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about php functions


isaac_cm

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If your function works when someone clicks an image on your site, then your function isn't a server-side php function, it's a client-side javascript function.

Care to explain what you're trying to do?
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here is the PHP function I trying to make

<?php
function copy_image_file($source_db_file){
if (!copy($source_db_file, "images/sample.jpg")){
copy("product_img/noimage_large.jpg", "images/sample.jpg");
}
}
?>

now can you tell me what is wrong so my function is executed every time the page is loaded ?
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PHP is executed before the final HTML is being sent to the user's browser. It's being executed on the server-side- your host or your computer.
Javascript is a client-side language, and that means it can be triggered by users click, when a user puts his mouse on something, when a user selects something etc'.

I dont know (but I dont think) Javascript can copy images, but you try and do something with php:
You said you want to copy an image to somewhere, when a user [b]clicks[/b] on a pic. So you can place a link to the pic, something like this (pay attention to bold part):
[b]<a href="copy.php?img=img1&ext=gif">[/b]<img src="img1.gif" border=0></a>
Now as you see, when the user clicks the image, he is directed to "copy.php" trasnfering data with the "GET" method (img=img1).
Now in copy.php, you can do this (your code):
[code]<?php
$img=$_GET['img'];
$img.=".";
$img.=$_GET['ext'];
if (!copy($img, "images/sample.jpg")){
copy("product_img/noimage_large.jpg", "images/sample.jpg");
}
?>[/code]

Then you can have link or a redirection to previous or other page.

Orio.
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That is the only way Javascript can "call" a PHP script in line and continue to execute. It's really not that hard to do. Look at the Yahoo User Interface Library, specifically the [a href=\"http://developer.yahoo.com/yui/connection/\" target=\"_blank\"]Connection Manager[/a].

Ken
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