fetching values from database and displaying selected value in dropdown list not working in php

Question

I am trying to fetch values from my database and display the selected value in a dropdown list with other values. I have seen many questions here regarding this issue but none of them is working. Please Help!

<select name= "indication_name" id= "update_indication_id" class="form-control" required>
  <?php
      $sql = "SELECT id,description From indications";
      $result=mysqli_query($con,$sql);
      while($row=mysqli_fetch_array($result)){
          echo "<option selected='selected'>" . $row['id']." - ". $row['description'];
      }

   ?>
</select>

So... where are you closing the option tag? pretty obvious here. — Funk Forty Niner
– Funk Forty Niner, Commented Oct 22, 2016 at 18:09
I want the value from the database to be the value of the value of my select — Nadim Obeid
– Nadim Obeid, Commented Oct 22, 2016 at 18:15

Marcelo43 · Accepted Answer · 2016-10-22 18:20:26Z

1

There are three main errors I can see:

You're making every option selected like that.
The options don't have a value.
And the option tags should be closed.

So it would be like this:

<select name= "indication_name" id= "update_indication_id" class="form-control" required>
  <?php
      $sql = "SELECT id,description From indications";
      $result=mysqli_query($con,$sql);
      while($row=mysqli_fetch_array($result))
          echo "<option value='" . $row['id'] . "'>" . $row['description'] . "</option>";
   ?>
</select>

Another thing, that isn't an error, but I personally think is good, is having a "unselected" option:

<select name= "indication_name" id= "update_indication_id" class="form-control" required>
  <option selected="selected" value="">-- Select an option --</option>
  <?php
      $sql = "SELECT id,description From indications";
      $result=mysqli_query($con,$sql);
      while($row=mysqli_fetch_array($result))
          echo "<option value='" . $row['id'] . "'>" . $row['description'] . "</option>";
   ?>
</select>

The value of that option should be null, so that required applies to it.

answered Oct 22, 2016 at 18:20

Marcelo43

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1 Comment

Nadim Obeid Over a year ago

but i need the value from my database. an that is my main problem. when my id=1 i want it to show me x. so x should in my textbox. wven if i have 7 ids

Prashant Kumar Singh · Accepted Answer · 2016-10-23 05:23:28Z

0

<?php 
$server="localhost";
$username="root";
$password="";
$link=@mysql_connect($server, $username, $password) or die ("Cannot connect to mysql server: ".@mysql_error());
$dbname = 'database_name';
@mysql_select_db($dbname, $link) or die ("Cannot connect to database: ".@mysql_error());

$query="SELECT id,description from indications";
$result = @mysql_query ($query);
echo "<select name=indication_name value=' '>";
while($drop=@mysql_fetch_array($result)){
echo "<option value=$drop[id]>$drop[description]</option>";
}
echo "</select>";
?>

edited Oct 23, 2016 at 5:23

answered Oct 22, 2016 at 20:16

Prashant Kumar Singh

619 bronze badges

Comments

baduker · Accepted Answer · 2018-05-07 11:00:14Z

0

Try using switch case and put the value in switch from the database and then set selected ='selected' if the case value matches.

edited May 7, 2018 at 11:00

baduker

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answered May 7, 2018 at 9:46

Vijay

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