show Image from URL with image ID -syntax question

[Anonymous]

Hi there,

Trying to parse an image using the image id .
The first line works and shows the image URL: https://www.kiva.org/img/2992239.jpg
The second line does not show the image and I am guessing that there is an error in the syntax. Could someone see it and make this work?

Thanks so much!
Sheaila

$output.= "Image Source: http://www.kiva.org/img/" .$loans['image']['id']. ".jpg/>";

$output.= "<img src='http://www.kiva.org/img/'".$loans['image']['id'].".jpg/>";

 

[Anonymous]

'http://www.kiva.org/img/'
 http://www.kiva.org/img/

You've closed the filename in the second line.

 

[Anonymous]

Guys, hanks so much for the description! I think I understood vital aspects which is indeed nice!