in my index.php file i used include function

<?php include ('ads/result/ads.php'); ?>

in ads.php I used to this code

<?php
$add="add";
$add .=mt_rand(1,2);
$add .=".jpg";
echo "<img border='0' src='$add'>";
?>

but it does not show any picture.

the <img> tag src value should be if random function generates value 1

<img src="ads/result/add1.jpg">
but the src value i see,

<img src="add1.jpg">

how i can solve this problem




Best Regard:
Md. Zakir Hossain (Raju)
www.rubd.net
www.xenex.rubd.net
www.forum.rubd.net

__________________
Best Regard:
Md. Zakir Hossain (Raju)

www.rajuru.xenexbd.com - my blog with PHP scripts, PHP Book Review and many more

dear as i think the following line should be
$add .=mt_rand(1,2);
$add .=".jpg";
--------------
this right syntax is
$add=.mt_rand(1,2);
$add=.".jpg";

try this
:)

what is wrong? $add="add";
use $add="/images/add"; or echo "<img border='0' src='/images/$add'>";
or whatever is ur images folder.

but, u r using echo "<img border='0' src='$add'>";($add is not a path)

Atually

  $add .=mt_rand(1,2);
  $add .=".jpg";

IS the correct syntax. I am not saying that the =. won't work, but it DIDN'T work on my system.

Paul Gardner
------------------
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Thanks Knight for your time. but your concatenation syntax is not right. (dot). must precede = sign
Peg is right

ahsanul did not understand my problem.
there are three images in my directory
add1.jpg
add2.jpg
add3.jpg

so i used the code
$add="add";
$add.=mt_rand(1,3);
$add.=".jpg";
if second line code generates a random value such as 2 then the final value of $add is add2.jpg
now i echo the variable with img tag
echo "<img src=$add>";
my codes are very fine if i run the ads.php file directly
but problem is when i include this file from another file (not in same path)

i think all of you can understand me now

Best Regard:
Md. Zakir Hossain (Raju)
www.rubd.net
www.xenex.rubd.net
www.forum.rubd.net

because your script runs relative to the scripts location, you must put the path to the images relative from the point where the script is located and not where the INCLUDED file is located. So anshul is correct (with the exception that his path is different then the one you desire.)
You could do this one of 2 ways: (again as anshul suggested)


  <?php
    $add="ads/result/add";
    $add .=mt_rand(1,2);
    $add .=".jpg";
    echo "<img border='0' src='$add'>";
  ?>


or


  <?php
    $add="add";
    $add .=mt_rand(1,2);
    $add .=".jpg";
    echo "<img border='0' src='ads/result/$add'>";
  ?>


Either one should give you the results you want.

Paul Gardner
------------------
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Via Web @ http://www.mnetweb.co.uk/irc
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room #PHP

yes peg110

i can do this if you use ads.php from only one location.

but i want to use it from different page at different location.


Best Regard:
Md. Zakir Hossain (Raju)
www.rubd.net
www.xenex.rubd.net
www.forum.rubd.net

then make the path ABSOLUTE instead of relative.

i.e.

<?php
    $add="/wwwroot/ads/result/add";
    $add .=mt_rand(1,2);
    $add .=".jpg";
    echo "<img border='0' src='$add'>";
  ?>


obviously my "Absolute" path is not necessarily your absolute path.


Paul Gardner
------------------
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Via Web @ http://www.mnetweb.co.uk/irc
Via IRC Client pgardner.net:6667
room #PHP

when you include files path will be taken from the file in which you included other file, not from the included file.

thanks peg110
i also decided this option

anshul, it is not matter while using absolute path, I think

Best Regard:
Md. Zakir Hossain (Raju)
www.rubd.net
www.xenex.rubd.net
www.forum.rubd.net