view my updating data from mysql to dropdown list

Question

I want to view some foods which I added through html and php code into database like food a, food b, food c, food d.

I want to view these food category in my dropdown list in another page of html as a updating food category .

<?php
$db_host = 'localhost'; // Server Name
$db_user = 'root'; // Username
$db_pass = ''; // Password
$db_name = 'rfid'; // Database Name

$conn = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if (!$conn) {
 die ('Failed to connect to MySQL: ' . mysqli_connect_error()); 
}

?>  
  <div class="sign_up">
  
    <div class="formtitle" align="center">View Food Items </div>
    <div class="top_section">
     <div class="section">
     <br>
      <div class="input-sign details2">
      <form name="form1" action="" method="post">
       &nbsp;&nbsp; View Foods: 
         <?php
        if ($result = @$dbconn->query("SELECT * FROM `food`")) 
        {
        //`food` represents the table name.
          echo "<select>";
          while ($row = $result->fetch_assoc()) 
    {
            echo "<option value=\"".$row["food"]."\">".$row["food"]."</option>";
            /*"id" and "foodname" represents the table column name.*/
          }
          echo "</select>";
        }
       ?>
       <select>
        <option><?php echo $row["food"];?></option>
       </select>
       
       </form>
      </div>      
     </div>    
     </div>
 <div class="clear"> </div>
  </div>

bro stil now I'm trying its my final year project at any cost I have to complete that's why I seeking help — shibin JOHN, Mar 10 '18 at 20:25
@shibinJOHN We are _still_ not your personal homework service. Perhaps you should do your own final year project work. — Tersosauros, Mar 16 '18 at 15:36

TheE · Answer 1 · 2018-03-11T09:01:51.813

0

So I'm assuming you know how to connect a db and use it.

Here is an example of what you want.

        if ($result = $dbconn->query("SELECT * FROM `food`")) {
          //`food` represents the table name.
          echo "<select>";
          while ($row = $result->fetch_assoc()) {
            echo "<option value=\"".row["id"]."\">".row["foodname"]."</option>";
            /*"id" and "foodname" represents the table column name.*/
          }
          echo "</select>";
        }

edited Mar 11 '18 at 09:01

answered Mar 10 '18 at 17:10

TheE

328
4
13

Using error suppressors in development environment is a terrible idea – Rotimi Mar 10 '18 at 17:14
@Akintunde007 What do you mean? – TheE Mar 10 '18 at 17:21
`@$dbconn` <--- – Rotimi Mar 10 '18 at 17:56
@Akintunde007 Oh yes, true. I'm just kinda used to it, because of previous work. – TheE Mar 10 '18 at 19:02
its not working bro, I used to do like same its not working – shibin JOHN Mar 10 '18 at 20:10
@shibinJOHN Check the php error logs and see what you get, I'd recommend as of now. – TheE Mar 10 '18 at 21:51
bro there is no error ... but what I want that one is not getting .... its simply showing my page .... dropdown not get any value – shibin JOHN Mar 11 '18 at 04:15
@shibinJOHN It would be cool if you showed what code you have and the result you are getting. – TheE Mar 11 '18 at 09:02

Xer Charles · Answer 2 · 2018-03-12T03:13:15.047

0

It might have something to do with your quotations, please try echo "<option value=\"".$row['food']."\">".$row["food"]."</option>"; In addition, I found this similar questions which may help you solve your problems. Credits to author @SpaceBeers

edited Mar 12 '18 at 03:13

answered Mar 12 '18 at 03:05

Xer Charles

251
1
6

view my updating data from mysql to dropdown list

2 Answers2