Create a PHP Dropdown menu from a for loop?

Question

I am trying to create a drop down menu with the options of 1,2,3 and 4.

The below code is what I am using just now and the dropdown is empty.

Any idea what I

Answer 1

This worked for me. It populates years as integers from the current year down to 1901:

        <select Name='ddlSelectYear'>
            <option value="">--- Select ---</option>

            <?php
            for ($x=date("Y"); $x>1900; $x--)
              {
                echo'<option value="'.$x.'">'.$x.'</option>'; 
              } 
            ?> 
        </select>

Answer 2

<select name="year">
    <?php for ($i = 0; $i <= 9; $i++) : ?>
         <option value='<?= $i; ?>' <?= $fetchData->year == $i ? 'selected' : '' ?>><?= $i; ?></option>
    <?php endfor; ?>
</select>

Answer 3

Just echo the <option> tag

echo "<option value=".$i.">".$i."</option>";

Answer 4

You are not outputting the option tags. Try it like this:

<select name="years">

<?php 

for($i=1; $i<=4; $i++)
{

    echo "<option value=".$i.">".$i."</option>";
}
?> 
     <option name="years"> </option>   
</select> 

<input type="submit" name="submitYears" value="Year" />

Answer 5

place an echo in your loop to output your options.

echo "<option value=".$i.">".$i."</option>";

Answer 6

You forgot something..

Add print / echo before "<option value=".$i.">".$i."</option>";