Populate a dropdown from mysql and have the chosen item display in the dropdown after page refreshes-CodePudding

I am using php with a dropdown to select data and display the selected item in the dropdown after the page refreshes so have used this code which works.

<select name="select" id="mysort" onchange="select(this.value);">
    <option value="5"  <?php if($selecting == '5'):?> selected="selected"<?php endif;?>>5 </option>
    <option value="6"  <?php if($selecting == '6'):?> selected="selected"<?php endif;?>>6 </option>
    <option value="7"  <?php if($selecting == '7'):?> selected="selected"<?php endif;?>>7 </option>
    <option value="8"  <?php if($selecting == '8'):?> selected="selected"<?php endif;?>>8 </option>
</select>

I then want to be able to populate the dropdown from a mysql database. This code works:

$sql = "SELECT DISTINCT number FROM files ORDER BY number";
$result = $mysqli->query($sql);
echo '<select name="choose" id="mychoice">';
    if ($result->num_rows > 0) {
        while($row = $result->fetch_assoc()) {
          echo '<option value='.$row["number"].'>'.$row,["number"].'</option>';
        }
        echo '</select>';
    }

I am having trouble with the syntax when I put the two blocks of code together so that the values are populated from the mysql and the selected option displays in the dropdown after the page refreshes. I would be grateful for any help with this.

CodePudding user response：

I think you want this -

    while($row = $result->fetch_assoc()) {
      echo '<option value="'.$row["number"].'"'; 
      if($selecting == $row["number"]) { echo " selected "; }
      echo '>'.$row["number"].'</option>';
    }