Javascript with PHP - problem

[Mortenjan]

Hi.

How can i achieve this?

 

<script language="JavaScript" type="text/javascript">

function CngPicture(bilde){ 
Movies= document.getElementById('movies');
    Index = Movies.selectedIndex; 
Value = Movies.options[index].value;

<?php
$query = "SELECT picture FROM `movies` WHERE id='"?> Value <?php "'"; 
$result = mysql_query($query);
$row = mysql_fetch_array($result);
$picture = $row['picture'];
?>

document.getElementById('bilde').src='<?php echo $picture; ?>';
}
</script>

 

Any help is appreciated.

Regards Morten.

[premiso]

How can I achieve this?

 

That has to be the worst question in the word.

 

Please post EXACTLY what you are trying to achieve in words, what you need help with, what you have tried and not tried and how is it "not working".

[kenrbnsn]

To do what I think you want to do you need to use AJAX techniques.

 

Ken

[Mortenjan]

Yes, i do think i have to use AJAX after what i've read on the web.

 

But i have never used AJAX before.

So any help on building that code would be appreciated.

 

Øhm.. How should i explain my issue..

I am trying to get a link(Picture) from a database where the link is changing depending to the Dropdown value into a javascript code.. The dropdown value is the same as the Movie ID in the database.

 

 

[rhodesa]

or, you need to use a PHP handler script to return the images....let me see if i can explain:

 

in your script, your JS would look like this:

<script language="JavaScript" type="text/javascript">
  function CngPicture(bilde){
    Movies = document.getElementById('movies');
    Index = Movies.selectedIndex;
    Value = Movies.options[index].value;
    document.getElementById('bilde').src = 'movieImage.php?id=' + Value;
  }
</script>

 

then, create a file called movieImage.php with the following code:

<?php
  //Put your MySQL Connection info here

  $id = mysql_real_escape_string($_GET['id']);
  $query = "SELECT picture FROM `movies` WHERE id='$id'";
  $result = mysql_query($query);
  $row = mysql_fetch_array($result);
  header('Location: '. $row['picture']);
  exit;
?>

[Mortenjan]

Oh damn!

It works!

 

Thanks alot!

 

 

 

Hm, the picture got the link: http://www.vinstrakino.co.cc/php/movieImage.php?id=100 on the first selected dropdown (the id is correct)

(the link returns the correct picture, so it has to be something with the javascript?)

 

It didnt return the link from the database.

 

Anything ive done wrong?

 

<script language="JavaScript" type="text/javascript">
  function CngPicture(bilde){
    Movies = document.getElementById('movies');
    Index = Movies.selectedIndex;
    Value = Movies.options[index].value;
    document.getElementById('bilde').src = 'php/movieImage.php?id=' + Value;
  }
</script>

 

 

<?php
  require 'tilkobling.php';

  $id = mysql_real_escape_string($_GET['id']);
  $query = "SELECT picture FROM `movies` WHERE id='$id'";
  $result = mysql_query($query);
  $row = mysql_fetch_array($result);
  header('Location: '. $row['picture']);
  exit;
?>

 

or, you need to use a PHP handler script to return the images....let me see if i can explain:

 

in your script, your JS would look like this:

<script language="JavaScript" type="text/javascript">
  function CngPicture(bilde){
    Movies = document.getElementById('movies');
    Index = Movies.selectedIndex;
    Value = Movies.options[index].value;
    document.getElementById('bilde').src = 'movieImage.php?id=' + Value;
  }
</script>

 

then, create a file called movieImage.php with the following code:

<?php
  //Put your MySQL Connection info here

  $id = mysql_real_escape_string($_GET['id']);
  $query = "SELECT picture FROM `movies` WHERE id='$id'";
  $result = mysql_query($query);
  $row = mysql_fetch_array($result);
  header('Location: '. $row['picture']);
  exit;
?>

[rhodesa]

That code looks fine. And I went to http://www.vinstrakino.co.cc/ and the dropdown seems to work fine. can you elaborate on your problem?

[Mortenjan]

Works fine:)

Was an issue with the database connection.

 

Thanks alot for your help!