CODEWITHSUNDEEP
php
Sushma
Sushma

Reputation: 1

Unable to retrieve image names correctly

Im trying to retrieve an array of images. All images are appearing fine. However,images with name "image_01" and "image_02" are being retrieved as image_1 and image_2. It is because of the "str_replace" Im using. Can someone tell me why this is happening?

$image=$policy->gallery;     
$image = trim($image,'*');
$temp = explode('*',$image );
echo $temp[5];
// $temp = array_filter($temp);
foreach($temp as $image)
{
  $images[]=base_url()."images/park_photos/".trim( str_replace( array('['*']') ,"" ,$image ) );
  //echo trim( str_replace( array('['*']') ,"" ,$image ));
}
foreach($images as $image)
{
  echo "<img src='{$image}' style='width:100px; height:100px; display:block; float:left;' />";
}
?>

Upvotes: 0

Views: 64

Answers (1)

TazGPL
TazGPL

Reputation: 3748

In order to fix it, you need to replace

$images[]=base_url()."images/park_photos/".trim( str_replace( array('['*']') ,"" ,$image ) );

with

$images[] = base_url() . trim($image);

or even with

$images[] = base_url() . $image;

What's happening? I guess that with this str_replace you wanted to make sure that * is removed from picture name. But you did it before already ($temp = explode('*',$image );) so no need to duplicate.

Now, why does 0 get removed from picture name? There's some "magic" happening in array('['*']'). The * character is interpreted as multiplication, because of the quotation marks sequence. To understand, try doing this:

$a = '['*']';
var_dump($a);
// outputs: int(0)

Why? PHP is guessing variables types from context. The context in this case is: multiply '[' by ']'. Since multiplication applies to numbers, PHP casts the two strings ('[' and ']') to integers, zeros specifically. (Read more about Type Juggling).
So, str_replace( array('['*']') ,"" ,$image ) equals str_replace( array(0) ,"" ,$image ) - that is "replace zeros with empty string.

Upvotes: 2

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