Update data from MYSQL, PHP to drop down button

Question

if(isset($_GET["id"])){
        $sql=mysql_query("SELECT * FROM aMovie WHERE aName= '{$_GET['id']}'"); 
        $row=mysql_fetch_object($sql);  
}

<input type = "text" name = "name" value = "<?php echo $row->aC; ?>"/> 
<select name = "name" >
        <option value = "" <?php echo ($row->aC== "Deadpool") ? 'selected = "selected"': '';?>">Deadpool</option>
        <option value = "" <?php echo ($row->aC == "BATMAN VS SUPERMAN") ? 'selected = "selected"': '';?>">BATMAN VS SUPERMAN</option>
</select>

Assume that aMovie is my table name, and in my table there are aName and and aC. However, I would want to display aName which matches aC ["Deadpool" or "Batman Vs Superman"] and display it in the drop down button. It only works for the input type but not the drop down button.

Answer 1

Your <select> should be like:

<select name = "name" >
<option value="Deadpool" <?=($rows->aC == "Deadpool" ? 'selected="selected"': '')?>>Deadpool</option>
<option value="BATMAN VS SUPERMAN" <?=($rows->aC == "BATMAN VS SUPERMAN" ? 'selected="selected"': '')?>>BATMAN VS SUPERMAN</option>
</select>

selected="selected" will use outside the value attribute.

UPDATE:

As @Maninderpreet-Singh mentioned, you also need to change $row to $rows.

Answer 2

try to change

<option <?php echo($row->aC== "Deadpool") ? 'selected = "selected"': '';?>  value="<?php echo $row->aC;?>">Deadpool</option>

Answer 3

try with this and you are using different variable in input

<input type = "text" name = "name" value = "<?php echo $rows->aC; ?>"/> 

$row and $rows are different

  <option value = "<?php echo $row->aC; ?>" <?php echo ($row->aC == "Deadpool") ? 'selected':'';?>">Deadpool</option>
  <option value = "<?php echo $row->aC; ?>" <?php echo ($row->aC == "BATMAN VS SUPERMAN") ? 'selected': '';?>">BATMAN VS SUPERMAN</option>