Help with form

noswad · Jul 11, 2006

Hi

I have this code which is for a drop down box

PHP:

<table border="0" cellspacing="0" cellpadding="0"> 
<tr> 
<td width="100%"><form name="mygallery"><p><select 
name="picture" size="1" onChange="showimage()"> 
<option selected value="../mobiles/images/handsets/6230i.jpg">Nokia 6230i</option> 
<option selected value="../mobiles/images/handsets/k750i.jpg">Sony Ericsson K750i</option> 
<option selected value="../mobiles/images/handsets/6280.jpg">Nokia 6280</option> 
</select></p> 
</form> 
</td> 
</tr> 
<tr> 
<td width="100%"><p align="center"><img src="images/handsets/1100.jpg" name="pictures" width="88" 
height="130"></td> 
</tr> 
</table>

As you can see, an image displays when a user selects an object from the drop down box.

My problem is that I don't want the image location to be the object value. Is it possible for me to choose my own values whilst still displaying the images?

Thanks for reading :hehe:

noswad · Jul 11, 2006

Anyone please. It's killing me

:hehe:

monaco · Jul 12, 2006

That's pretty easy. Set the value to whatever you want and pass the value to the function. Then check that value against a db table or whatever storage method ya use to display the image.

noswad · Jul 13, 2006

monaco said:
That's pretty easy. Set the value to whatever you want and pass the value to the function. Then check that value against a db table or whatever storage method ya use to display the image.

Thanks for the reply.

Is it not possible to do this using a similar style to above?

i.e.

PHP:

<value="Nokia 6280" display image="../mobiles/images/handsets/6280.jpg">Nokia 6280</option>

...because I won't have a clue how to do this with a db.

:hehe:

monaco · Jul 13, 2006

Well, you could hardcode it into the dsplay function:

Code:

function foo( value ) {
  if( value == "something" ) {
    // use image a
  }
  if( value == "something else" ) {
    // use image b
  }
}

noswad · Jul 13, 2006

Hi

Thanks for your help.

Im now receiving the following error:

Parse error: parse error, unexpected T_STRING, expecting ')' in /homepages/5/d170332235/htdocs/gimp.php on line 15

Cheers.

CreedFeed · Jul 13, 2006

What's line 15?

noswad · Jul 13, 2006

Hi

Its the top line...

function foo( value ) {

Cheers

Flubber · Jul 13, 2006

Its a variable aye? All instances should be $value.

Flubb.

noswad · Jul 13, 2006

Hi.

Still no luck guys. I'm bamboozled.

This code has NO errors but doesn't show the images

PHP:

<?  
  function foo( $phone ) { 
if( $phone == "Nokia 6230i" ) {
  echo '<img src="/mobiles/images/handsets/6230i.jpg" />';  
} elseif( $phone == "Nokia 6280" ) { 
  echo '<img src="/mobiles/images/handsets/6280.jpg" />'; 
  } 
} 
  ?>

Thanks for trying

Flubber · Jul 13, 2006

Are you actually calling the function at the options?

Flubb.

noswad · Jul 13, 2006

I don't follow you. I just have that code in the body which produces the table, and when I select an option (using my browser), no pictures display.

Sorry - Im still quite new to PHP :td:

Thanks for helping :hehe:

edmarriner · Jul 13, 2006

Is this corrent.

There your page with a drop down on it.

There are are say 5 options, each phones.

When the user chooses one of the phones you want the picture to kind of pop up or somthing?

Is that what your looking for?

noswad · Jul 13, 2006

Yes that is correct but I don't want the value of each item to be the image location as what is shown in my first post at the top of this page.

I need to be able to name the values what I want.

Thanks :hehe:

noswad · Jul 14, 2006

I've done it! WOOOOOOOOO :hehe:

The code I used was:

In the head:

PHP:

<script language="javascript">
<!--
function showimage()
{
if (!document.images)
return
document.images.pictures.src=
document.form.gift.options[document.form.gift.selectedIndex].imagelocation
}
//-->
</script>

and in the body...

PHP:

<img src="../mobiles/images/handsets/6230i.jpg" name="pictures" width="84"
height="130">                    <span class="gift_sub_name">
                    <P align="center"> 
                      
                      <select name="gift" size="1" id="gift" onChange="showimage()">
                        <option selected imagelocation="../mobiles/images/handsets/6230i.jpg" value="Nokia 6230i">Nokia 6230i</option>
                        <option imagelocation="../mobiles/images/handsets/6030.jpg" value="Nokia 6030">Nokia 6030</option>
                        <option imagelocation="../mobiles/images/handsets/6101.jpg" value="Nokia 6101">Nokia 6101</option>
                      </select>

:sold:

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