why my image path is not taken?

Question

Hi

I am getting the image path from database in this foreach

foreach($image as $row){ 
  $value = $row['dPath'];
  $imgpath =base_url()."images/".$value;//this is not taken
  $imgpath =  base_url()."images/con_icon.jpg";//this$imgpath is taken

  echo $value;

when i give $imgpath as $imgpath = base_url()."images/con_icon.jpg"; it is accepted in

<img src="<?php echo $imgpath; ?>" and image is displayed

But when i give $imgpath as $imgpath =base_url()."images/".$value; but echo $value; results con_icon.jpg The image is not displayed what is the problem

EDIT:

echo $imgpath =base_url()."images"."/".$value;
 echo $img =  base_url()."images/con_icon.jpg";

gave me this

http://localhost/ssit/images/con_icon.jpg
http://localhost/ssit/images/con_icon.jpg 

then why cant i get this in my <img>

<img src="<?php echo $imgpath; ?>" name=b1 width=90 height=80 
 border=0 onmouseover=mouseOver() onmouseout=mouseOut()>

Answer 1

make sure your $value does not contain extra whitespace at the front or end. use

$value = trim($value);

to remove whitespace. also echo is not the best way to quick-debug variables, use var_dump instead.

and please make sure to escape your imagepath to prevent XSS

edit

you cannot say <img src="<?php echo $imgpath; ?>" name=b1 width=90 height=80 border=0 onmouseover=mouseOver() onmouseout=mouseOut()> because you have whitespace at the end of your string. use <img src="<?php echo trim($imgpath); ?> … /> if you have to use it this way.

apart from that, quote your attributes: onmouseover="mouseOver", don't use parentheses after your event handler names (unless mouseOver() returns a function—i don't think you are doing that …). and you should use urlencode for your imagepath, to lock out all those malicious hackers who want to harm your users

Answer 2

Make sure that $value is not coming empty:

var_dump($value);

Also, you may try this instead:

$imgpath = get_bloginfo('template_url') . "/images/" . $value;