如何在PHP中的动态下拉列表中显示选定的值

Question

I am trying to retain the value selected in a drop down menu. Everything is working, but I don't know how to show and retain the selected value. How can I do this?

I've got this working using another way:

<?php if($_POST['selClass'] == $row1['class']) echo 'selected="selected"' ?>

but this leads to other problems, i.e. a blank option in my drop down menu.

<form action="" method="POST" name="form1" id="form1">
<select name="selClass" size="1" id="selClass" onchange="form1.submit()">
<option value="">Select a class</option>
<?php
echo "<option value='". "All records". "' . >" . "all records". "</option>";
while ($row1 = mysqli_fetch_array($rs5)) {
echo "<option value='".$row1["class"] ."'>" . $row1["class"]. "</option>";
}
?>
</select>
</form>

Answer 1

You can approach this as

<?php
$selectedOption = '';
if($_POST){
 $selectedOption = $_POST['selClass'];
}
?>
<form action="" method="POST" name="form1" id="form1">
<select name="selClass" size="1" id="selClass" onchange="form1.submit()">
<option value="">Select a class</option>
<?php
    echo "<option value='". "All records". "' . >" . "all records". "</option>";
    while ($row1 = mysqli_fetch_array($rs5)) {
        if($row1["class"] == $selectedOption)
            echo "<option value='".$row1["class"] ."' selected='selected'>" . $row1["class"]. "</option>";
        else
            echo "<option value='".$row1["class"] ."'>" . $row1["class"]. "</option>";
    }
?>
 </select>
</form>