php - How to get data from the database and display it in a combobox in html - 


<tr>     <td>subject:</td>     <td>         <select name="subject">             <option value="<?php echo $subject;?>"></option>         </select>     </td> </tr> <?php  $connection = mysqli_connect("localhost", "root", "", "phobiamming") or die (mysqli_error());  $sql = "select * section";  $result = mysqli_query($connection, $sql);   while ($row= mysqli_fetch_array($result)) {     echo $row['subject'];  }   $connection->close();  ?>

i can't display data db combobox. don't know if code right.

    <tr>     <td>subject:</td>     <td><select name="subject">         <option value="subject"><?php $connection = mysqli_connect("localhost", "root", "", "phobiamming") or die (mysqli_error());    // sql query   $strsql = "select *  sections";    // execute query (the recordset $rs contains result)   $rs = mysqli_query($connection, $strsql);    // loop recordset $rs   // each row made array ($row) using mysql_fetch_array   while($row = mysqli_fetch_array($rs)) {       // write value of column firstname (which in array $row)     echo $row['subject']. "<br>";        }    // close database connection   $connection->close(); ?></option>     </select></td>    </tr>

this updated code, can display data db problem rows in db display, can't create line break

put option value in while loop

<tr>     <td>subject:</td>     <td>     <select name="subject"> <?php $connection = mysqli_connect("localhost", "root", "", "phobiamming") or die  (mysqli_error()); $sql = "select * section"; $result = mysqli_query($connection, $sql);  while ($row= mysqli_fetch_array($result)) { ?>         <option value="<?php echo $row['$subject']; ?>"><?php echo $row['$subject']; ?></option> <?php } $connection->close(); ?>     </select>     </td> </tr>

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