PHP array with different amounts (Undefined offset)

Question

I have two arrays one can have example 1 element inside and other array can have 2 elements. So main problem is i got "Undefined offset" error i think its because my arrays elements are different amount file_data loop goes two times and other with only single element goes together and whats why i got this error. How can i fix this problem?

first array is in caption variable

array (size=1)
  0 => string 'Some string' (length=9)

My second array is file_data

array (size=2)
  'userfile_1' => string './uploads/images/131/Untitled.jpg' (length=33)
  'userfile_15' => string './uploads/images/131/unnamed.jpg' (length=32)

$i = 0;

foreach ($file_data as $file)
{
    $image['hotel_id'] = $last_id;
    $image['image']    = $file;
    $image['caption']  = $caption[$i++];

    var_dump($image);
}

In my result i got

array (size=3)
  'hotel_id' => int 132
  'image' => string './uploads/images/132/Untitled.jpg' (length=33)
   'caption' => string 'Some string' (length=9)

 // and here i got this error: Message: Undefined offset: 1


array (size=3)
  'hotel_id' => int 132
  'image' => string './uploads/images/132/unnamed.jpg' (length=32)
  'caption' => null

user2560539 · Accepted Answer · 2017-01-18 11:26:47Z

1

You get an undefined offset error due to this

$image['caption']  = $caption[$i++];

To avoid the error:

if(!empty($caption[$i])){
   $image['caption']  = $caption[$i];
} else {
   $image['caption']  = "";
}
$i++;

answered Jan 18, 2017 at 11:26

user2560539

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Comments

CETb · Accepted Answer · 2017-01-18 11:25:36Z

0

May-be that more correct ?

foreach ($file_data as $file)
{
    $image[] =['hotel_id' => $last_id,
               'image'    => $file,
               'caption'  => $caption];
}
var_dump($image);

or replace line:

$image['caption']  = isset($caption[$i++])?$caption[$i++]:"";

answered Jan 18, 2017 at 11:25

CETb

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Comments

sujivasagam · Accepted Answer · 2017-01-18 11:28:37Z

0

$i = 0;

foreach ($file_data as $file)
{
    $image['hotel_id'] = $last_id;
    $image['image']    = $file;
    if($caption[$i++]!='')
        $image['caption']  = $caption[$i++];
    else
        $image['caption']  = '';

    var_dump($image);
}

answered Jan 18, 2017 at 11:28

sujivasagam

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