[SOLVED] looping over array

[muckv]

I have some problems with displaying the contents of an array

 

<?php

$test= array();

$test=return_images();

for($i=-1;$i<count($test);$i++){
echo $test[$i]. '<br>';


}

 

Output = the First element in the array (B2.jpg)

 

print_r(return_images());

 

Give complete array

 

output = Array ( [2] => B2.jpg [3] => akso.gif [4] => sample.jpg )

 

 

Why does the array start at the third element?

Shouldn't it be Array ( [0] => B2.jpg [1] => akso.gif [2] => sample.jpg )

 

How to loop over it ?

 

[Andy-H]

Can't you just use

 

foreach($test as $img):

   echo $img . '<br>';

Endforeach;

 

[JasonLewis]

There is something wrong with return_images() function if the keys are starting at 2. And you shouldn't start the counter in a for() loop at -1. Not unless you have an element in your array that has a key of -1, which I would doubt since they start at 0.

[muckv]

Well this is the function

 

function return_images()
{
    $images = array();

    $files = scandir('C:\wamp\www\IPTC\main\jan');

    for($i = 0;$i < sizeof($files);$i++) {
        $ext = substr($files[$i], strrpos($files[$i], '.') + 1);

        if ($ext == 'jpg' || $ext == 'jpeg' || $ext == 'gif') {
            $images[$i] = $files[$i];
        }
    }

    return $images;
}

[JasonLewis]

Change this:

 

$images[$i] = $files[$i];

 

To this:

 

$images[] = $files[$i];

 

By defining $i in the square brackets you are giving it a key. If you provide nothing then it will start at 0 and count its way up.

[Andy-H]

$test = array(); <-- Un-necessary

[JasonLewis]

$test = array(); <-- Un-necessary

 

In this instance yes, but it can be a habit for people to declare variables beforehand all the time. In this case, no matter what the return is it'll at least be an array. I admit to doing this sometimes as well.

[ghostdog74]

Well this is the function

 

function return_images()
{
    $images = array();

    $files = scandir('C:\wamp\www\IPTC\main\jan');

    for($i = 0;$i < sizeof($files);$i++) {
        $ext = substr($files[$i], strrpos($files[$i], '.') + 1);

        if ($ext == 'jpg' || $ext == 'jpeg' || $ext == 'gif') {
            $images[$i] = $files[$i];
        }
    }

    return $images;
}

that's because of your if{} statement inside the for loop. You only check for jpg|jpeg|gif but what about the rest? If the file is not jpg or jpeg or gif, the counter $i will get incremented as well, that's why your array index doesn't start at 1( or 0) but at 2.

 

[Andy-H]

Because he needs to narrow down to .jp(e)g and .gif formats...

[ghostdog74]

he's initial question is why its starting at the third element. That's because the first few files are not what he wants AND the $i counter also increase. He need to somehow decrease(or remain?) the $i counter if a file is not what he wants.

[JasonLewis]

I've already answered the question, I believe. Removing $i will eliminate the count, since using [] will start at 0 and count up.

 

Change this:

 

$images[$i] = $files[$i];

 

To this:

 

$images[] = $files[$i];

 

By defining $i in the square brackets you are giving it a key. If you provide nothing then it will start at 0 and count its way up.

[muckv]

thanks projectfear

[kenrbnsn]

I would use the glob() function here:

<?php
function return_images() {
      return(array_map('basename', glob('C:/wamp/www/IPTC/main/jan/*.{jpg,jpeg,gif}',GLOB_BRACE)));
}
?>

 

Note: untested.

 

Ken

[prexep]

Or you can just use reset().