Fetching data from MySQL database to html drop-down list

Question

I search for answers here but haven\'t found a solution. I have also added picture of the error. I want the data to go to the first drop-down list ( above the error) I think th

Answer 1

<div class="row form-group">
<div class="col col-md-3">
<label for="email-input" class=" form-control-label"> Vehicle</label>
</div>
<div class="col-12 col-md-9">

<select name="car_id" id="car_id" class="form-control-label" >
<?php
$list = mysqli_query($conn,"SELECT * FROM `vehicle_registration` where `status`='0' ");
while ($row_ah = mysqli_fetch_assoc($list)) {
?>
<option value="<?php echo $row_ah['id']; ?>"><?php echo $row_ah['car_no']; ?></option>
<?php } ?>
</select>

</div>
</div>

Answer 2

In MySQLi, the first parameter of a query needs to be the database connection. Also, there's no need to add a \ before the statement.

$sql = \mysqli_query("SELECT name From users"); should be $sql = mysqli_query($con, "SELECT name From users");

Note: replace $con with your database connection variable!

As you mentioned that you wanted the result from the database to go inside the select form, simply adjust your code to look like this:

<select name="to_user" class="form-control">
<option value="pick">בחר מהרשימה</option>
<?php
$sql = mysqli_query($con, "SELECT name From users");
$row = mysqli_num_rows($sql);
while ($row = mysqli_fetch_array($sql)){
echo "<option value='". $row['name'] ."'>" .$row['name'] ."</option>" ;
}
?>
</select>

Answer 3

<label><b>Select Steam: </b></label>
<select id="study">
    <option value="" selected="selected" disabled="">---Selected---</option>
    <?php
    $query = "SELECT study FROM details";
    $query_run = mysqli_query($con, $query);

    while ($row = mysqli_fetch_array($query_run)) {
        echo "<option value='".$row['study']."'>".$row['study']."</option>";
    }

?>
</select>