Cannot Fetch Dropdown Value From Database

Question

So I want to fetch my data from Database and be able to display and update it. Im on the part where I have to display a value from the db on a Dropdown but only the first option of the dropdown is being displayed.

Below is complete code.

    <?php

  mysqli_connect('localhost', 'root', '');
  mysqli_select_db('storm');

  $_GET['id'];
  $ssh = $_GET['ssh'];
  $_GET['provi'];
  $_GET['impact'];
  $_GET['advice'];
  $_GET['date'];
  $_GET['typhoon'];
  $_GET['warning']; 

?>

<html>

<body>

  <div class="container">
    <form action="edit.php" method="post">

              <div class="form-group">
                <label for="prov">Provinces</label><br>
                <select id="prov" class="form-control" type="text" name="provi1">
                  <option value="Isabela"><?php echo $_GET['provi'];?></option>
                  <option value="La Union"><?php echo $_GET['provi'];?></option>
                  <option value="Pangasinan"><?php echo $_GET['provi'];?></option>
                  <option value="Ilocos Sur"><?php echo $_GET['provi'];?></option>
                  <option value="Ilocos Norte"><?php echo $_GET['provi'];?></option>
                </select>
              </div>

              <div class="form-group">
                <label>Date</label><br>
                <input class="w3-input w3-border form-control" type="date" name="date" value="">
              </div>

            <div class="col-md-6">
              <div class="form-group">
                <label>Typhoon Name</label><br>
                <input type="text" name="typhoon" value="<?php echo $_GET['typhoon']; ?>" class="form-control">
                <input type="hidden" name="id" value="">
              </div>
              <div class="form-group">
                <label>Warning #</label><br>
                <input type="text" name="warning" value="<?php echo $_GET['warning']; ?>" class="form-control">
              </div>
            </div>

          <div class="col">
            <div class="form-group">
              <input class="btnSubmit" type="submit" value="Update" name="submit" style="background-color: #408cff;">
              <input class="btnSubmit" type="reset" value="Cancel" style="background-color: #de5959;">
           </div>
          </div>

</form>

<?php
if(isset($_GET['submit'])){
  $id = $_GET['id'];
  $ssh = $_GET['ssh'];
  $muni = $_GET['muni'];
  $impact = $_GET['impact'];
  $advice = $_GET['advice'];
  $date = $_GET['date'];
  $typhoon = $_GET['typhoon'];
  $warning = $_GET['warning'];

  $query = "UPDATE twothree SET ssh='$ssh', muni='$muni', impact='$impact', advice='$advice', date='$date', typhoon='$typhoon', warning='$warning' WHERE id='$id'";
  $data = mysqli_query($conn, $query);
  if($data){
    echo "Record Updated Successfully!";
  }else{
    echo "Record Not Updated.";
  }
}
?>

I'm pretty sure that I am doing something wrong. Hope you guys figure it out for me. I'm new to this and I hope that I can learn from you guys. Thanks.

What you are trying to achieve?can you you explain bit more about your problem? — navjot singh
– navjot singh, Commented Feb 18, 2019 at 5:54
what will be in dropdown what you fetch? is there any ajax call? what is purpose of code? where is database code? is this core php? — Vasim Shaikh
– Vasim Shaikh, Commented Feb 18, 2019 at 6:01
please provide full code and also explain what you are trying to achieve. — Sayed Mohd Ali
– Sayed Mohd Ali, Commented Feb 18, 2019 at 6:05
if you provide parameter like... URL?provi = "desteen"; then... $_GET['provi'] will display desteen... there is nothing related with you database. for database you need to make database connection then you need to write a fetch query. to get data from database. — Sayed Mohd Ali
– Sayed Mohd Ali, Commented Feb 18, 2019 at 6:08
See about sql injection and the importance of prepared and bound queries — Strawberry
– Strawberry, Commented Feb 18, 2019 at 7:11

Lahiru Madhusanka · Accepted Answer · 2019-02-18 06:02:34Z

2

You can use something like this at basic level. this will retrieve all data from table and add a dropdown option to it.

    <select class="" name="" required>
      <option value="" selected disabled>Select a option</option>
      <?php
      $select_1 = $db->query("SELECT * FROM table");
      while ($row_1 = $select_1->fetch_assoc()) {
      ?>
         <option value="<?php echo $row_1['value']; ?>">
             <?php echo $row_1['name']; ?>
         </option>
      <?php } ?>
   </select>

where $db is your database connection and it must be mysqli connection.

answered Feb 18, 2019 at 6:02

Lahiru Madhusanka

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4 Comments

desteen Over a year ago

I don't quite understand your codes. I edited my post please refer to it. Thank you.

Lahiru Madhusanka Over a year ago

first of all you submitting the form as POST method and you trying to retrieve those from GET method. you should use $_POST or $_REQUEST for that. Then your database connection must assign to a variable like $db = new mysqli(host,user,password,database); Then in my code use select query to retrieve data that you want to retrieve changing table and adding where condition to it.

desteen Over a year ago

what does $select_1 and $row_1 stands for?

Lahiru Madhusanka Over a year ago

$select_1 is the variable that store the output object of the mysql query and the $row_1 is the variable that store the fetch array data of looping $select_1 data.

McBern · Accepted Answer · 2019-02-19 01:40:07Z

You should do like this, to display the list of provinces.

<?php
$host = "localhost";
$user = "root";
$pwd = "";
$db = "storm";
$db_connection = new mysqli($host, $user, $pwd, $db);

if ($db_connection->connect_errno) {
    printf("Connect failed: %s\n", $db_connection->connect_error);
    exit();
}

//select provinces here
$provinces = $db_connection->query("Select * from provinces"); //change provinces according to your table name that you want to query.



?>
             <div class="form-group">
                <label for="prov">Provinces</label><br>
                <select id="prov" class="form-control" type="text" name="provi1">
                  <?php
                  while ($row = $provinces->fetch_object()){
                      echo '<option value="'.$row->province_name.'">'.$row->province_name.'</option>'; //change province_name according to your fieldname.
                  }
                  ?>
                </select>
              </div>

<?php
$db_connection->close();
?>

ahamba solomon · Accepted Answer · 2019-02-19 02:07:30Z

bro you need to loop data form your database to get continuous data out of your database you could use this.

 <?php
$host = "localhost";
$user = "root";
$pwd = "";
$db = "storm";
$db_connection = new mysqli($host, $user, $pwd, $db);

if ($db_connection->connect_errno) {
    printf("Connect failed: %s\n", $db_connection->connect_error);
    exit();
}

//select provinces here
$query = mysqli_query($db_connection,"Select * from provinces");



?>
             <div class="form-group">
                <label for="prov">Provinces</label><br>
                <select id="prov" class="form-control" type="text" name="provi1">
                  <?php
                  while ($row = mysqli_fetch_assoc($query)){
                      echo '<option value="'.$row["province_name"].'">'.$row["province_name"].'</option>'; //change province_name according to your fieldname.
                  }
                  ?>
                </select>
              </div>

<?php
$db_connection->close();
?>

you code should look like this

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