
} elseif($Tacktiek == '5-4-1') {
$aantalDef = 5;
$aantalMid = 4;
$aantalAtt = 1;
} elseif($Tacktiek == '5-3-2') {
$aantalDef = 5;
$aantalMid = 3;
$aantalAtt = 2;
}

?>

<!--
After this run, the next code must be run and made a choose of text
-->

<tr>
<td><p class="formulier">
<?php if($aantalDef == '3') {
echo "3 defs";   //Choosen text if aantalDef == 3
}
elseif($aantalDef == '4') {
echo "4 defs";   //Choosen text if aantalDef == 4
}
elseif($aantalDef == '5') {
echo "5 defs";   //Choosen text if aantalDef == 5
}
?>
</p></td>

<!--
Other choose of options, but always the same options.
-->
<td>
<select size="1" name="$Speler4">
<option selected><?php echo $NaamDef1;?></option>
<option><?php echo $NaamDef2;?></option>
<option><?php echo $NaamDefMid1;?></option>
<option><?php echo $NaamDefMid2;?></option>
<option><?php echo $NaamDefMid3;?></option>
<option><?php echo $NaamDefMid4;?></option>
<option><?php echo $NaamDefMid5;?></option>
<option><?php echo $NaamDefMid6;?></option>
<option><?php echo $NaamDefMid7;?></option>
<option><?php echo $NaamDefMid8;?></option>
<option><?php echo $NaamDefMidAtt1;?></option>
<option><?php echo $NaamDefMidAtt2;?></option>
<option><?php echo $NaamDefMidAtt3;?></option>
</select>
</td>
</tr>


<?php
} 
else{
echo 'U dient in te loggen.';
}
?>


</div>

<?php get_footer(); ?>

If I have a variable and a hard coded select dropdown, what is the easiest way to compare the variable with the value and set that particular option's selected attributed to selected?

I'm trying to create a dynamic option menu with one alert selected based on the first query to the db. Any help would be greatly appreciated. 

Code: [Select]
//function to get alerts and create select menu with current alerts pre-selected
function getALERTS1($id){
    require('db.php');
    $alert = mysqli_query($conn, "SELECT alert1 FROM visit_data WEHERE patientid = $id AND discharged IS NULL");
    $row = mysqli_fetch_array($alert);
   
    $selects=null;
    $query = mysqli_query($conn, "SELECT alertid, name FROM alerts");

while($row1 = mysqli_fetch_array($query)) {
    $selects .= "<option value=\"" . $row1['alertid'] . "\">
    if($row1['alertid']==$row['alert1'])
            {
                echo ' selected';
            }
   
    ".$row1['name']."</option>";
  }
  return $selects;
}

Not sure exactly where this problem is coming from, but php seems the most likely candidate to fix it.  My php loads, from a MySQL database, a list of names (students in a class) and a drop-down for each student with attendance options.  Based on the information in the database, the appropriate attendance option is selected (<option selected></option>).  Everything works fine on the initial page load. 

Next, I want to update the status, so I select a new option from the drop-down list and click submit.  The php code updates the database correctly, but the selected option (in HTML) is still the original one.  If I change the option to something different and submit again, I get the previously selected option (that I selected the first time) - in essence, the HTML is always one step behind the database.

I've attached the snippet of relevant code - again, I know that the code works when the page loads initially and that the database is getting updated correctly.  Any help or comments are appreciated.

<?php  
                              if(mysqli_num_rows($result) )  
                          {  
                               while($row = mysqli_fetch_array($result))  
                               {  
                          ?>  
                          <form method="POST">
                              <select name="inv"> 
                               <option><?php echo $row["rizikos_lygis"];?></option>
                               <option value=<?php echo $row["sugeneruoja"]; ?>><?php echo $row["pavadinimas"];?>&nbsp<?php echo $row["sugeneruoja"]; ?></option>
                              </select>
                              <input type="submit" name="submit" value="Apskaičiuoti">    
                          </form>
                          <?php  
                               }  
                          }  
  ?> 
<?php
if(isset($_POST['submit'])){
 echo $row["sugeneruoja"];
}
?>

Edited January 27 by Lukeishen

I have a form that contains 3 text fields and 7 populated lists (that are dependent on the choice of the previous lists).  When I post the information to MySQL and to my processing page, it posts the table "ids" instead of the selected option. 

Here's the php code for the form:

<form action="gccsuccess_form.php" method="post" enctype="multipart/form-data" name="gcc" id="gcc">
  <table width="700" border="0" align="center" cellpadding="3" cellspacing="3">
    <tr>
      <td><label for="date">Today's Date:</label>
 <input type="text" name="date" id="date"></td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>Week Number:
       <select name="week" size="1" id="week">
         <option value="0" selected>Week 0 - 3/26 - 4/1</option>
         <option value="1">Week 1 - 4/2 - 4/8</option>
         <option value="2">Week 2 - 4/9 - 4/15</option>
         <option value="3">Week 3 - 4/16 - 4/22</option>
         <option value="4">Week 4 - 4/23 - 4/29</option>
         <option value="5">Week 5 - 4/30 - 5/6</option>
         <option value="6">Week 6 - 5/7 - 5/13</option>
      </select></td>
      <td><label for="location"></label>
      </td>
    </tr>
    <tr>
      <td>Location:
        <select id='locations' name='location'>
      </select></td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td><label for="label8">Team Name</label>
<select id='team_names' name='team_name'>
</select></td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td><em>*Please enter each team member's first and last name:</em></td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td><label for="mem_1"> Team <strong>Captain:</strong></label>
        <select id='team_members 1' name='capt'>
        </select></td>
      <td><label for="label4"><strong>Steps:</strong></label>
        <input type="text" name="steps_1" id="steps_1"></td>
    </tr>
    <tr>
      <td><label for="mem_2">Team Member #2 <strong>Name: 
        <select id='team_members 2' name='mem_2'>
        </select>
      </strong></label></td>
      <td><label for="label5"><strong>Steps:</strong></label>
        <input type="text" name="steps_2" id="steps_2"></td>
    </tr>
    <tr>
      <td><label for="mem_3">Team Member #3 <strong>Name:
            <select id='team_members 3' name='mem_3'>
        </select>
      </strong></label></td>
      <td><label for="label6"><strong>Steps:</strong></label>
        <input type="text" name="steps_3" id="steps_3"></td>
    </tr>
    <tr>
      <td><label for="mem_4">Team Member #4<strong> Name</strong>:
        <select id='team_members 4' name='mem_4'>
        </select>
      </label></td>
      <td><label for="label7"><strong>Steps:</strong></label>
        <input type="text" name="steps_4" id="steps_4"></td>
    </tr>
    <tr>
      <td><label for="label">Team Member #5<strong> Name</strong>:
        <select id='team_members 5' name='mem_5'>
        </select>
      </label></td>
      <td><label for="label7"><strong>Steps:</strong></label>
          <input type="text" name="steps_5" id="steps_5"></td>
    </tr>
    <tr>
      <td>&nbsp;</td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>&nbsp;</td>
      <td><em>* Maximum steps for each team member per week is <strong>105,000. </strong></em></td>
    </tr>
    <tr>
      <td>&nbsp;</td>
      <td><input type="submit" name="Submit Form" id="Submit Form" value="Submit"></td>
    </tr>
  </table>
  <label></label>
  <div align="center"><br>
    If you have questions about the competition or reporting please contact KC   WELLNESS, INC&nbsp;toll free at <SPAN id="lw_1237248942_0">877-634-1412.</SPAN><br>
    <br>
  </div>
</form>


And here's the code for the page it is posted in:

<?php
      
      $date = $_POST['date'];
      $week = $_POST['week'];
      $location = $_POST['location'];
      $team_name = $_POST['team_name'];
      $capt = $_POST['capt'];
      $mem_2 = $_POST['mem_2'];
      $mem_3 = $_POST['mem_3'];
      $mem_4 = $_POST['mem_4'];
      $mem_5 = $_POST['mem_5'];
      $steps_1= $_POST['steps_1'];
      $steps_2= $_POST['steps_2'];
      $steps_3= $_POST['steps_3'];
      $steps_4 = $_POST['steps_4'];
      $steps_5 = $_POST['steps_5'];
      
      mysql_connect("mysql1.myregisteredsite.com", "49200_kandace", "Leonardo56")
      or die(mysql_error());
      mysql_select_db("49200_GCC") or die(mysql_error());
      mysql_query("INSERT INTO `walkathon` VALUES ('$date', '$week', '$location', '$team_name', '$capt','$mem_2','$mem_3',  '$mem_4', '$mem_5', '$steps_1','$steps_2', '$steps_3','$steps_4', '$steps_5' )"); 

      Print "Thank you for your entry. <br>
      Report summary: <br/><br/>";    
      
         echo "<strong> Date: </strong>   {$date} 
         <br/> <br/><strong> Week: </strong>   {$week}
         <br/> <br/><strong> Location: </strong>  {$location}
         <br/> <br/><strong> Team Name: </strong>   {$team_name}
         <br/> <br/><strong> Team Captain: </strong>  {$capt} / <strong> Steps: </strong> {$steps_1} 
         <br /> <br/><strong> Team Member 2 Name: </strong>  {$mem_2} / <strong> Steps: </strong> {$steps_2} 
         <br /> <br/><strong> Team Member 3 Name: </strong>  {$mem_3} / <strong> Steps: </strong> {$steps_3}
         <br /> <br/><strong> Team Member 4 Name: </strong>  {$mem_4} / <strong> Steps: </strong> {$steps_4}
         <br /> <br/><strong> Team Member 5 Name: </strong>  {$mem_5} / <strong> Steps: </strong> {$steps_5}"
         
               ?>

How do I get the processing page to echo the selected option instead of the id?

Hi, I have an html form in "send.htm" that sends data to "retrieve.php" with the following structu

send.htm:
Code: [Select]
<html>
<body>

<form method="post" action="retrieve.php"
<select id='select' name='select' >
                        <option>option1</option>
                        <option>option2</option>
                        <option>option3</option>
                        <option>etc...</option>
</select>
<input type="submit">
</form>

</body>
</html>
retrieve.php:
Code: [Select]
<?php

$selectedoption        = $_POST['select']; 

print"

<html>
<body>
<select id='select' name='select' >
                        <option>option1</option>
                        <option>option2</option>
                        <option>option3</option>
                        <option>etc...</option>
</select>
</body>
</html>
";
?>
How do I specify that the select menu in retrieve.php should have the option selected that was chosen from the select menu in send.htm?

I'm guessing this is probably quite straightforward, but I can't quite work it out. Please help!
Thanks.

Hi I have a list of states using the array method in a form. The drop down menu works fine.  I want to save the user choice,if the form is re-displayed due to a blank field or pattern mismatch.  I know I can use the selected=selected, but don't know wher to put the statement:
My array is:
state_province = array ("list of states", "provinces")
Var in my labels array is "state"=>"state"
Here is my code for the select/option statement:

{
 if($field == "state")
   {
      echo "<div class='province_state'><label for='state' size='10'>* Province/State</label><select>";
            
      foreach($state_province as $state)
         {
         echo "\n<option value='$state_province' /> ";   
          echo $state ;
          echo "</option>";
         }
      echo "</select></div>\n";   

   }
?Is this the correct code to add and where would I add  it?
          if(@$_POST['state'] == $value)
     {
         echo "selected='selected'  ";
     }          

The rest of my coding works however this part does not and I'm trying to figure out why. I'm sure my syntax isn't right so I hope someone can correct my mistake.


$contentpageID = $_GET['id'];
    $query = "SELECT  
        contentpages.contentpage, 
        contentpages.shortname,
        contentpages.contentcode,
        contentpages.linebreaks, 
        contentpages.backlink,
        contentpages.showheading,
        contentpages.visible,
        contentpages.template_id
    FROM 
        contentpages
    WHERE 
        contentpages.id = '" . $contentpageID . "'";

<label for="template">Template</label>
                <select class="dropdown" name="template" id="template" title="Template">
                <option value="0">- Select -</option>
                <?php
                $query = 'SELECT * FROM templates';
                $result = mysql_query ( $query );
                while ( $template_row = mysql_fetch_assoc ( $result ) ) {
                        print "<option value=\"".$template_row['id']."\" ";
if($template_row['id'] == $row['template_id']) {
     print " SELECTED";
}
                        print ">".$template_row['templatename']."</option>\r";
                }
                ?>
                </select>

Hi

I have been trying to get a value to be selected in a mysql populated dropdown list but can't get it to work and was hoping someone could help

I have a database with user info in it and this is an update page where they can update their details. The code i have (which doesn't work) is:

<select name="agency">
<? $query1 = mysql_query("SELECT * FROM agents ORDER BY agent ASC",$connect);
while($myrow = mysql_fetch_assoc($query1)){
$agent = $myrow['agent'];
echo "<option"; if ($agent == $agency) { echo "selected='selected'"; } echo ">$agent</option>";
}
?>
</select>

The $agency value is the current agency which is stored in the users profile and the value does exist in the list which is being populated (also, i have define $agency further up in my code)  so i don't know why the selected value won't display. No value is displayed in the dropdown list on the page - but the values are in the list if i remove the selected='selected' part of the code.

Any help yould be greatly appreciated.

Merry Christmas
Andy

I must be missing something simple. I've got this little script that pulls rows from the database to populate a dropdown. If one of them matches a predetermined variable, then I want it to show selected. It's... almost working.

The dropdown prints, and shows the options. But the selected item shows separate, printed just below the dropdown?

Here's what I've got:

Code: [Select]
<?php
$quer3=mysql_query("SELECT discount_id, discount_name,discount_amount FROM tbl_discount order by discount_amount");
echo "&nbsp; &nbsp; <select name='discount_id'><option value=''>Select</option>";
while($row = mysql_fetch_array($quer3)) { 
if($row[discount_id]==$discount_id){echo "<option selected value='".$row[discount_id]."'>".$row[discount_name]."&nbsp; / &nbsp; $".$row[discount_amount]."</option>";}
else{echo  "<option value='$row[discount_id]'>$row[discount_amount]</option>";}
echo "</select>";
} ?>

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

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

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?

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?