Getting variable from database and putting to image link

Question

<?php
  if($skin > 0 && $skin < 100)
  {
    $imagename = "www.xxxxxxxxxxxxx.com/pcp/Skins/Skin_0".$skin.".png";
  }
  else
  {
    $imagename = "www.xxxxxxxxxxxxx.com/pcp/Skins/Skin_".$skin.".png";
  }
  echo '<img src='.$imagename.' class="floatRight">';
?>

I tried this, but it doesn't seem to work, I viewed the source code and this appaered:

"www.xxxxxxxxxxxxx.com/pcp/Skins/Skin_.png class="floatRight">"

score 0 · Answer 1 · edited May 23 '17 at 11:49

0

use double quotes:

echo "<img src='$imagename' class='floatRight'>";

read this answer on the difference between single-quoted and double-quoted strings in PHP.

edited May 23 '17 at 11:49

Community

1
1

answered Nov 10 '14 at 08:28

Nitsan Baleli

5,393
3
30
52

IT does not solve the issue unless you add the single quotes around `$imagename` as well like around `floatRight`. – fejese Nov 10 '14 at 08:52

score 0 · Answer 2 · answered Nov 10 '14 at 08:33

0

image link add http:// prefix

$imagename = "http://www.xxxxxxxxxxxxx.com/pcp/Skins/Skin_".$skin.".png";

or

don't write domain name

$imagename = "/pcp/Skins/Skin_".$skin.".png";

answered Nov 10 '14 at 08:33

masouk

576
4
8

score 0 · Answer 3 · answered Nov 10 '14 at 08:33

if you are getting this:

"www.xxxxxxxxxxxxx.com/pcp/Skins/Skin_.png class="floatRight">"

in your source code then i guess the problem is with $skin variable please check is this an array or not if so then use $skin[index].

here:

$imagename = "www.xxxxxxxxxxxxx.com/pcp/Skins/Skin_0".$skin.".png";

score 0 · Answer 4 · answered Nov 10 '14 at 08:47

0

use like :-

echo '<img src="'.$imagename.'" class="floatRight">';

this may help you.

answered Nov 10 '14 at 08:47

Param Kumar

123
3
18

Getting variable from database and putting to image link

4 Answers4