根据组合框选择显示结果

Question

I have a dynamic combobox and I have my Fetch button. When a user selects a value from combobox and clicks fetch button, all the other related values are displayed in a textbox for the user to edit and update records. And that works fine.

<form id="form1" method="post" action="edit.php">   
  <select name="ID" id="select">
     <?php display_Id();?>
  </select> 
  <input type="submit" name="Fetch" id="Fetch" value="Fetch" />
</form>

function display_Id() {
    $query = "SELECT * FROM Flight";

    $result = mysql_query($query) or die("Failed to fetch records");
    confirm_query($result);

    while($rows = mysql_fetch_array($result)) {
        $flightNum = $rows['FlightNo'];
        echo "<option value=\"$flightNum\" ";
        echo " selected";
        echo "> $flightNum </option>";
    }       
}

The problem is in the Fetch button. When user clicks Fetch, other values are displaying but the selected value from combobox is refreshing. How to make the values remain selected even after pressing the Fetch button?

Answer 1

Your question is incomplete in the sense, that you don't have your dislay_Id() code shown here. However, Generally speaking, you should add selected after <option value="something" programmatically,

Code should be something like this:

function displayId(){
  if($value[x]== $currentValue) {
    echo "<option value='$value[x]' selected>sth</option>";
  } 
  else
  {
    echo "<option value='$value[x]'>sth</option>";
  }
}

EDIT:: Your code adds a "selected" to each of the values, you must only add a "selected" to a current value.

So, your code must look like this:

echo "<option value=\"$flightNum\" ";

if($_POST['ID'] == $flightNum)
{        
  echo " selected";
}
echo "> $flightNum </option>";

Answer 2

while($rows = mysql_fetch_array($result))
{
    $flightNum = $rows['FlightNo'];
    echo "<option value=\"$flightNum\" ";
    if($_POST['ID'] == $flightNum)
    {
        echo " selected";
    }
    echo "> $flightNum </option>";
}