如何从数据库中获取值并在下拉列表中显示

Question

I already try many ways but the value didn't show in dropdown list
Here, this is my code. can you suggest me anything that i was wrong

<?php  
$result = mysqli_query($con,"SELECT * FROM project"); 

    if( mysqli_num_rows( $result )==0){
    echo "<tr><td>No Rows Returned</td></tr>";
  }else{
    $row = mysqli_fetch_assoc( $result );
      $pos = 0;
      echo "<select name=Pname >"; 
        while($pos <= count ($row)){ 
      echo "<option value="$row["project_no"]">"$row["project_name"]"</option>";
           $pos++;
          }
echo "</select>";?>

And i write as .php file. Thanks for your help.

Answer 1

Try this out:

$output = '';
if(mysqli_num_rows($result) == 0){
    // echo error;
} else {
    while($row = mysqli_fetch_assoc($result)){
        $project_no = $row['project_no'];
        $project_name = $row['project_name'];

        $output .= '<option value="' . $project_no . '">' . $project_name . '</option>";
    }
}

Then inside of your HTML, print your $output variable inside of your <select> element:

<select>
<?php
    print("$output");
?>
</select>

It should print all options for every row that you have requested from the database.

Hope this helps :)