How to make a dropdown stay when processing different If/Else blocks? - php

Question

I'm new to PHP-to-Mysql so I'm using the mysql_* function which others have stated to be unsupported anymore. But I find this place the best to ask my question since I can't find anyone with a similar question.
So basically, my problem is how to make a dropdown list stay which is inside an IF/ELSE block. When it is passes a value (submit) through another IF/ELSE block, it basically disappears so I'm trying a code where I don't have to re-write the entire dropdown code inside the next IF/ELSE block. Since what I'm trying is to show a textbox depending on what option is selected inside the dropdown (it has an auto-submit function). Is this possible? Or I have to resort to other libraries (jQuery,javascript etc.)?
Here's a part of the code if it helps:
elseif($_POST['question'] == edit ){
$res = mysql_query("SELECT questionID FROM questions");
echo "<form action='' method='post'>
Edit Question No.
<select name='question_select' onchange=this.form.submit()>
<option value=null selected>--</option>";
while($row = mysql_fetch_array($res))
{
echo "<option value=\"".$row['questionID']."\">".$row['questionID']."</option>";
}
echo "</select></form>";
}
//TEXTBOX TO BE DISPLAYED WHEN A NUMBER IS SELECTED IN EDIT EXISTING
if (isset($_POST['question_select'])){
$res = mysql_query("SELECT question FROM questions WHERE questionID='{$_POST['question_select']}'");
$row = mysql_fetch_assoc($res);
echo "Editing Question ",$_POST['question_select'],"<br>
<form action='' method='post'>
<textarea name='edited_question' rows='4' cols='50'>",$row['question'],"</textarea><br>
<input type='submit' name='save' value='Save'>
<input type='hidden' name='question_num' value='{$_POST['question_select']}'>
<input type='submit' name='cancel' value='Cancel'>
</form>";
}
//PASSING OF VALUES WHEN SAVE IS PRESSED
if (isset($_POST['save'])){
$edited_question = trim($_POST['edited_question']);
mysql_query("UPDATE questions SET question='$edited_question' WHERE questionID='{$_POST['question_num']}'") or die(mysql_error);
header("Location:admin_questions.php");
}

Answer 1

use jquery for that.it is easy and simple to use.
for depanding dropdown,jquery code is:
$(document).ready(function(e) {
$("#abc").change(function()
{
var firstfeild= $("#firstfeild").val();
$.post("getdata.php",{"firstfeild":firstfeild},function(data)
{
$("#abc").html(data);
});
});
});
and in getdata.php, just write mysql query.

Answer 2

function renderDropdown($res)
{
echo "
<select name='question_select' onchange=this.form.submit()>
<option value=null selected>--</option>";
while($row = mysql_fetch_array($res))
{
echo "<option value=\"".$row['questionID']."\">".$row['questionID']."</option>";
}
echo "</select>";
}
if(1==1)
{
echo "<form action='' method='post'>Edit If Question No.";
renderDropdown($res);
echo "</form>";
}
else
{
// $someOtherRes
echo "<form action='' method='post'>Edit Else Question No.";
renderDropdown($someOtherRes);
echo "</form>";
}