how to call a php function in html? working wit thumbnails?

rekik

Dear all,I am so not good in programming I need help to finish my assignment asap. I have tried to write with my bad programming skill a function in php to get thumbnail size of the pictures from database when I upload image files and and a default image Notepad.gif if i upload non-image files.here is the function and underneath it is I am trying to call the function in a separate html template inside a table data and it is not showing any result. the name of the php file where the function is edit.php please tell me how i should write the function and why function calling is not working.

function defaultThumb($file_types_array){
$file_types_array=array("txt","jpg","doc","php","css","pdf","gif","JPG");
if(!(($file_types_array=="jpg")&&($file_types_array=="gif"))){
$thumbnail="<img_src='./pictures/Notepad.gif' border='0' height='30' width='30' align='right'>";
return $thumbnail;
}else{
$thumbnail="<img src='./attachments/$relationId.'call'.$id/$attachmentData->filename' border='0' height='30' width='30' align='right'>";
return $thumbnail;
}

}

  <td>

<?php
include("edit.php");
$thumbimage=defaultThumb($file_types_array);
?>
</td>

leatherback

Hi,

A few points:
- Unless you have set your webserver to process .HTML as php files, you should give your file a php extention

here: if(!(($file_types_array=="jpg")&&($file_types_array=="gif"))){ you are checking whether file_types_array is both jpg and gif, which will never be true. Two options: You either have one variable, and you use:
if(!(($file_types_array=="jpg")||($file_types_array=="gif"))){ (|| = or)
Or you have the option of passiong an array, and use in_array:
if(!((in_ARRAY($file_types_array, "jpg"))&&($file_types_array=="gif"))){
when this gives you a result: $thumbimage=defaultThumb($file_types_array); you need to use the result:
echo thumbimage;