Re: Problem in passing the necessary variable.

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Ciao..hi, i would suggest to embed the html code in php like this:

<?php
$_your_variable=$row["imgid"];
echo
        "<a href=\"../images/image.php?img=$_your_variable\"
target=\"_blank\"
         onClick=\"window.open(this.href, this.target,
\'width=200,height=250\');
         return false;\"><img SRC=\"../preview.jpg\" ALT=\"Click to look at
the picture
         you are charging\" border=\"0\"></a>";
?>

If it works let me know , bye

Francesco Basile
______________________________________________________________________

"Alessandro Folghera" <info@xxxxxxxxxxx> ha scritto nel messaggio
news:40A257AF0010BE1E@xxxxxxxxxxxxxxxxxxxxx
> Hi,
>
> I have developed a news system .. I can select to insert an image into the
> articles.
>
> The function to extract the image calling them by the "imgid" of images
> table is the following:
>
> function getImm() {
> ?>
>       <select name="image">
>         <?
> connect();
> $sql = "select * from images";
> $result = mysql_query($sql);
> while($row = mysql_fetch_array($result)) {
> printf("<option value=\"%s\">%s</option>", $row["id_img"], $row["imgid"]);
> }
> ?>
>       </select>
>     <?
>
> I'd like to be able to see the image before to publish the piece of news.
>
> I think about a window pop up with java in order to create a preview of
the
> image. The code follows:
>
> <a href="../images/image.php?img=XXXXXXXXXX" target="_blank"
> onClick="window.open(this.href, this.target, 'width=200,height=250');
> return false;"><img SRC="../preview.jpg" ALT="Click to look at the picture
> you are charging" border="0"></a>
>
> The problem is that I can't pass the "imgid" to the href inspite of the
> XXXXXXXX
>
> May someone tell me how to pass the id number in order to show me the
image
> or how may I solve the trouble?
>
> I hope someone will help me ...
>
> Sincerely,
> Alexander

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