Not returning selected pciture name - php

Question

I have a database with pictures that i can delete from or use them in different parts of the the website by choosing one of the radio buttons in the popup.Problem is everytime I click I click on a picture the name of the first picture in the database comes up not the one from the picture i clicked . What's wrong? I used PHP 5.3 and HeidiSQL
<?php
// ...
$result=mysql_query("SELECT * FROM imagini WHERE menu_id=6");
while($data=mysql_fetch_row($result)){ ?>
<div class="tag">
<div id='container_poze_originale'>
Alege alt rol
<div id="my_popup" style="display:none;border:1px dotted gray;padding:.3em;background-color:white;position:absolute;width:auto;left:100px;top:100px">
close
<form action="popup.php" method="post" >
<input type="text" name="alt-rol" value="<?php echo $data[1];?>" /> //always returns the name of the first picture in the database not the one i select
<input type="submit" value="Adauga imagine" class="buton_imagine" />
<div class="radio">
<input type="radio" name="tip_imagine" value="0"/><label for="tip_imagine" class="radio2">Logo</label>
<input type="radio" name="tip_imagine" value="1"/><label for="tip_imagine" class="radio2">Slider</label>
<input type="radio" name="tip_imagine" value="2"/><label for="tip_imagine" class="radio2">Hot destinations</label>
<input type="radio" name="tip_imagine" value="3"/><label for="tip_imagine" class="radio2">Pachete</label>
<input type="radio" name="tip_imagine" value="4"/><label for="tip_imagine" class="radio2">Reclama</label>
<input type="radio" name="tip_imagine" value="5"/><label for="tip_imagine" class="radio2">Background</label>
</div>
</form>
</div>
<div class="imagine_originala">
<img src= "../upload/original/<?php echo $data[1];?>" /></a>
</div>
<div class="Btag" style="display:none;">
<div id="buton_slide4" >
Sterge</td>
</div>
</div>
</div>
</div>
<?php
}
?>

Answer 1

I think I understand your question. You are clicking on
Alege alt rol
and you are surprised that the same <div> always appears?
Well, it is simply because an id should be unique. And if you cycle through your images you create a new <div> each time with the same id: my_popup. When clicking on the link (no matter which one), JavaScript will then just take the first <div> it finds with the id my_popup. So you should just assign an unique id to each div so your JavaScript knows which one to open. You could do this by appending your image id to the div id (assuming you have a unique image id in $data[1]).
So change it to (shortened):
<? while($data=mysql_fetch_row($result)) { ?>
<div class="tag">
<div id="container_poze_originale">
Alege alt rol
<div id="my_popup<?=$data[1]?>" style="display:none;border:1px dotted gray;padding:.3em;background-color:white;position:absolute;width:auto;left:100px;top:100px">
...
</div>
</div>
</div>
<? } ?>
Oh and it might be good to put all those huge inline CSS attributes into an external stylesheet, so it will be easier to maintain/change your webpage later on. And it makes things a bit more readable... :-)

Answer 2

I think your form is ok when you click on the radio but not when you click on the label.
The for attribute of the label need an ID as reference like this :
<input type="radio" name="tip_imagine" value="0" id="tip0" /><label for="tip0" class="radio2">Logo</label>
<input type="radio" name="tip_imagine" value="1" id="tip1"/><label for="tip1" class="radio2">Slider</label>
<input type="radio" name="tip_imagine" value="2" id="tip2"/><label for="tip2" class="radio2">Hot destinations</label>
<input type="radio" name="tip_imagine" value="3" id="tip3"/><label for="tip3" class="radio2">Pachete</label>
<input type="radio" name="tip_imagine" value="4" id="tip4"/><label for="tip4" class="radio2">Reclama</label>
<input type="radio" name="tip_imagine" value="5" id="tip5"/><label for="tip5" class="radio2">Background</label>
With your actual code, the FOR is always the same so you get always the first.